Class 12 physics, NCERT chapter 11. Attempt each question, check your answer, then open the step-by-step solution.
The energy of a photon of wavelength $400\ \text{nm}$ is approximately (take $hc = 1240\ \text{eV nm}$)
Answer: (B) $3.1\ \text{eV}$
Concept. Photon energy $E = \dfrac{hc}{\lambda}$.
Step 1. Substitute. $\dfrac{1240}{400} = 3.1$ eV.
Answer. $3.1$ eV (Option B).
Solution by Sreeraj P, M.Sc Physics
A beam of light falls on a metal surface such that photo-electrons are generated. If power of the light source starts to decrease linearly with time $t$, then variation of the photocurrent $I$ and magnitude of the stopping potential $|V|$ with time is best represented by :
Answer: (A) see figure
Concept. Photocurrent $\propto$ photons per second (power); stopping potential depends only on frequency: $e|V| = h\nu - \phi$.
Step 1. Current. The power falls linearly, so $I$ falls linearly.
Step 2. Stopping potential. Frequency unchanged, so $|V|$ stays constant: graph (1).
Answer. Graph (1) (Option A).
Solution by Sreeraj P, M.Sc Physics
A ray of light with wavelength $\lambda$ is incident on three different photoelectric cells namely 1, 2 and 3. The threshold wavelength of these photoelectric cells are $\lambda_1$, $\lambda_2$, and $\lambda_3$, respectively and the magnitude of stopping potentials of these cells are $V_1$, $V_2$ and $V_3$, respectively. The relation between $\lambda$ and threshold wavelengths are $\lambda_1 < \lambda$, $\lambda_2 > \lambda$ and $\lambda_3 >> \lambda$. The correct option is :
Answer: (A) $V_1 = 0,\ V_2 < V_3$
Concept. Emission needs $\lambda < \lambda_0$; then $eV = hc\left(\dfrac1\lambda - \dfrac{1}{\lambda_0}\right)$.
Step 1. Cell 1. $\lambda > \lambda_1$: no emission, $V_1 = 0$.
Step 2. Cells 2 and 3. Both emit; a larger $\lambda_0$ gives a larger $V$, and $\lambda_3 > \lambda_2$, so $V_3 > V_2$.
Answer. $V_1 = 0,\ V_2 < V_3$ (Option A).
Solution by Sreeraj P, M.Sc Physics
A photon and an electron, each of $20$ eV energy, move in free space. The ratio of linear momentum of electron $p_e$ to that of photon $p_{Ph}$, $\dfrac{p_e}{p_{Ph}}$ is :
(Take speed of light $= 3 \times 10^8\ \text{m s}^{-1}$, charge of electron $= -1.6 \times 10^{-19}$ C and mass of electron $= 9 \times 10^{-31}$ kg)
Answer: (C) $225$
Concept. Photon $p = \dfrac{E}{c}$; electron (non-relativistic) $p = \sqrt{2mE}$.
Step 1. Energy. $E = 20 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-18}$ J.
Step 2. Electron momentum. $\sqrt{2 \times 9 \times 10^{-31} \times 3.2 \times 10^{-18}} = 2.4 \times 10^{-24}$ kg m s$^{-1}$.
Step 3. Ratio.
$$\frac{p_e}{p_{Ph}} = \frac{2.4 \times 10^{-24} \times 3 \times 10^8}{3.2 \times 10^{-18}} = 225$$
Answer. $225$ (Option C).
Solution by Sreeraj P, M.Sc Physics
A photon and an electron (mass $m$) have the same energy $E$. The ratio ($\lambda_{\text{photon}}/\lambda_{\text{electron}}$) of their de Broglie wavelengths is: ($c$ is the speed of light)
Answer: (C) $c\sqrt{\dfrac{2m}{E}}$
Concept. Photon $\lambda = \dfrac{hc}{E}$; electron $\lambda = \dfrac{h}{\sqrt{2mE}}$.
Step 1. Ratio.
$$\frac{\lambda_{\text{photon}}}{\lambda_{\text{electron}}} = \frac{c\sqrt{2mE}}{E} = c\sqrt{\frac{2m}{E}}$$
Answer. $c\sqrt{\dfrac{2m}{E}}$ (Option C).
Solution by Sreeraj P, M.Sc Physics
Which of the following options represent the variation of photoelectric current with property of light shown on the x-axis?
Answer: (A) A only
Concept. Photocurrent is proportional to the number of photons per second (intensity), not to frequency.
Step 1. A and B. Current against intensity is a line through the origin: A correct, B wrong.
Step 2. C and D. Above threshold the current does not rise linearly with frequency (D describes the stopping potential); C starts at zero frequency, also wrong.
Answer. A only (Option A).
Solution by Sreeraj P, M.Sc Physics
If $c$ is the velocity of light in free space, the correct statements about photon among the following are :
A. The energy of a photon is $E = h\nu$.
B. The velocity of a photon is $c$.
C. The momentum of a photon, $p = \dfrac{h\nu}{c}$.
D. In a photon-electron collision, both total energy and total momentum are conserved.
E. Photon possesses positive charge.
Choose the correct answer from the options given below :
Answer: (D) A, B, C and D only
Concept. Photon properties: $E = h\nu$, speed $c$, $p = \dfrac{h\nu}{c}$, neutral.
Step 1. Check. A ✓, B ✓, C ✓, D (Compton: energy and momentum conserved) ✓, E photons are neutral ✗.
Answer. A, B, C and D only (Option D).
Solution by Sreeraj P, M.Sc Physics
The graph which shows the variation of $\left(\dfrac{1}{\lambda^2}\right)$ and its kinetic energy, $E$ is (where $\lambda$ is de Broglie wavelength of a free particle) :
Answer: (B) see figure
Concept. $\lambda = \dfrac{h}{\sqrt{2mE}}$.
Step 1. Relation. $\dfrac{1}{\lambda^2} = \dfrac{2m}{h^2}E$: a straight line through the origin, graph (2).
Answer. Graph (2) (Option B).
Solution by Sreeraj P, M.Sc Physics
The minimum wavelength of $X$-rays produced by an electron accelerated through a potential difference of $V$ volts is proportional to:
Answer: (B) $\dfrac{1}{V}$
Concept. The shortest wavelength comes when the whole energy $eV$ goes into one photon.
Step 1. Relation. $eV = \dfrac{hc}{\lambda_{\min}} \Rightarrow \lambda_{\min} \propto \dfrac1V$.
Answer. $\dfrac1V$ (Option B).
Solution by Sreeraj P, M.Sc Physics
The work functions of Caesium (Cs), Potassium (K) and Sodium (Na) are $2.14$ eV, $2.30$ eV and $2.75$ eV respectively. If incident electromagnetic radiation has an incident energy of $2.20$ eV, which of these photosensitive surfaces may emit photoelectrons?
Answer: (A) Cs only
Concept. Emission needs photon energy greater than the work function.
Step 1. Compare. $2.20 > 2.14$ (Cs) but $< 2.30$ (K) and $< 2.75$ (Na).
Answer. Cs only (Option A).
Solution by Sreeraj P, M.Sc Physics
The number of photons per second on an average emitted by the source of monochromatic light of wavelength $600$ nm, when it delivers the power of $3.3 \times 10^{-3}$ watt will be : ($h = 6.6 \times 10^{-34}$ J s)
Answer: (D) $10^{16}$
Concept. Photons per second $= \dfrac{P}{hc/\lambda}$.
Step 1. Photon energy. $\dfrac{6.6 \times 10^{-34} \times 3 \times 10^8}{600 \times 10^{-9}} = 3.3 \times 10^{-19}$ J.
Step 2. Number. $\dfrac{3.3 \times 10^{-3}}{3.3 \times 10^{-19}} = 10^{16}$.
Answer. $10^{16}$ (Option D).
Solution by Sreeraj P, M.Sc Physics
An electromagnetic wave of wavelength '$\lambda$' is incident on a photosensitive surface of negligible work function. If 'm' mass is of photoelectron emitted from the surface has de-Broglie wavelength $\lambda_d$, then :
Answer: (D) $\lambda = \left(\dfrac{2mc}{h}\right)\lambda_d^2$
Concept. With zero work function, the photon energy becomes the electron's KE $= \dfrac{h^2}{2m\lambda_d^2}$.
Step 1. Equation. $\dfrac{hc}{\lambda} = \dfrac{h^2}{2m\lambda_d^2}$.
Step 2. Solve. $\lambda = \left(\dfrac{2mc}{h}\right)\lambda_d^2$.
Answer. $\lambda = \left(\dfrac{2mc}{h}\right)\lambda_d^2$ (Option D).
Solution by Sreeraj P, M.Sc Physics
Light of frequency $1.5$ times the threshold frequency is incident on a photosensitive material. What will be the photoelectric current if the frequency is halved and intensity is doubled ?
Answer: (D) zero
Concept. Below the threshold frequency there is no emission at any intensity.
Step 1. New frequency. $\dfrac{1.5\nu_0}{2} = 0.75\nu_0 < \nu_0$.
Step 2. Current. Zero.
Answer. Zero (Option D).
Solution by Sreeraj P, M.Sc Physics
An electron is accelerated from rest through a potential difference of V volt. If the de Broglie wavelength of the electron is $1.227 \times 10^{-2}$ nm, the potential difference is :
Answer: (D) $10^4$ V
Concept. For an electron through $V$ volts: $\lambda = \dfrac{1.227}{\sqrt V}$ nm.
Step 1. Solve. $\sqrt V = \dfrac{1.227}{1.227 \times 10^{-2}} = 100 \Rightarrow V = 10^4$ V.
Answer. $10^4$ V (Option D).
Solution by Sreeraj P, M.Sc Physics
Statement I: A metallic surface is irradiated by a monochromatic light of frequency $\nu > \nu_0$ (the threshold frequency). The maximum kinetic energy and the stopping potential are $K_{max}$ and $V_0$ respectively. If the frequency incident on the surface is doubled, both $K_{max}$ and $V_0$ are also doubled.
Statement II: The maximum kinetic energy and the stopping potential of photoelectrons emitted from a surface are linearly dependent on the frequency of incident light.
Answer: (D) Statement I is false, Statement II is true
Concept. $K_{max}=h\nu-\phi$ and $eV_0=K_{max}$: both are linear in $\nu$ but not proportional to it, because of the $-\phi$ term.
Step 1. Statement I. Doubling $\nu$ gives $2h\nu-\phi$, which is more than double $h\nu-\phi$: false.
Step 2. Statement II. Linear dependence: true.
Answer. Statement I false, Statement II true (Option D).
Solution by Sreeraj P, M.Sc Physics
Statement I: Davisson–Germer experiment established the wave nature of electrons.
Statement II: If electrons have wave nature, they can interfere and show diffraction.
Answer: (A) Statement I is true, Statement II is true; Statement II is a correct explanation of Statement I
Concept. Davisson and Germer observed electron diffraction from a nickel crystal.
Step 1. Check. Diffraction is a wave property, so both are true and II explains why the experiment shows wave nature.
Answer. Statement I is true, Statement II is true; Statement II is a correct explanation of Statement I (Option A).
Solution by Sreeraj P, M.Sc Physics
Statement I: When ultraviolet light is incident on a photocell, its stopping potential is $V_0$ and the maximum kinetic energy of the photoelectrons is $K_{max}$. When the ultraviolet light is replaced by X-rays, both $V_0$ and $K_{max}$ increase.
Statement II: Photoelectrons are emitted with speeds ranging from zero to a maximum value because of the range of frequencies present in the incident light.
Answer: (C) Statement I is true, Statement II is false
Concept. $K_{max}=h\nu-\phi$ grows with frequency. Electrons leave with a range of speeds because those from deeper inside lose energy on the way out, even for one frequency.
Step 1. Statement I. X-rays have a higher frequency than UV, so $K_{max}$ and $V_0$ increase: true.
Step 2. Statement II. The spread of speeds is not due to a range of frequencies (it occurs with monochromatic light too): false.
Answer. Option C.
Solution by Sreeraj P, M.Sc Physics
The work functions of metals A and B are in the ratio $1 : 2$. If light of frequencies $f$ and $2f$ are incident on metal surfaces A and B respectively, the ratio of the maximum kinetic energies of the photoelectrons emitted is
Answer: (B) $1 : 2$
Concept. $K_{\max} = h\nu - \phi$.
Step 1. Values. $K_A = hf - \phi$, $K_B = 2hf - 2\phi = 2K_A$.
Answer. $1 : 2$ (Option B).
Solution by Sreeraj P, M.Sc Physics
The de Broglie wavelength of an electron having $80$ eV of energy is nearly ($1$ eV $= 1.6 \times 10^{-19}$ J, mass of electron $= 9 \times 10^{-31}$ kg, Planck's constant $= 6.6 \times 10^{-34}$ J s)
Answer: (D) $1.4$ Å
Concept. $\lambda = \dfrac{h}{\sqrt{2mE}}$.
Step 1. Substitute.
$$\lambda = \frac{6.6 \times 10^{-34}}{\sqrt{2 \times 9 \times 10^{-31} \times 80 \times 1.6 \times 10^{-19}}} = \frac{6.6 \times 10^{-34}}{4.8 \times 10^{-24}} \approx 1.4\ \text{Å}$$
Answer. $1.4$ Å (Option D).
Solution by Sreeraj P, M.Sc Physics
Electrons accelerated through a potential difference $V$ are diffracted from a crystal whose planes are $d = 1$ Å apart. A strong diffraction peak is observed when the electrons are incident at an angle $i = 30°$ from the normal to the crystal planes, with the condition $2d\cos i = n\lambda$ ($n = 1$). $V$ should be about ($h = 6.6 \times 10^{-34}$ J s, $m = 9.1 \times 10^{-31}$ kg, $e = 1.6 \times 10^{-19}$ C)
Answer: (B) $50$ V
Concept. The Bragg-type condition gives $\lambda$; then $V=\dfrac{h^2}{2me\lambda^2}$.
Step 1. Wavelength. $\lambda=2\times1\times\cos30^\circ=\sqrt3$ Å $=1.73\times10^{-10}$ m.
Step 2. Voltage. $V=\dfrac{(6.6\times10^{-34})^2}{2\times9.1\times10^{-31}\times1.6\times10^{-19}\times3\times10^{-20}}=50$ V.
Answer. Option B.
Solution by Sreeraj P, M.Sc Physics
Light of wavelength $300$ nm falls on a metal of work function $2.14$ eV. The maximum kinetic energy of the photoelectrons is ($hc = 1240$ eV nm)
Answer: (A) $2.0$ eV
Concept. $K_{\max} = \dfrac{hc}{\lambda} - \phi$.
Step 1. Photon energy. $\dfrac{1240}{300} \approx 4.13$ eV.
Step 2. KE. $4.13 - 2.14 \approx 2.0$ eV.
Answer. $2.0$ eV (Option A).
Solution by Sreeraj P, M.Sc Physics
The threshold wavelength of a metal is $620$ nm. Its work function is ($hc = 1240$ eV nm)
Answer: (B) $2$ eV
Concept. $\phi = \dfrac{hc}{\lambda_0}$.
Step 1. Substitute. $\dfrac{1240}{620} = 2$ eV.
Answer. $2$ eV (Option B).
Solution by Sreeraj P, M.Sc Physics
The stopping potentials for light of frequencies $\nu_1$ and $\nu_2$ on a metal are $V_1$ and $V_2$. Planck's constant is given by
Answer: (C) $\dfrac{e(V_1 - V_2)}{\nu_1 - \nu_2}$
Concept. Einstein's equation at two frequencies: $eV = h\nu - \phi$.
Step 1. Subtract. $e(V_1 - V_2) = h(\nu_1 - \nu_2)$.
Step 2. Solve. $h = \dfrac{e(V_1 - V_2)}{\nu_1 - \nu_2}$.
Answer. $\dfrac{e(V_1 - V_2)}{\nu_1 - \nu_2}$ (Option C).
Solution by Sreeraj P, M.Sc Physics
The slope of the graph of stopping potential against frequency of incident light, for any metal, is
Answer: (D) $\dfrac{h}{e}$
Concept. $V_0 = \dfrac{h}{e}\nu - \dfrac{\phi}{e}$.
Step 1. Slope. $\dfrac{h}{e}$, the same for every metal.
Answer. $\dfrac{h}{e}$ (Option D).
Solution by Sreeraj P, M.Sc Physics
In a photoelectric experiment the intensity of the incident light is doubled, its frequency being unchanged. Then
Answer: (A) the saturation current doubles and the stopping potential is unchanged
Concept. Intensity sets the number of photons; frequency sets each photon's energy.
Step 1. Double intensity. Twice the photoelectrons: saturation current doubles; $K_{\max}$ and $V_0$ unchanged.
Answer. The saturation current doubles and the stopping potential is unchanged (Option A).
Solution by Sreeraj P, M.Sc Physics
A $100$ W source emits light of wavelength $600$ nm. The number of photons it emits per second is ($h = 6.6 \times 10^{-34}$ J s)
Answer: (B) $3 \times 10^{20}$
Concept. Photons per second $= \dfrac{P}{hc/\lambda}$.
Step 1. Photon energy. $3.3 \times 10^{-19}$ J.
Step 2. Number. $\dfrac{100}{3.3 \times 10^{-19}} \approx 3 \times 10^{20}$.
Answer. $3 \times 10^{20}$ (Option B).
Solution by Sreeraj P, M.Sc Physics
The momentum of a photon of energy $3$ eV is ($1$ eV $= 1.6 \times 10^{-19}$ J)
Answer: (C) $1.6 \times 10^{-27}$ kg m/s
Concept. Photon momentum $p = \dfrac{E}{c}$.
Step 1. Substitute. $\dfrac{4.8 \times 10^{-19}}{3 \times 10^8} = 1.6 \times 10^{-27}$ kg m/s.
Answer. $1.6 \times 10^{-27}$ kg m/s (Option C).
Solution by Sreeraj P, M.Sc Physics
A proton and an alpha particle are accelerated through the same potential difference. The ratio of their de Broglie wavelengths $\lambda_p : \lambda_\alpha$ is
Answer: (D) $2\sqrt{2} : 1$
Concept. $\lambda = \dfrac{h}{\sqrt{2mqV}}$.
Step 1. Ratio. Alpha: $4m$, $2e$:
$$\frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{4m \times 2e}{m \times e}} = 2\sqrt2$$
Answer. $2\sqrt2 : 1$ (Option D).
Solution by Sreeraj P, M.Sc Physics
If the kinetic energy of a free electron is doubled, its de Broglie wavelength becomes
Answer: (A) $\dfrac{1}{\sqrt{2}}$ times
Concept. $\lambda = \dfrac{h}{\sqrt{2mK}} \propto \dfrac{1}{\sqrt K}$.
Step 1. Double $K$. $\lambda$ becomes $\dfrac{1}{\sqrt2}$ times.
Answer. $\dfrac{1}{\sqrt2}$ times (Option A).
Solution by Sreeraj P, M.Sc Physics
An electron and a proton have the same de Broglie wavelength. Which has the greater kinetic energy?
Answer: (B) the electron
Concept. Same $\lambda$ means same momentum; $K = \dfrac{p^2}{2m}$.
Step 1. Compare. Smaller mass, larger $K$: the electron.
Answer. The electron (Option B).
Solution by Sreeraj P, M.Sc Physics
The de Broglie wavelength of a $1$ kg ball moving at $1$ m/s is of the order of ($h = 6.6 \times 10^{-34}$ J s)
Answer: (C) $10^{-33}$ m
Concept. $\lambda = \dfrac{h}{mv}$.
Step 1. Substitute. $6.6 \times 10^{-34}$ m, of order $10^{-33}$ m (far too small to detect).
Answer. $10^{-33}$ m (Option C).
Solution by Sreeraj P, M.Sc Physics
Photons of energy $5$ eV fall on a metal surface and the fastest photoelectrons are stopped by a potential of $2$ V. Find the work function of the metal in eV.
Numerical value type. Enter your answer.
Answer: 3
Concept. Einstein's equation: $\phi = E - eV_0$.
Step 1. Substitute. $5 - 2 = 3$ eV.
Answer. $3$.
Solution by Sreeraj P, M.Sc Physics
An electron is accelerated from rest through $150$ V. Find its de Broglie wavelength in ångström, to one decimal place.
Numerical value type. Enter your answer.
Answer: 1.0
Concept. For an electron through $V$ volts: $\lambda = \dfrac{12.27}{\sqrt V}$ Å.
Step 1. Substitute. $\dfrac{12.27}{\sqrt{150}} = \dfrac{12.27}{12.25} \approx 1.0$ Å.
Answer. $1.0$.
Solution by Sreeraj P, M.Sc Physics
A monochromatic source of light operating at $15$ kW emits $2.5 \times 10^{22}$ photons/s. The region of an electromagnetic spectrum to which the emitted electromagnetic radiation belongs to ______. (Take $h = 6.6 \times 10^{-34}\ \text{J.s}$ and $c = 3 \times 10^8\ \text{m/s}$).
Answer: (D) Ultraviolet
Concept. Energy per photon $=\dfrac{P}{n}$; then $\lambda=\dfrac{hc}{E}$ tells the band.
Step 1. Energy. $\dfrac{15\times10^3}{2.5\times10^{22}}=6\times10^{-19}$ J.
Step 2. Wavelength. $\dfrac{6.6\times10^{-34}\times3\times10^8}{6\times10^{-19}}=3.3\times10^{-7}$ m $=330$ nm, below the visible.
Answer. Ultraviolet (Option D).
Solution by Sreeraj P, M.Sc Physics
$K_1$ and $K_2$ be the maximum kinetic energies of photoelectrons emitted from a surface of a given material for the light of wavelength $\lambda_1$ and $\lambda_2$, respectively. If $\lambda_1 = 2\lambda_2$ then the work function of material is given by :
Answer: (D) $K_2 - 2K_1$
Concept. Write Einstein's equation for both; $\lambda_1=2\lambda_2$ means the second photon has twice the energy $E$ of the first.
Step 1. Equations. $K_1=E-\phi$, $K_2=2E-\phi$.
Step 2. Eliminate $E$. $K_2-2K_1=-\phi+2\phi=\phi$.
Answer. $\phi=K_2-2K_1$ (Option D).
Solution by Sreeraj P, M.Sc Physics
The de Broglie wavelength for an electron accelerated through the potential difference of $V_1$ volt is $\lambda_1$. When the potential difference is changed to $V_2$ volt, the associated de Broglie wavelength is increased by $50\%$. If $(V_1/V_2) = (9/\alpha)$, then the value of $\alpha$ is ______.
Numerical value type. Enter your answer.
Answer: 4
Concept. $\lambda \propto \dfrac{1}{\sqrt V}$.
Step 1. Ratio. $\dfrac{V_1}{V_2} = \left(\dfrac{\lambda_2}{\lambda_1}\right)^2 = 1.5^2 = \dfrac94$.
Step 2. Compare. $\alpha = 4$.
Answer. $4$.
Solution by Sreeraj P, M.Sc Physics
An electron of mass $m$ is moving in an electric field $\vec{E} = -2E_0\hat{i}$ ($E_0 =$ constant $> 0$), with an initial velocity $\vec{V} = v_0\hat{i}$ ($v_0 =$ constant $> 0$). If $\lambda_0 = \dfrac{h}{4mv_0}$, its de Broglie wavelength at time $t$ is ______. ($e =$ charge of electron)
Answer: (C) $\dfrac{4\lambda_0}{\left[1 + \frac{2E_0e}{m}\frac{t}{v_0}\right]}$
Concept. The force $-e\vec E = 2eE_0\hat i$ is along the motion, so the electron speeds up; $\lambda = \dfrac{h}{mv}$.
Step 1. Speed. $v = v_0 + \dfrac{2eE_0}{m}t$.
Step 2. Wavelength. With $\dfrac{h}{mv_0} = 4\lambda_0$:
$$\lambda = \frac{4\lambda_0}{1 + \frac{2E_0e}{m}\frac{t}{v_0}}$$
Answer. $\dfrac{4\lambda_0}{\left[1 + \frac{2E_0e}{m}\frac{t}{v_0}\right]}$ (Option C).
Solution by Sreeraj P, M.Sc Physics
Light source having wavelength $331$ nm is used to generate photo-electrons whose stopping potential is $0.2$ V. The work function of the used metal in the experiment is $\alpha \times 10^{-19}$ J. The value of $\alpha$ is ______. ($h = 6.62 \times 10^{-34}$ J s, $e = 1.6 \times 10^{-19}$ C and $c = 3 \times 10^8$ m/s)
Answer: (C) $5.68$
Concept. $\phi=\dfrac{hc}{\lambda}-eV_0$, in joules.
Step 1. Photon energy. $\dfrac{6.62\times10^{-34}\times3\times10^8}{331\times10^{-9}}=6.0\times10^{-19}$ J.
Step 2. Electron energy. $1.6\times10^{-19}\times0.2=0.32\times10^{-19}$ J.
Step 3. Work function. $5.68\times10^{-19}$ J.
Answer. Option C.
Solution by Sreeraj P, M.Sc Physics
An electron is travelling with a velocity $v$ in free space and when it enters a medium, its velocity is reduced by $20\%$. The de Broglie wavelength of electron in the medium is $\alpha\lambda_0$, where $\lambda_0$ is its de Broglie wavelength in free space. The value of $\alpha$ is ______.
Answer: (C) $1.25$
Concept. $\lambda=\dfrac{h}{mv}\propto\dfrac1v$.
Step 1. New speed. $0.8v$, so $\lambda=\dfrac{\lambda_0}{0.8}=1.25\lambda_0$.
Answer. Option C.
Solution by Sreeraj P, M.Sc Physics
The de Broglie wavelength associated with an electron accelerated through a potential difference $V$ is $\lambda_e$ and the de Broglie wavelength associated with a proton accelerated through the same potential difference is $\lambda_p$. If their corresponding masses are $m_e$ and $m_p$, respectively, then the ratio of their de Broglie wavelengths $\left(\dfrac{\lambda_e}{\lambda_p}\right)$ is ______.
Answer: (A) $\sqrt{\dfrac{m_p}{m_e}}$
Concept. Same accelerating voltage and same charge size: $\lambda=\dfrac{h}{\sqrt{2meV}}\propto\dfrac{1}{\sqrt m}$.
Answer. $\dfrac{\lambda_e}{\lambda_p}=\sqrt{\dfrac{m_p}{m_e}}$ (Option A).
Solution by Sreeraj P, M.Sc Physics
For a certain metal, when monochromatic light of wavelength $\lambda$ is incident, the stopping potential for photoelectrons is $3V_0$. When the same metal is illuminated by light of wavelength $2\lambda$, then the stopping potential becomes $V_0$. The threshold wavelength for photoelectric emission for the given metal is $\alpha\lambda$. The value of $\alpha$ is ______.
Answer: (B) $4$
Concept. Two Einstein equations; eliminate the photon energy $E$ at $\lambda$.
Step 1. Equations. $3eV_0=E-\phi$ and $eV_0=\dfrac E2-\phi$.
Step 2. Subtract. $2eV_0=\dfrac E2\Rightarrow E=4eV_0$, so $\phi=eV_0=\dfrac E4$.
Step 3. Threshold. $\lambda_0=\dfrac{hc}{\phi}=4\lambda$.
Answer. $\alpha=4$ (Option B).
Solution by Sreeraj P, M.Sc Physics
Number of photons of equal energy emitted per second by a 6 mW laser source operating at 663 nm is ______.
(Given : $h = 6.63\times10^{-34}$ J.s and $c = 3\times10^8$ m/s)
Answer: (D) $2\times10^{16}$
Concept. Photons per second $=\dfrac{\text{power}}{\text{energy per photon}}=\dfrac{P\lambda}{hc}$.
Step 1. Value. $\dfrac{6\times10^{-3}\times663\times10^{-9}}{6.63\times10^{-34}\times3\times10^8}=\dfrac{3.98\times10^{-9}}{1.99\times10^{-25}}=2\times10^{16}$.
Answer. Option D.
Solution by Sreeraj P, M.Sc Physics
The ratio of de Broglie wavelength of a deutron with kinetic energy $E$ to that of an alpha particle with kinetic energy $2E$, is $n : 1$. The value of $n$ is ______ .
(Assume mass of proton $=$ mass of neutron) :
Numerical value type. Enter your answer.
Answer: 2
Concept. $\lambda\propto\dfrac1{\sqrt{mK}}$. Deuteron mass $2m$, alpha $4m$.
Step 1. Ratio. $\dfrac{\lambda_d}{\lambda_\alpha}=\sqrt{\dfrac{4m\times2E}{2m\times E}}=2$.
Answer. $n=2$.
Solution by Sreeraj P, M.Sc Physics
When a light of a given wavelength falls on a metallic surface the stopping potential for photoelectrons is 3.2 V . If a second light having wavelength twice of first light is used, the stopping potential drops to 0.7 V . The wavelength of first light is ______ m .
$(\text{h} = 6.63\times10^{-34}\ \text{J.s}, \text{e} = 1.6\times10^{-19}\text{C}, \text{c} = 3\times10^8\ \text{m/s})$
Answer: (B) $2.5\times10^{-7}$
Concept. $eV=E-\phi$; doubling $\lambda$ halves $E$.
Step 1. Equations (eV). $3.2=E-\phi$, $0.7=\dfrac E2-\phi$.
Step 2. Solve. $\dfrac E2=2.5\Rightarrow E=5$ eV.
Step 3. Wavelength. $\lambda=\dfrac{1243}{5}$ nm $\approx2.5\times10^{-7}$ m.
Answer. Option B.
Solution by Sreeraj P, M.Sc Physics
The work functions of cesium (Cs) and lithium (Li) metals are $1.9\ \text{eV}$ and $2.5\ \text{eV}$, respectively. If light of wavelength $550\ \text{nm}$ is incident on these two metal surfaces, then photoelectric effect is possible for the case of
Answer: (C) Cs only
Concept. Emission happens only if $\dfrac{hc}{\lambda}\ge\phi$.
Step 1. Photon energy. $\dfrac{1240}{550}=2.25$ eV.
Step 2. Compare. $2.25>1.9$ (Cs emits); $2.25<2.5$ (Li does not).
Answer. Cs only (Option C).
Solution by Sreeraj P, M.Sc Physics
A light source of wavelength $\lambda$ illuminates a metal surface and electrons are ejected with maximum kinetic energy of $2\ \text{eV}$. If the same surface is illuminated by a light source of wavelength $\lambda/2$, then the maximum kinetic energy of ejected electrons will be (the work function of the metal is $1\ \text{eV}$)
Answer: (D) $5\ \text{eV}$
Concept. Halving $\lambda$ doubles the photon energy; the work function stays.
Step 1. First photon. $2+1=3$ eV.
Step 2. Second. $6-1=5$ eV.
Answer. Option D.
Solution by Sreeraj P, M.Sc Physics
A sub-atomic particle of mass $10^{-30}\ \text{kg}$ is moving with a velocity $2.21\times10^6\ \text{m/s}$. Under the matter wave consideration, the particle will behave closely like ($h = 6.63\times10^{-34}\ \text{J s}$)
Answer: (D) X-rays
Concept. $\lambda=\dfrac{h}{mv}$, then compare with the spectrum.
Step 1. Wavelength. $\dfrac{6.63\times10^{-34}}{10^{-30}\times2.21\times10^6}=3\times10^{-10}$ m $=0.3$ nm.
Answer. X-rays (Option D).
Solution by Sreeraj P, M.Sc Physics
In the photoelectric effect, an em-wave is incident on a metal surface and electrons are ejected from the surface. If the work function of the metal is $2.14\ \text{eV}$ and the stopping potential is $2\ \text{V}$, what is the wavelength of the em-wave? (Given $hc = 1242\ \text{eV nm}$, where $h$ is Planck's constant and $c$ is the speed of light in vacuum.)
Answer: (A) $300\ \text{nm}$
Concept. $\dfrac{hc}{\lambda} = \phi + eV_0$.
Step 1. Photon energy. $2.14 + 2 = 4.14$ eV.
Step 2. Wavelength. $\dfrac{1242}{4.14} = 300$ nm.
Answer. $300$ nm (Option A).
Solution by Sreeraj P, M.Sc Physics
An electron of mass $m$ with an initial velocity $\vec v = v_0\hat i$ ($v_0 > 0$) enters an electric field $\vec E = -E_0\hat k$. If the initial de Broglie wavelength is $\lambda_0$, the value after time $t$ would be
Answer: (A) $\dfrac{\lambda_0}{\sqrt{1 + \dfrac{e^2E_0^2t^2}{m^2v_0^2}}}$
Concept. Here the force $eE_0\hat k$ is perpendicular to $v_0\hat i$: $v_x$ stays, $v_z=\dfrac{eE_0}{m}t$.
Step 1. Speed. $v=v_0\sqrt{1+\dfrac{e^2E_0^2t^2}{m^2v_0^2}}$.
Step 2. Wavelength. $\lambda=\dfrac{\lambda_0}{\sqrt{1+\frac{e^2E_0^2t^2}{m^2v_0^2}}}$.
Answer. Option A.
Solution by Sreeraj P, M.Sc Physics
In photoelectric effect, the stopping potential $(V_0)$ vs frequency $(\nu)$ curve is plotted ($h$ is the Planck's constant and $\phi_0$ is the work function of the metal).
(A) $V_0$ vs $\nu$ is linear.
(B) The slope of the $V_0$ vs $\nu$ curve $= \dfrac{\phi_0}{h}$.
(C) $h$ is related to the slope of the $V_0$ vs $\nu$ line.
(D) The value of the electric charge of the electron is not required to determine $h$ using the $V_0$ vs $\nu$ curve.
(E) The work function can be estimated without knowing the value of $h$.
Choose the correct answer from the options given below:
Answer: (B) (A), (C) and (E) only
Concept. $V_0=\dfrac he\nu-\dfrac{\phi}{e}$.
Step 1. Check. A: linear, true. B: slope is $\dfrac he$, false. C: true. D: we get $\dfrac he$, so $e$ is needed for $h$: false. E: the intercept on the $V_0$ axis is $-\dfrac\phi e$, giving $\phi$ in eV without $h$: true.
Answer. (A), (C) and (E) only (Option B).
Solution by Sreeraj P, M.Sc Physics
The energy $E$ and momentum $p$ of a moving body of mass $m$ are related by some equation. Given that $c$ represents the speed of light, identify the correct equation.
Answer: (D) $E^2 = p^2c^2 + m^2c^4$
Concept. Every term in a valid equation has the same dimensions. $E^2$ has the dimensions of (energy)$^2$; $pc$ and $mc^2$ are energies.
Step 1. Check terms. $p^2c^2=(pc)^2$ and $m^2c^4=(mc^2)^2$ are both (energy)$^2$; $pc^2$ and $m^2c^2$ are not.
Step 2. Equation. $E^2=p^2c^2+m^2c^4$ (it gives $E=mc^2$ at rest and $E=pc$ for a photon).
Answer. Option D.
Solution by Sreeraj P, M.Sc Physics
The ratio of the power of a light source $S_1$ to that of the light source $S_2$ is 2. $S_1$ is emitting $2\times10^{15}$ photons per second at $600\ \text{nm}$. If the wavelength of the source $S_2$ is $300\ \text{nm}$, then the number of photons per second emitted by $S_2$ is ______ $\times10^{14}$.
Numerical value type. Enter your answer.
Answer: 5
Concept. $n=\dfrac{P\lambda}{hc}$, so $n\propto P\lambda$.
Step 1. Ratio. $\dfrac{n_2}{n_1}=\dfrac{P_2}{P_1}\cdot\dfrac{\lambda_2}{\lambda_1}=\dfrac12\cdot\dfrac12=\dfrac14$.
Step 2. Value. $\dfrac{2\times10^{15}}{4}=5\times10^{14}$.
Answer. $5$.
Solution by Sreeraj P, M.Sc Physics
A proton of mass $m_p$ has the same energy as that of a photon of wavelength $\lambda$. If the proton is moving at non-relativistic speed, then the ratio of its de Broglie wavelength to the wavelength of the photon is
Answer: (D) $\dfrac{1}{c}\sqrt{\dfrac{E}{2m_p}}$
Concept. Proton: $\lambda_p=\dfrac{h}{\sqrt{2m_pE}}$. Photon: $\lambda=\dfrac{hc}{E}$.
Step 1. Ratio. $\dfrac{\lambda_p}{\lambda}=\dfrac{E}{c\sqrt{2m_pE}}=\dfrac1c\sqrt{\dfrac{E}{2m_p}}$.
Answer. Option D.
Solution by Sreeraj P, M.Sc Physics
Which of the following phenomena can not be explained by wave theory of light?
Answer: (A) Compton effect
Concept. In the Compton effect an X-ray photon collides with an electron like a particle, losing energy and momentum. Waves cannot explain this; reflection, refraction and diffraction are wave effects.
Answer. Compton effect (Option A).
Solution by Sreeraj P, M.Sc Physics
If $\lambda$ and $K$ are the de Broglie wavelength and kinetic energy, respectively, of a particle with constant mass, the correct graphical representation for the particle will be
Answer: (B) see figure
Concept. $\lambda=\dfrac{h}{\sqrt{2mK}}$, so $\dfrac1K=\dfrac{2m}{h^2}\lambda^2$: $\dfrac1K$ grows as $\lambda^2$.
Step 1. Graph. A parabola through the origin, curving upward.
Answer. Option B.
Solution by Sreeraj P, M.Sc Physics
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): Emission of electrons in the photoelectric effect can be suppressed by applying a sufficiently negative electric potential to the photoemissive substance.
Reason (R): A negative electric potential which stops the emission of electrons from the surface of a photoemissive substance varies linearly with the frequency of incident radiation.
In the light of the above statements, choose the most appropriate answer from the options given below:
Answer: (D) Both (A) and (R) are true but (R) is not the correct explanation of (A)
Concept. $eV_0 = h\nu - \phi$.
Step 1. Assertion. A large enough retarding potential stops even the fastest electrons → true.
Step 2. Reason. $V_0$ is linear in $\nu$ → true, but this describes the frequency dependence, not why emission can be suppressed.
Answer. Both (A) and (R) are true but (R) is not the correct explanation of (A) (Option D).
Solution by Sreeraj P, M.Sc Physics
In an experiment with the photoelectric effect, the stopping potential
Answer: (D) is $\left(\dfrac{1}{e}\right)$ times the maximum kinetic energy of the emitted photoelectrons
Concept. The stopping potential just stops the fastest electrons: $eV_0 = K_{\max} = \dfrac{hc}{\lambda} - \phi$.
Step 1. Check. Independent of intensity; decreases with wavelength; $V_0 = \dfrac{K_{\max}}{e}$.
Answer. Is $\left(\dfrac1e\right)$ times the maximum kinetic energy of the emitted photoelectrons (Option D).
Solution by Sreeraj P, M.Sc Physics
The de Broglie wavelengths of a proton and an $\alpha$ particle are $\lambda$ and $2\lambda$ respectively. The ratio of the velocities of the proton and the $\alpha$ particle will be
Answer: (D) $8:1$
Concept. $\lambda=\dfrac{h}{mv}$, so $v=\dfrac{h}{m\lambda}$.
Step 1. Ratio. $\dfrac{v_p}{v_\alpha}=\dfrac{m_\alpha\lambda_\alpha}{m_p\lambda_p}=\dfrac{4\times2}{1\times1}=8$.
Answer. $8:1$ (Option D).
Solution by Sreeraj P, M.Sc Physics
Monochromatic light of frequency $6\times10^{14}\ \text{Hz}$ is produced by a laser. The power emitted is $2\times10^{-3}\ \text{W}$. How many photons per second, on average, are emitted by the source? (Given $h = 6.63\times10^{-34}\ \text{J s}$)
Answer: (C) $5\times10^{15}$
Concept. $n=\dfrac{P}{h\nu}$.
Step 1. Photon energy. $6.63\times10^{-34}\times6\times10^{14}=3.98\times10^{-19}$ J.
Step 2. Rate. $\dfrac{2\times10^{-3}}{3.98\times10^{-19}}=5\times10^{15}$ per second.
Answer. Option C.
Solution by Sreeraj P, M.Sc Physics
Which figure shows the correct variation of the applied potential difference ($V$) with photoelectric current ($i$) at two different intensities of light ($I_1 < I_2$) of the same wavelength?
Answer: (B) see figure
Concept. Same wavelength means the same $K_{max}$, so the same stopping potential $-V_0$. Higher intensity means more electrons, so a larger saturation current.
Step 1. Graph. Both curves start at the same $-V_0$; the $I_2$ curve saturates higher.
Answer. Option B.
Solution by Sreeraj P, M.Sc Physics
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: The number of photons increases with increase in frequency of light.
Reason R: The maximum kinetic energy of emitted electrons increases with the frequency of incident radiation.
In the light of the above statements, choose the most appropriate answer from the options given below:
Answer: (C) **A** is not correct but **R** is correct.
Concept. At a given power $P$, the number of photons per second is $\dfrac{P}{h\nu}$, which falls as $\nu$ rises. $K_{max}=h\nu-\phi$ rises with $\nu$.
Step 1. Assertion. False (fewer photons at higher frequency for the same power).
Step 2. Reason. True.
Answer. A is not correct but R is correct (Option C).
Solution by Sreeraj P, M.Sc Physics
Given below are two statements:
Statement I: The figure shows the variation of stopping potential with frequency $(\nu)$ for two photosensitive materials $M_1$ and $M_2$. The slope gives the value of $\dfrac{h}{e}$, where $h$ is Planck's constant and $e$ is the charge of an electron.
Statement II: $M_2$ will emit photoelectrons of greater kinetic energy for incident radiation of the same frequency.
In the light of the above statements, choose the most appropriate answer from the options given below.
Answer: (D) Statement I is correct and Statement II is incorrect
Concept. $V_0=\dfrac he\nu-\dfrac\phi e$: every metal gives a line of slope $\dfrac he$; the line cuts the $\nu$-axis at $\nu_0$.
Step 1. Statement I. The slope is $\dfrac he$: correct.
Step 2. Statement II. $M_2$ has the higher threshold, so the larger $\phi$; at the same $\nu$ its electrons have less kinetic energy: incorrect.
Answer. Option D.
Solution by Sreeraj P, M.Sc Physics
Which of the following statements is not true about the stopping potential $(V_0)$?
Answer: (B) It increases with increase in intensity of the incident light.
Concept. $eV_0=K_{max}=h\nu-\phi$: the stopping potential depends on the frequency and the metal, not on the intensity.
Step 1. Check. A ($V_0=K_{max}/e$), C (depends on the metal) and D (depends on frequency) are true. B (rises with intensity) is false.
Answer. Option B.
Solution by Sreeraj P, M.Sc Physics
In a photoelectric experiment, light of energy $2.48\ \text{eV}$ irradiates a photosensitive material. The stopping potential was measured to be $0.5\ \text{V}$. The work function of the photosensitive material is
Answer: (C) $1.98\ \text{eV}$
Concept. Einstein's equation $h\nu=\phi+eV_0$.
Step 1. Work function. $\phi=2.48-0.5=1.98$ eV.
Answer. Option C.
Solution by Sreeraj P, M.Sc Physics
When UV light of wavelength $300\ \text{nm}$ is incident on a metal surface having work function $2.13\ \text{eV}$, electron emission takes place. The stopping potential is (Given $hc = 1240\ \text{eV nm}$)
Answer: (C) $2\ \text{V}$
Concept. $eV_0=\dfrac{hc}{\lambda}-\phi$.
Step 1. Photon energy. $\dfrac{1240}{300}=4.13$ eV.
Step 2. Stopping potential. $4.13-2.13=2$ V.
Answer. Option C.
Solution by Sreeraj P, M.Sc Physics
A convex lens of focal length $40\ \text{cm}$ forms an image of an extended source of light on a photoelectric cell. A current $I$ is produced. The lens is replaced by another convex lens having the same diameter but focal length $20\ \text{cm}$. The photoelectric current now is:
Answer: (D) $I$
Concept. The photocurrent depends on the total light power reaching the cell. A lens of the same diameter collects the same light from the source; a shorter focal length only makes the image smaller and brighter.
Answer. $I$ (Option D).
Solution by Sreeraj P, M.Sc Physics
The threshold frequency of a metal with work function $6.63\ \text{eV}$ is:
Answer: (D) $1.6\times10^{15}\ \text{Hz}$
Concept. $\nu_0 = \dfrac{\phi}{h}$.
Step 1. Substitute.
$$\nu_0 = \frac{6.63 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}} = 1.6 \times 10^{15}\ \text{Hz}$$
Answer. $1.6 \times 10^{15}$ Hz (Option D).
Solution by Sreeraj P, M.Sc Physics
The de-Broglie wavelength of an electron is the same as that of a photon. If velocity of electron is $25\%$ of the velocity of light, then the ratio of K.E. of electron and K.E. of photon will be:
Answer: (B) $\dfrac18$
Concept. Same wavelength means same momentum $p$. Electron: $K=\dfrac{p^2}{2m}=\dfrac{pv}{2}$. Photon: $E=pc$.
Step 1. Ratio. $\dfrac{K}{E}=\dfrac{v}{2c}=\dfrac{0.25}{2}=\dfrac18$.
Answer. Option B.
Solution by Sreeraj P, M.Sc Physics
Two sources of light emit with a power of $200\ \text{W}$. The ratio of number of photons of visible light emitted by each source having wavelengths $300\ \text{nm}$ and $500\ \text{nm}$ respectively, will be:
Answer: (D) $3:5$
Concept. Same power: $n=\dfrac{P\lambda}{hc}\propto\lambda$.
Step 1. Ratio. $\dfrac{300}{500}=\dfrac35$.
Answer. $3:5$ (Option D).
Solution by Sreeraj P, M.Sc Physics
A proton and an electron are associated with the same de-Broglie wavelength. The ratio of their kinetic energies is: (Assume $h = 6.63\times10^{-34}\ \text{J s}$, $m_e = 9.0\times10^{-31}\ \text{kg}$ and $m_p = 1836$ times $m_e$)
Answer: (D) $1 : 1836$
Concept. Same wavelength means same momentum, and $K=\dfrac{p^2}{2m}\propto\dfrac1m$.
Step 1. Ratio. $\dfrac{K_p}{K_e}=\dfrac{m_e}{m_p}=\dfrac1{1836}$.
Answer. $1:1836$ (Option D).
Solution by Sreeraj P, M.Sc Physics
A proton and an electron have the same de Broglie wavelength. If $K_p$ and $K_e$ be the kinetic energies of proton and electron respectively, then choose the correct relation:
Answer: (B) $K_p < K_e$
Concept. Equal wavelengths mean equal momenta; $K = \dfrac{p^2}{2m}$.
Step 1. Compare. $m_p > m_e$, so $K_p < K_e$.
Answer. $K_p < K_e$ (Option B).
Solution by Sreeraj P, M.Sc Physics
A proton, an electron and an alpha particle have the same energies. Their de-Broglie wavelengths will be compared as:
Answer: (A) $\lambda_\alpha < \lambda_p < \lambda_e$
Concept. Same energy: $\lambda \propto \dfrac{1}{\sqrt m}$.
Step 1. Order. $m_\alpha > m_p > m_e \Rightarrow \lambda_\alpha < \lambda_p < \lambda_e$.
Answer. $\lambda_\alpha < \lambda_p < \lambda_e$ (Option A).
Solution by Sreeraj P, M.Sc Physics
UV light of $4.13\ \text{eV}$ is incident on a photosensitive metal surface having work function $3.13\ \text{eV}$. The maximum kinetic energy of ejected photoelectrons will be:
Answer: (C) $1\ \text{eV}$
Concept. $K_{\max} = h\nu - \phi$.
Step 1. Substitute. $4.13 - 3.13 = 1$ eV.
Answer. $1$ eV (Option C).
Solution by Sreeraj P, M.Sc Physics
The work function of a substance is $3.0\ \text{eV}$. The longest wavelength of light that can cause the emission of photoelectrons from this substance is approximately:
Answer: (B) $414\ \text{nm}$
Concept. Longest wavelength (threshold): photon energy just equals $\phi$, $\lambda_0=\dfrac{hc}{\phi}$.
Step 1. Value. $\dfrac{1240}{3.0}=413$ nm.
Answer. About $414$ nm (Option B).
Solution by Sreeraj P, M.Sc Physics
For the photoelectric effect, the maximum kinetic energy $E_k$ of the photoelectrons is plotted against the frequency $\nu$ of the incident photons. The graph is a straight line cutting the frequency axis at the threshold frequency. The slope of the graph gives
Answer: (D) Planck's constant
Concept. $K_{max}=h\nu-\phi$ is a straight line in $\nu$ with slope $h$ and $\nu$-intercept $\nu_0$.
Step 1. Slope. $h$, Planck's constant. (The $V_0$–$\nu$ graph would have slope $h/e$.)
Answer. Option D.
Solution by Sreeraj P, M.Sc Physics
When a metal surface is illuminated by light of wavelength $\lambda$, the stopping potential is $8\ \text{V}$. When the same surface is illuminated by light of wavelength $3\lambda$, the stopping potential is $2\ \text{V}$. The threshold wavelength for this surface is:
Answer: (C) $9\lambda$
Concept. Same method: $eV=E-\phi$ with $E\propto\dfrac1\lambda$.
Step 1. Equations (in eV). $8=E-\phi$, $2=\dfrac E3-\phi$.
Step 2. Solve. $6=\dfrac{2E}{3}\Rightarrow E=9$, $\phi=1$.
Step 3. Threshold. $\phi=\dfrac E9$, so $\lambda_0=9\lambda$.
Answer. Option C.
Solution by Sreeraj P, M.Sc Physics
In a photoelectric effect experiment, light of frequency $1.5$ times the threshold frequency is made to fall on the surface of a photosensitive material. Now if the frequency is halved and the intensity is doubled, the number of photoelectrons emitted will be:
Answer: (C) Zero
Concept. Emission needs $\nu\ge\nu_0$, whatever the intensity.
Step 1. New frequency. $\dfrac{1.5\nu_0}{2}=0.75\nu_0<\nu_0$.
Answer. Zero (Option C).
Solution by Sreeraj P, M.Sc Physics
The threshold wavelength for photoelectric emission from a material is $5500\ \text{Å}$. Photoelectrons will be emitted, when this material is illuminated with monochromatic radiation from a
A. $75\ \text{W}$ infra-red lamp
B. $10\ \text{W}$ infra-red lamp
C. $75\ \text{W}$ ultra-violet lamp
D. $10\ \text{W}$ ultra-violet lamp
Choose the correct answer from the options given below:
Answer: (D) C and D only
Concept. Emission depends on photon energy, i.e. on wavelength, not on the lamp's power. The threshold is $5500$ Å.
Step 1. Infrared. Wavelength longer than $5500$ Å: no emission at any power.
Step 2. Ultraviolet. Shorter than $5500$ Å: emission from both lamps (the brighter one gives more electrons).
Answer. C and D only (Option D).
Solution by Sreeraj P, M.Sc Physics
The ratio of de-Broglie wavelength of an $\alpha$-particle and a proton accelerated from rest by the same potential is $\dfrac{1}{\sqrt m}$. The value of $m$ is
Answer: (C) 8
Concept. Accelerated through $V$: $K=qV$, $\lambda=\dfrac{h}{\sqrt{2mqV}}\propto\dfrac{1}{\sqrt{mq}}$.
Step 1. Ratio. $\dfrac{\lambda_\alpha}{\lambda_p}=\sqrt{\dfrac{m\cdot e}{4m\cdot2e}}=\dfrac1{\sqrt8}$.
Answer. $m=8$ (Option C).
Solution by Sreeraj P, M.Sc Physics
An electron accelerated through a potential difference $V_1$ has a de-Broglie wavelength of $\lambda$. When the potential is changed to $V_2$, its de-Broglie wavelength increases by $50\%$. The value of $\left(\dfrac{V_1}{V_2}\right)$ is equal to
Answer: (B) $\dfrac94$
Concept. $\lambda\propto\dfrac1{\sqrt V}$, so $\dfrac{V_1}{V_2}=\left(\dfrac{\lambda_2}{\lambda_1}\right)^2$.
Step 1. Value. $(1.5)^2=\dfrac94$.
Answer. Option B.
Solution by Sreeraj P, M.Sc Physics
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: The beam of electrons shows wave nature and exhibits interference and diffraction.
Reason R: Davisson Germer experimentally verified the wave nature of electrons.
In the light of the above statements, choose the most appropriate answer from the options given below:
Answer: (D) Both A and R are correct and R is the correct explanation of A
Concept. Electron beams diffract from crystals and give interference patterns, which is wave behaviour. Davisson and Germer's experiment was the first direct verification.
Step 1. Assertion. Correct.
Step 2. Reason. Correct, and it is the experimental basis of the assertion.
Answer. Option D.
Solution by Sreeraj P, M.Sc Physics
If a source of electromagnetic radiation having power $15\ \text{kW}$ produces $10^{16}$ photons per second, the radiation belongs to a part of spectrum is (Take Planck constant $h=6\times10^{-34}\ \text{J s}$)
Answer: (C) Gamma rays
Concept. Energy per photon $=\dfrac Pn$, then compare with the spectrum.
Step 1. Energy. $\dfrac{15\times10^3}{10^{16}}=1.5\times10^{-12}$ J $\approx9$ MeV.
Step 2. Wavelength. $\dfrac{6\times10^{-34}\times3\times10^8}{1.5\times10^{-12}}=1.2\times10^{-13}$ m: nuclear scale.
Answer. Gamma rays (Option C).
Solution by Sreeraj P, M.Sc Physics
From the photoelectric effect experiment, following observations are made. Identify which of these are correct.
A. The stopping potential depends only on the work function of the metal.
B. The saturation current increases as the intensity of incident light increases.
C. The maximum kinetic energy of a photo electron depends on the intensity of the incident light.
D. Photoelectric effect can be explained using wave theory of light.
Choose the correct answer from the options given below:
Answer: (C) B only
Concept. $eV_0 = h\nu - \phi$; intensity sets the current, not $K_{\max}$.
Step 1. Check.
Answer. B only (Option C).
Solution by Sreeraj P, M.Sc Physics
Electron beam used in an electron microscope, when accelerated by a voltage of $20\ \text{kV}$ has a de-Broglie wavelength of $\lambda_0$. If the voltage is increased to $40\ \text{kV}$ then the de-Broglie wavelength associated with the electron beam would be
Answer: (D) $\dfrac{\lambda_0}{\sqrt2}$
Concept. For an accelerated electron $\lambda=\dfrac{h}{\sqrt{2meV}}\propto\dfrac1{\sqrt V}$.
Step 1. Doubling $V$. $\lambda\to\dfrac{\lambda_0}{\sqrt2}$.
Answer. Option D.
Solution by Sreeraj P, M.Sc Physics
Given below are two statements:
Statement I: Stopping potential in photoelectric effect does not depend on the power of the light source.
Statement II: For a given metal, the maximum kinetic energy of the photoelectron depends on the wavelength of the incident light.
In the light of above statements, choose the most appropriate answer from the options given below.
Answer: (D) Both Statement I and Statement II are correct
Concept. $eV_0=\dfrac{hc}{\lambda}-\phi$: neither side contains the power of the source.
Step 1. Statement I. $V_0$ does not depend on power: correct.
Step 2. Statement II. For a given metal $K_{max}$ depends on $\lambda$: correct.
Answer. Both correct (Option D).
Solution by Sreeraj P, M.Sc Physics
Given below are two statements:
Statement I: Out of microwaves, infrared rays and ultraviolet rays, ultraviolet rays are the most effective for the emission of electrons from a metallic surface.
Statement II: Above the threshold frequency, the maximum kinetic energy of photoelectrons is inversely proportional to the frequency of the incident light.
In the light of above statements, choose the correct answer from the options given below.
Answer: (B) Statement I is true but Statement II is false
Concept. Emission needs $h\nu>\phi$; ultraviolet has the highest frequency of the three. $K_{max}=h\nu-\phi$ increases with $\nu$.
Step 1. Statement I. UV is the most effective: true.
Step 2. Statement II. $K_{max}$ is not inversely proportional to $\nu$ (it rises linearly): false.
Answer. Option B.
Solution by Sreeraj P, M.Sc Physics
The de Broglie wavelength of an electron having kinetic energy $E$ is $\lambda$. If the kinetic energy of electron becomes $\dfrac E4$, then its de-Broglie wavelength will be
Answer: (D) $2\lambda$
Concept. $\lambda=\dfrac{h}{\sqrt{2mK}}\propto\dfrac{1}{\sqrt K}$.
Step 1. New value. $K\to\dfrac K4$ gives $\lambda\to2\lambda$.
Answer. Option D.
Solution by Sreeraj P, M.Sc Physics
The variation of stopping potential $V_0$ as a function of the frequency ($\nu$) of the incident light for a metal is shown in figure. The work function of the surface is
Answer: (B) $2.07\ \text{eV}$
Concept. The intercept on the frequency axis is the threshold: $\phi = h\nu_0$.
Step 1. Threshold. $\nu_0 = 5 \times 10^{14}$ Hz.
Step 2. Work function.
$$\phi = \frac{6.63 \times 10^{-34} \times 5 \times 10^{14}}{1.6 \times 10^{-19}} \approx 2.07\ \text{eV}$$
Answer. $2.07$ eV (Option B).
Solution by Sreeraj P, M.Sc Physics
A metallic surface is illuminated with radiation of wavelength $\lambda$, the stopping potential is $V_0$. If the same surface is illuminated with radiation of wavelength $2\lambda$, the stopping potential becomes $\dfrac{V_0}4$. The threshold wavelength for this metallic surface will be
Answer: (A) $3\lambda$
Concept. Einstein's equation at the two wavelengths.
Step 1. Equations. $eV_0 = \dfrac{hc}{\lambda} - \phi$ and $\dfrac{eV_0}{4} = \dfrac{hc}{2\lambda} - \phi$.
Step 2. Eliminate $V_0$. Multiply the second by 4 and subtract: $3\phi = \dfrac{hc}{\lambda}$.
Step 3. Threshold. $\lambda_0 = \dfrac{hc}{\phi} = 3\lambda$.
Answer. $3\lambda$ (Option A).
Solution by Sreeraj P, M.Sc Physics
The ratio of the de-Broglie wavelengths of proton and electron having same kinetic energy is (Assume $m_p=m_e\times1849$)
Answer: (A) $1:43$
Concept. Same KE: $\lambda \propto \dfrac{1}{\sqrt m}$.
Step 1. Ratio. $\dfrac{1}{\sqrt{1849}} = \dfrac{1}{43}$.
Answer. $1 : 43$ (Option A).
Solution by Sreeraj P, M.Sc Physics
A proton and an $\alpha$-particle are accelerated from rest by $2\ \text{V}$ and $4\ \text{V}$ potentials, respectively. The ratio of their de-Broglie wavelength is
Answer: (C) $4:1$
Concept. $\lambda\propto\dfrac1{\sqrt{mqV}}$.
Step 1. Products. Proton: $1\times1\times2=2$. Alpha: $4\times2\times4=32$.
Step 2. Ratio. $\dfrac{\lambda_p}{\lambda_\alpha}=\sqrt{\dfrac{32}{2}}=4$.
Answer. $4:1$ (Option C).
Solution by Sreeraj P, M.Sc Physics
The difference between threshold wavelengths for two metal surfaces $A$ and $B$ having work function $\phi_A=9\ \text{eV}$ and $\phi_B=4.5\ \text{eV}$ in nm is (Given $hc=1242\ \text{eV nm}$)
Answer: (D) 138
Concept. $\lambda_0=\dfrac{hc}{\phi}$ for each metal.
Step 1. Values. $\dfrac{1242}{4.5}=276$ nm and $\dfrac{1242}{9}=138$ nm.
Step 2. Difference. $138$ nm.
Answer. Option D.
Solution by Sreeraj P, M.Sc Physics
The kinetic energy of an electron, $\alpha$-particle and a proton are given as $4K$, $2K$ and $K$ respectively. The de-Broglie wavelength associated with electron ($\lambda_e$), $\alpha$-particle ($\lambda_\alpha$) and the proton ($\lambda_p$) are as follows:
Answer: (B) $\lambda_\alpha<\lambda_p<\lambda_e$
Concept. $\lambda=\dfrac{h}{\sqrt{2mK}}$: the larger $mK$, the shorter $\lambda$.
Step 1. Products. Electron $m_e\cdot4K$ (tiny); $\alpha$: $4m_p\cdot2K=8m_pK$; proton: $m_pK$.
Step 2. Order. $\lambda_\alpha\lt\lambda_p\lt\lambda_e$.
Answer. Option B.
Solution by Sreeraj P, M.Sc Physics
The work functions of Aluminium and Gold are $4.1\ \text{eV}$ and $5.1\ \text{eV}$ respectively. The ratio of the slope of the stopping potential versus frequency plot for Gold to that of Aluminium is
Answer: (C) 1
Concept. $V_0=\dfrac he\nu-\dfrac\phi e$: the slope $\dfrac he$ is the same for every metal.
Answer. $1$ (Option C).
Solution by Sreeraj P, M.Sc Physics
Proton (P) and electron (e) will have same de-Broglie wavelength when the ratio of their momentum is (assume $m_p=1849\,m_e$)
Answer: (A) $1:1$
Concept. $\lambda = \dfrac{h}{p}$.
Step 1. Equal wavelengths. Equal momenta: $1 : 1$.
Answer. $1 : 1$ (Option A).
Solution by Sreeraj P, M.Sc Physics
In photoelectric effect
A. The photocurrent is proportional to the intensity of the incident radiation.
B. Maximum kinetic energy with which photoelectrons are emitted depends on the intensity of incident light.
C. Max K.E. with which photoelectrons are emitted depends on the frequency of incident light.
D. The emission of photoelectrons requires a minimum threshold intensity of incident radiation.
E. Max K.E. of the photoelectrons is independent of the frequency of the incident light.
Choose the correct answer from the options given below:
Answer: (C) A and C only
Concept. Intensity controls the number of electrons; frequency controls their maximum energy; there is a threshold frequency, not a threshold intensity.
Step 1. Statements. A true (current $\propto$ intensity). B false. C true. D false (threshold frequency). E false.
Answer. A and C only (Option C).
Solution by Sreeraj P, M.Sc Physics
The de Broglie wavelength of a molecule in a gas at room temperature $300$ K is $\lambda_1$. If the temperature of the gas is increased to $600$ K, then the de Broglie wavelength of the same gas molecule becomes
Answer: (C) $\dfrac1{\sqrt2}\lambda_1$
Concept. Thermal energy $\tfrac32kT$ gives $\lambda=\dfrac{h}{\sqrt{3mkT}}\propto\dfrac{1}{\sqrt T}$.
Step 1. Doubling $T$. $\lambda=\dfrac{\lambda_1}{\sqrt2}$.
Answer. Option C.
Solution by Sreeraj P, M.Sc Physics
A proton moving with one tenth of velocity of light has a certain de Broglie wavelength of $\lambda$. An alpha particle having certain kinetic energy has the same de Broglie wavelength $\lambda$. The ratio of kinetic energy of proton and that of alpha particle is
Answer: (B) $4:1$
Concept. Equal wavelengths mean equal momenta; $K = \dfrac{p^2}{2m}$.
Step 1. Ratio. $\dfrac{K_p}{K_\alpha} = \dfrac{m_\alpha}{m_p} = 4$.
Answer. $4 : 1$ (Option B).
Solution by Sreeraj P, M.Sc Physics
The threshold frequency of a metal is $f_0$. When the light of frequency $2f_0$ is incident on the metal plate, the maximum velocity of photoelectron is $v_1$. When the frequency of incident radiation is increased to $5f_0$, the maximum velocity of photoelectrons emitted is $v_2$. The ratio of $v_1$ to $v_2$ is
Answer: (A) $\dfrac{v_1}{v_2}=\dfrac12$
Concept. $\dfrac12mv^2 = h(f - f_0)$.
Step 1. Values. $v_1^2 \propto f_0$, $v_2^2 \propto 4f_0$.
Step 2. Ratio. $\dfrac{v_1}{v_2} = \dfrac12$.
Answer. $\dfrac{v_1}{v_2} = \dfrac12$ (Option A).
Solution by Sreeraj P, M.Sc Physics
When light of frequency twice the threshold frequency is incident on a metal plate, the maximum velocity of emitted electrons is $v_1$. When the frequency of incident radiation is increased to five times the threshold value, the maximum velocity of emitted electrons becomes $v_2$. If $v_2 = xv_1$, the value of $x$ will be ______.
Numerical value type. Enter your answer.
Answer: 2
Concept. $\tfrac12mv^2=h(\nu-\nu_0)$, so $v^2\propto\nu-\nu_0$.
Step 1. Values. $2\nu_0\to\nu_0$; $5\nu_0\to4\nu_0$.
Step 2. Ratio. $\dfrac{v_2}{v_1}=\sqrt4=2$.
Answer. $x=2$.
Solution by Sreeraj P, M.Sc Physics
Light of two different frequencies whose photons have energies $3.8\ \text{eV}$ and $1.4\ \text{eV}$ respectively illuminate a metallic surface whose work function is $0.6\ \text{eV}$ successively. The ratio of maximum speeds of emitted electrons for the two frequencies respectively will be:
Answer: (A) $2 : 1$
Concept. $K_{\max} = E - \phi$ and $v \propto \sqrt{K}$.
Step 1. KEs. $3.2$ eV and $0.8$ eV.
Step 2. Ratio. $\sqrt{4} = 2$.
Answer. $2 : 1$ (Option A).
Solution by Sreeraj P, M.Sc Physics
Given below are two statements:
Statement I: Davisson-Germer experiment establishes the wave nature of electrons.
Statement II: If electrons have wave nature, they can interfere and show diffraction.
In the light of the above statements choose the correct answer from the options given below:
Answer: (A) Both Statement I and Statement II are true.
Concept. Davisson and Germer saw sharp diffraction peaks when electrons were scattered from a nickel crystal, at the angle Bragg's law predicts for $\lambda=h/p$. Only waves interfere and diffract.
Step 1. Statement I. True.
Step 2. Statement II. True.
Answer. Both true (Option A).
Solution by Sreeraj P, M.Sc Physics
A proton, a neutron, an electron and an $\alpha$-particle have the same energy. If $\lambda_p$, $\lambda_n$, $\lambda_e$ and $\lambda_\alpha$ are the de Broglie wavelengths of proton, neutron, electron and $\alpha$-particle respectively, then choose the correct relation from the following
Answer: (B) $\lambda_\alpha < \lambda_n < \lambda_p < \lambda_e$
Concept. Same KE: $\lambda \propto \dfrac{1}{\sqrt m}$.
Step 1. Masses. $m_\alpha > m_n > m_p > m_e$ (neutron slightly heavier).
Step 2. Order. $\lambda_\alpha < \lambda_n < \lambda_p < \lambda_e$.
Answer. $\lambda_\alpha < \lambda_n < \lambda_p < \lambda_e$ (Option B).
Solution by Sreeraj P, M.Sc Physics
An electron with speed $v$ and a photon with speed $c$ have the same de Broglie wavelength. If the kinetic energy and momentum of the electron are $E_e$ and $P_e$ and that of the photon are $E_{ph}$ and $P_{ph}$ respectively, which of the following is correct?
Answer: (B) $\dfrac{E_e}{E_{ph}} = \dfrac{v}{2c}$
Concept. Same wavelength means same momentum $p$; $E_e = \dfrac{pv}{2}$, $E_{ph} = pc$.
Step 1. Ratio. $\dfrac{E_e}{E_{ph}} = \dfrac{v}{2c}$ (and $\dfrac{P_e}{P_{ph}} = 1$).
Answer. $\dfrac{E_e}{E_{ph}} = \dfrac{v}{2c}$ (Option B).
Solution by Sreeraj P, M.Sc Physics
A metal surface is illuminated by radiation of wavelength $4500\ \text{Å}$. The ejected photoelectron enters a constant magnetic field of $2\ \text{mT}$ making an angle of $90^\circ$ with the magnetic field. If it starts revolving in a circular path of radius $2\ \text{mm}$, the work function of the metal is approximately
Answer: (A) $1.36\ \text{eV}$
Concept. The circle radius gives the electron's speed, $r=\dfrac{mv}{eB}$; then $\phi=\dfrac{hc}{\lambda}-\tfrac12mv^2$.
Step 1. Speed. $v=\dfrac{eBr}{m}=\dfrac{1.6\times10^{-19}\times2\times10^{-3}\times2\times10^{-3}}{9.1\times10^{-31}}=7.0\times10^5$ m/s.
Step 2. Kinetic energy. $\tfrac12\times9.1\times10^{-31}\times(7.0\times10^5)^2=2.25\times10^{-19}$ J $=1.41$ eV.
Step 3. Work function. $\dfrac{12400}{4500}-1.41=2.76-1.41=1.35$ eV.
Answer. About $1.36$ eV (Option A).
Solution by Sreeraj P, M.Sc Physics
The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength $6630\ \text{Å}$ is $0.42\ \text{V}$. If the threshold frequency is $x\times10^{13}\ \text{s}^{-1}$, where $x$ is (nearest integer): (Given, speed of light $= 3\times10^8\ \text{m s}^{-1}$, Planck's constant $= 6.63\times10^{-34}\ \text{J s}$)
Numerical value type. Enter your answer.
Answer: 35
Concept. $h\nu_0=\dfrac{hc}{\lambda}-eV_0$.
Step 1. Photon energy. $\dfrac{1243}{663}=1.875$ eV.
Step 2. Work function. $1.875-0.42=1.455$ eV $=2.33\times10^{-19}$ J.
Step 3. Threshold frequency. $\dfrac{2.33\times10^{-19}}{6.63\times10^{-34}}=3.5\times10^{14}$ Hz $=35\times10^{13}$ Hz.
Answer. $35$.
Solution by Sreeraj P, M.Sc Physics
A metal exposed to light of wavelength $800\ \text{nm}$ emits photoelectrons with a certain kinetic energy. The maximum kinetic energy of photo-electrons doubles when light of wavelength $500\ \text{nm}$ is used. The work function of the metal is (Take $hc = 1230\ \text{eV nm}$)
Answer: (C) $0.615\ \text{eV}$
Concept. $K=\dfrac{hc}{\lambda}-\phi$ for both wavelengths, with $K_2=2K_1$.
Step 1. Photon energies. $\dfrac{1230}{800}=1.5375$ eV, $\dfrac{1230}{500}=2.46$ eV.
Step 2. Equation. $2.46-\phi=2(1.5375-\phi)\Rightarrow\phi=3.075-2.46=0.615$ eV.
Answer. Option C.
Solution by Sreeraj P, M.Sc Physics
The ratio of wavelengths of proton and deuteron accelerated by potentials $V_p$ and $V_d$ is $1 : \sqrt2$. Then the ratio of $V_p$ to $V_d$ will be
Answer: (D) $4 : 1$
Concept. $\lambda = \dfrac{h}{\sqrt{2mqV}}$ with equal charges and $m_d = 2m_p$.
Step 1. Ratio.
$$\frac{\lambda_p}{\lambda_d} = \sqrt{\frac{2m_pV_d}{m_pV_p}} = \frac{1}{\sqrt2} \;\Rightarrow\; \frac{2V_d}{V_p} = \frac12$$
Step 2. Solve. $\dfrac{V_p}{V_d} = 4$.
Answer. $4 : 1$ (Option D).
Solution by Sreeraj P, M.Sc Physics
A parallel beam of light of wavelength $900\ \text{nm}$ and intensity $100\ \text{W m}^{-2}$ is incident on a surface perpendicular to the beam. The number of photons crossing $1\ \text{cm}^2$ area perpendicular to the beam in one second is
Answer: (B) $4.5\times10^{16}$
Concept. Energy per second through the area $=IA$; divide by the photon energy.
Step 1. Power through 1 cm$^2$. $100\times10^{-4}=10^{-2}$ W.
Step 2. Photon energy. $\dfrac{1240}{900}=1.38$ eV $=2.2\times10^{-19}$ J.
Step 3. Count. $\dfrac{10^{-2}}{2.2\times10^{-19}}=4.5\times10^{16}$.
Answer. Option B.
Solution by Sreeraj P, M.Sc Physics
A nucleus of mass $M$ at rest splits into two parts having masses $\dfrac{M'}{3}$ and $\dfrac{2M'}{3}$ $(M' < M)$. The ratio of de Broglie wavelengths of the two parts will be
Answer: (C) $1 : 1$
Concept. A nucleus at rest that splits gives the two parts equal and opposite momenta; $\lambda=\dfrac hp$.
Answer. $1:1$ (Option C).
Solution by Sreeraj P, M.Sc Physics
An $\alpha$ particle and a carbon $12$ atom has same kinetic energy $K$. The ratio of their de-Broglie wavelengths $(\lambda_\alpha : \lambda_{C12})$ is
Answer: (A) $\sqrt3 : 1$
Concept. Equal KE: $\lambda \propto \dfrac{1}{\sqrt m}$.
Step 1. Ratio. $\sqrt{\dfrac{12}{4}} = \sqrt3$.
Answer. $\sqrt3 : 1$ (Option A).
Solution by Sreeraj P, M.Sc Physics
An electron (mass $m$) with an initial velocity $\vec v = v_0\hat i$ $(v_0 > 0)$ is moving in an electric field $\vec E = -E_0\hat i$ $(E_0 > 0)$ where $E_0$ is constant. If at $t = 0$, de-Broglie wavelength is $\lambda_0 = \dfrac{h}{mv_0}$, then its de-Broglie wavelength after time $t$ is given by
Answer: (D) $\dfrac{\lambda_0}{\left(1 + \dfrac{eE_0t}{mv_0}\right)}$
Concept. The electron's force is $-e\vec E=+eE_0\hat i$, along its velocity, so it speeds up: $v=v_0+\dfrac{eE_0}{m}t$.
Step 1. Wavelength. $\lambda=\dfrac{h}{mv}=\dfrac{h}{mv_0\left(1+\frac{eE_0t}{mv_0}\right)}$.
Answer. $\dfrac{\lambda_0}{1+\frac{eE_0t}{mv_0}}$ (Option D).
Solution by Sreeraj P, M.Sc Physics
With reference to the observations in photo-electric effect, identify the correct statements from below:
A. The square of maximum velocity of photoelectrons varies linearly with frequency of incident light.
B. The value of saturation current increases on moving the source of light away from the metal surface.
C. The maximum kinetic energy of photo-electrons decreases on decreasing the power of LED (light emitting diode) source of light.
D. The immediate emission of photo-electrons out of metal surface can not be explained by particle nature of light/electromagnetic waves.
E. Existence of threshold wavelength can not be explained by wave nature of light/electromagnetic waves.
Choose the correct answer from the options given below:
Answer: (B) A and E only
Concept. $\tfrac12mv_{\max}^2=h\nu-\phi$: linear in $\nu$, independent of intensity.
Step 1. Check. A: true. B: moving the source away lowers intensity, so the saturation current falls: false. C: power changes intensity, not $K_{\max}$: false. D: instant emission is exactly what the photon picture explains: false. E: true; waves would give emission at any frequency if intense enough.
Answer. A and E only (Option B).
Solution by Sreeraj P, M.Sc Physics
The equation $\lambda = \dfrac{1.227}{x}$ nm can be used to find the de-Broglie wavelength of an electron. In this equation $x$ stands for :
Where, $m$ = mass of electron, $P$ = momentum of electron, $K$ = Kinetic energy of electron, $V$ = Accelerating potential in volts for electron
Answer: (D) $\sqrt V$
Concept. An electron accelerated through $V$ volts has $\lambda=\dfrac{h}{\sqrt{2meV}}$. Putting in the constants gives $\dfrac{1.227}{\sqrt V}$ nm.
Step 1. Identify. $x=\sqrt V$.
Answer. Option D.
Solution by Sreeraj P, M.Sc Physics
A particle of mass $9.1\times10^{-31}$ kg travels in a medium with a speed of $10^6\ \text{m s}^{-1}$ and a photon of radiation of linear momentum $10^{-27}\ \text{kg m s}^{-1}$ travels in a vacuum. The wavelength of the photon is ______ times the wavelength of the particle.
Numerical value type. Enter your answer.
Answer: 910
Concept. Both wavelengths are $\dfrac hp$, so their ratio is the inverse ratio of the momenta.
Step 1. Particle's momentum. $9.1\times10^{-31}\times10^6=9.1\times10^{-25}$ kg m s$^{-1}$.
Step 2. Ratio. $\dfrac{\lambda_{ph}}{\lambda}=\dfrac{9.1\times10^{-25}}{10^{-27}}=910$.
Answer. $910$.
Solution by Sreeraj P, M.Sc Physics
A moving proton and electron have the same de-Broglie wavelength. If $K$ and $P$ denote the K.E. and momentum respectively, then choose the correct option:
Answer: (B) $K_p < K_e$ and $P_p = P_e$
Concept. $\lambda = \dfrac hP$ and $K = \dfrac{P^2}{2m}$.
Step 1. Momenta. Equal $\lambda$: $P_p = P_e$.
Step 2. Energies. $m_p > m_e \Rightarrow K_p < K_e$.
Answer. $K_p < K_e$ and $P_p = P_e$ (Option B).
Solution by Sreeraj P, M.Sc Physics
Consider two separate ideal gases of electrons and protons having same number of particles. The temperature of both the gases are same. The ratio of the uncertainty in determining the position of an electron to that of a proton is proportional to:
Answer: (A) $\sqrt{\dfrac{m_p}{m_e}}$
Concept. The momentum spread of a gas particle is of the order of its thermal momentum, $\Delta p\sim\sqrt{mkT}$; the uncertainty principle gives $\Delta x\sim\dfrac{h}{\Delta p}\propto\dfrac1{\sqrt{mT}}$.
Step 1. Same temperature. $\dfrac{\Delta x_e}{\Delta x_p}=\sqrt{\dfrac{m_p}{m_e}}$.
Answer. Option A.
Solution by Sreeraj P, M.Sc Physics
Given below are two statements:
Statement I: Two photons having equal linear momenta have equal wavelengths.
Statement II: If the wavelength of the photon is decreased, then the momentum and energy of a photon will also decrease.
In the light of the above statements, choose the correct answer from the options given below.
Answer: (C) Statement I is true but Statement II is false
Concept. A photon has $p=\dfrac h\lambda$ and $E=\dfrac{hc}\lambda$.
Step 1. Statement I. Equal momenta give equal wavelengths: true.
Step 2. Statement II. A smaller $\lambda$ gives a larger $p$ and a larger $E$: false.
Answer. Statement I true, Statement II false (Option C).
Solution by Sreeraj P, M.Sc Physics
An X-ray tube is operated at 1.24 million volt. The shortest wavelength of the produced photon will be:
Answer: (A) $10^{-3}$ nm
Concept. The shortest X-ray wavelength comes when one electron gives all its energy $eV$ to one photon: $\lambda_{\min}=\dfrac{hc}{eV}$.
Step 1. Value. $\dfrac{1240\ \text{eV nm}}{1.24\times10^6\ \text{eV}}=10^{-3}$ nm.
Answer. Option A.
Solution by Sreeraj P, M.Sc Physics
The de Broglie wavelength of a proton and α-particle are equal. The ratio of their velocities is
Answer: (A) 4 : 1
Concept. $\lambda = \dfrac{h}{mv}$: equal $\lambda$ means $m_pv_p = m_\alpha v_\alpha$.
Step 1. Ratio. $\dfrac{v_p}{v_\alpha} = \dfrac{m_\alpha}{m_p} = 4$.
Answer. $4 : 1$ (Option A).
Solution by Sreeraj P, M.Sc Physics
An $\alpha$ particle and a proton are accelerated from rest by a potential difference of 200 V. After this, their de Broglie wavelengths are $\lambda_\alpha$ and $\lambda_p$ respectively. The ratio $\dfrac{\lambda_p}{\lambda_\alpha}$ is:
Answer: (C) 2.8
Concept. $\lambda = \dfrac{h}{\sqrt{2mqV}}$.
Step 1. Ratio.
$$\frac{\lambda_p}{\lambda_\alpha} = \sqrt{4 \times 2} = 2\sqrt2 \approx 2.8$$
Answer. $2.8$ (Option C).
Solution by Sreeraj P, M.Sc Physics
The speed of electrons in a scanning electron microscope is $1\times10^7\ \text{m s}^{-1}$. If the protons having the same speed are used instead of electrons, then the resolving power of scanning proton microscope will be changed by a factor of:
Answer: (A) 1837
Concept. A microscope resolves details of the order of the wavelength used, so resolving power $\propto\dfrac1\lambda\propto p=mv$.
Step 1. Same speed. The factor is $\dfrac{m_p}{m_e}=1837$.
Answer. Increased $1837$ times (Option A).
Solution by Sreeraj P, M.Sc Physics
The radiation corresponding to $3 \to 2$ transition of a hydrogen atom falls on a gold surface to generate photoelectrons. These electrons are passed through a magnetic field of $5\times10^{-4}$ T. Assume that the radius of the largest circular path followed by these electrons is 7 mm, the work function of the metal is:
(Mass of electron $= 9.1\times10^{-31}$ kg)
Answer: (D) 0.82 eV
Concept. Photon energy from the hydrogen transition; electron's $K_{max}$ from its circle radius ($p=eBr$); then $\phi=E-K_{max}$.
Step 1. Photon. $E=13.6\left(\dfrac14-\dfrac19\right)=1.89$ eV.
Step 2. Electron. $p=eBr=1.6\times10^{-19}\times5\times10^{-4}\times7\times10^{-3}=5.6\times10^{-25}$ kg m s$^{-1}$; $K=\dfrac{p^2}{2m}=\dfrac{3.14\times10^{-49}}{1.82\times10^{-30}}=1.72\times10^{-19}$ J $=1.08$ eV.
Step 3. Work function. $1.89-1.08=0.81$ eV.
Answer. $0.82$ eV (Option D).
Solution by Sreeraj P, M.Sc Physics
An electron having de-Broglie wavelength $\lambda$ is incident on a target in a $X$-ray tube. Cut-off wavelength of emitted $X$-ray is:
Answer: (C) $\frac{2mc\lambda^2}{h}$
Concept. The electron's kinetic energy $K=\dfrac{p^2}{2m}=\dfrac{h^2}{2m\lambda^2}$ can all go into one X-ray photon: $\lambda_c=\dfrac{hc}{K}$.
Step 1. Cut-off. $\lambda_c=\dfrac{hc\cdot2m\lambda^2}{h^2}=\dfrac{2mc\lambda^2}{h}$.
Answer. Option C.
Solution by Sreeraj P, M.Sc Physics
A certain metallic surface is illuminated by monochromatic radiation of wavelength $\lambda$. The stopping potential for photoelectric current for this radiation is $3V_0$. If the same surface is illuminated with a radiation of wavelength $2\lambda$, the stopping potential is $V_0$. The threshold wavelength of this surface for photoelectric effect is ______ $\lambda$.
Numerical value type. Enter your answer.
Answer: 4
Concept. Einstein's equation at both wavelengths.
Step 1. Equations. $3eV_0 = \dfrac{hc}{\lambda} - \phi$, $eV_0 = \dfrac{hc}{2\lambda} - \phi$.
Step 2. Subtract. $2eV_0 = \dfrac{hc}{2\lambda} \Rightarrow eV_0 = \dfrac{hc}{4\lambda}$, so $\phi = \dfrac{hc}{4\lambda}$.
Step 3. Threshold. $\lambda_0 = 4\lambda$.
Answer. $4$.
Solution by Sreeraj P, M.Sc Physics
An electron of mass $m_e$ and a proton of mass $m_p$ are accelerated through the same potential difference. The ratio of the de-Broglie wavelength associated with the electron to that with the proton is
Answer: (C) $\sqrt{\frac{m_p}{m_e}}$
Concept. $\lambda = \dfrac{h}{\sqrt{2mqV}}$ with equal $q$ and $V$.
Step 1. Ratio. $\dfrac{\lambda_e}{\lambda_p} = \sqrt{\dfrac{m_p}{m_e}}$.
Answer. $\sqrt{\dfrac{m_p}{m_e}}$ (Option C).
Solution by Sreeraj P, M.Sc Physics
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: An electron microscope can achieve better resolving power than an optical microscope.
Reason R: The de Broglie's wavelength of the electrons emitted from an electron gun is much less than wavelength of visible light.
In the light of the above statements, choose the correct answer from the options given below:
Answer: (B) Both A and R are true and R is the correct explanation of A
Concept. Resolving power improves as the wavelength falls. Electrons accelerated through kilovolts have $\lambda\sim0.01$ nm, far below visible wavelengths ($\sim500$ nm).
Step 1. Assertion and reason. Both true, and the short wavelength is the reason.
Answer. Option B.
Solution by Sreeraj P, M.Sc Physics
Two streams of photons, possessing energies equal to twice and ten times the work function of metal are incident on the metal surface successively. The value of ratio of maximum velocities of the photoelectrons emitted in the two respective cases is $x : 3$. The value of $x$ is
Numerical value type. Enter your answer.
Answer: 1
Concept. $K_{\max} = E - \phi$ and $v \propto \sqrt K$.
Step 1. KEs. $K_1 = \phi$, $K_2 = 9\phi$.
Step 2. Ratio. $\dfrac{v_1}{v_2} = \dfrac13$, so $x = 1$.
Answer. $1$.
Solution by Sreeraj P, M.Sc Physics
When radiation of wavelength $\lambda$ is incident on a metallic surface, the stopping potential of ejected photoelectrons is $4.8$ V. If the same surface is illuminated by radiation of double the previous wavelength, then the stopping potential becomes $1.6$ V. The threshold wavelength of the metal is:
Answer: (B) $4\lambda$
Concept. Einstein's equation at both wavelengths.
Step 1. Equations. $\dfrac{hc}{\lambda} - \phi = 4.8$ eV, $\dfrac{hc}{2\lambda} - \phi = 1.6$ eV.
Step 2. Subtract. $\dfrac{hc}{2\lambda} = 3.2$ eV, so $\dfrac{hc}{\lambda} = 6.4$ eV and $\phi = 1.6$ eV.
Step 3. Threshold. $\phi = \dfrac14\dfrac{hc}{\lambda} \Rightarrow \lambda_0 = 4\lambda$.
Answer. $4\lambda$ (Option B).
Solution by Sreeraj P, M.Sc Physics
A light beam of wavelength $500$ nm is incident on a metal having work function of $1.25$ eV, placed in a magnetic field of intensity $B$. The electrons emitted perpendicular to the magnetic field $B$, with maximum kinetic energy are bent into a circular arc of radius $30$ cm. The value of $B$ is ______ $\times10^{-7}$ T. Given $hc = 20\times10^{-26}$ J m, the mass of the electron $= 9\times10^{-31}$ kg.
Numerical value type. Enter your answer.
Answer: 125
Concept. $K_{\max} = \dfrac{hc}{\lambda} - \phi$; in the field $r = \dfrac{p}{eB}$.
Step 1. Photon energy. $\dfrac{20 \times 10^{-26}}{500 \times 10^{-9}} = 4 \times 10^{-19}$ J $= 2.5$ eV.
Step 2. KE. $1.25$ eV $= 2 \times 10^{-19}$ J; $p = \sqrt{2 \times 9 \times 10^{-31} \times 2 \times 10^{-19}} = 6 \times 10^{-25}$.
Step 3. Field.
$$B = \frac{6 \times 10^{-25}}{1.6 \times 10^{-19} \times 0.3} = 1.25 \times 10^{-5}\ \text{T} = 125 \times 10^{-7}\ \text{T}$$
Answer. $125$.
Solution by Sreeraj P, M.Sc Physics
In a photoelectric experiment, ultraviolet light of wavelength $280$ nm is used with lithium cathode having work function $\phi = 2.5$ eV. If the wavelength of incident light is switched to $400$ nm, find out the change in the stopping potential. ($h = 6.63\times10^{-34}$ J s, $c = 3\times10^8$ m s$^{-1}$)
Answer: (C) $1.3$ V
Concept. $e\Delta V_s = hc\left(\dfrac{1}{\lambda_1} - \dfrac{1}{\lambda_2}\right)$ (the work function cancels).
Step 1. Substitute. $hc \approx 1243$ eV nm:
$$\Delta V_s = 1243\left(\frac{1}{280} - \frac{1}{400}\right) = 4.44 - 3.11 \approx 1.3\ \text{V}$$
Answer. $1.3$ V (Option C).
Solution by Sreeraj P, M.Sc Physics
The de-Broglie wavelength of a particle having kinetic energy $E$ is $\lambda$. How much extra energy must be given to this particle so that the de-Broglie wavelength reduces to $75\%$ of the initial value?
Answer: (B) $\frac{7E}{9}$
Concept. $\lambda\propto\dfrac1{\sqrt K}$, so $K\propto\dfrac1{\lambda^2}$.
Step 1. New energy. $\lambda'=\dfrac34\lambda$ gives $K'=\dfrac{16}9E$.
Step 2. Extra. $\dfrac{16E}9-E=\dfrac{7E}9$.
Answer. Option B.
Solution by Sreeraj P, M.Sc Physics
In a photoelectric experiment, increasing the intensity of incident light:
Answer: (C) increases the number of photons incident and the K.E. of the ejected electrons remains unchanged.
Concept. At fixed frequency, intensity sets the photon count; each photon still carries $hf$.
Step 1. Effect. More photons per second; $K_{\max} = hf - \phi$ unchanged.
Answer. Increases the number of photons incident and the K.E. of the ejected electrons remains unchanged (Option C).
Solution by Sreeraj P, M.Sc Physics
A monochromatic neon lamp with wavelength of $670.5$ nm illuminates a photo-sensitive material which has a stopping voltage of $0.48$ V. What will be the stopping voltage if the source light is changed with another source of wavelength of $474.6$ nm?
Answer: (C) $1.25$ V
Concept. Find $\phi$ from the first source, then use it for the second; $hc \approx 1240$ eV nm.
Step 1. Work function. $\dfrac{1240}{670.5} = 1.85$ eV, so $\phi = 1.37$ eV.
Step 2. New stopping voltage. $\dfrac{1240}{474.6} = 2.61$ eV, so $V_s = 2.61 - 1.37 \approx 1.25$ V.
Answer. $1.25$ V (Option C).
Solution by Sreeraj P, M.Sc Physics
An electron (mass $m$) with initial velocity $\vec{v} = v_0\hat{i} + v_0\hat{j}$ is in an electric field $\vec{E} = -E_0\hat{k}$. If $\lambda_0$ is initial de-Broglie wavelength of electron, its de-Broglie wavelength at time $t$ is given by:
Answer: (C) $\dfrac{\lambda_0}{\sqrt{1 + \dfrac{e^2E_0^2t^2}{2m^2v_0^2}}}$
Concept. The force $-e\vec E = eE_0\hat k$ adds a $z$-velocity; $\lambda = \dfrac{h}{mv}$.
Step 1. Speed. $v^2 = 2v_0^2 + \dfrac{e^2E_0^2t^2}{m^2}$.
Step 2. Wavelength. $\lambda_0 = \dfrac{h}{\sqrt2mv_0}$:
$$\lambda = \frac{\lambda_0\sqrt2v_0}{\sqrt{2v_0^2 + \frac{e^2E_0^2t^2}{m^2}}} = \frac{\lambda_0}{\sqrt{1 + \dfrac{e^2E_0^2t^2}{2m^2v_0^2}}}$$
Answer. $\dfrac{\lambda_0}{\sqrt{1 + \dfrac{e^2E_0^2t^2}{2m^2v_0^2}}}$ (Option C).
Solution by Sreeraj P, M.Sc Physics
Radiation, with wavelength $6561$ Å falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of $3\times10^{-4}$ T. If the radius of the largest circular path followed by the electrons is $10$ mm, the work function of the metal is close to:
Answer: (C) $1.1$ eV
Concept. The largest circle is for the fastest electrons, $p = eBr$; then $\phi = E - K_{\max}$.
Step 1. Photon energy. $\dfrac{12400}{6561} \approx 1.89$ eV.
Step 2. Momentum. $1.6 \times 10^{-19} \times 3 \times 10^{-4} \times 10^{-2} = 4.8 \times 10^{-25}$ kg m/s.
Step 3. KE. $\dfrac{(4.8 \times 10^{-25})^2}{2 \times 9.1 \times 10^{-31}} = 1.27 \times 10^{-19}$ J $\approx 0.79$ eV.
Step 4. Work function. $1.89 - 0.79 \approx 1.1$ eV.
Answer. $1.1$ eV (Option C).
Solution by Sreeraj P, M.Sc Physics
A particle moving with kinetic energy $E$ has de Broglie wavelength $\lambda$. If energy $\Delta E$ is added to its energy, the wavelength becomes $\dfrac{\lambda}{2}$. Value of $\Delta E$, is:
Answer: (C) $3E$
Concept. $\lambda \propto \dfrac{1}{\sqrt E}$.
Step 1. Halve $\lambda$. Energy must become $4E$.
Step 2. Extra. $\Delta E = 3E$.
Answer. $3E$ (Option C).
Solution by Sreeraj P, M.Sc Physics
An electron of mass $m$ and magnitude of charge $e$ at rest, gets accelerated by a constant electric field $E$. The rate of change of de-Broglie wavelength of this electron at a time $t$ is (ignore relativistic effects)
Answer: (D) $\dfrac{d\lambda}{dt} = -\dfrac{h}{eEt^2}$
Concept. From rest in a constant field, $p=eEt$, so $\lambda=\dfrac{h}{eEt}$.
Step 1. Differentiate. $\dfrac{d\lambda}{dt}=-\dfrac{h}{eEt^2}$.
Answer. Option D.
Solution by Sreeraj P, M.Sc Physics
The surface of a metal is illuminated alternately with photons of energies $E_1 = 4$ eV and $E_2 = 2.5$ eV respectively. The ratio of the maximum speeds of the photoelectrons emitted in the two cases is $2$. The work function of the metal (in eV) is ______.
Numerical value type. Enter your answer.
Answer: 2
Concept. $\tfrac12mv_{max}^2=E-\phi$, so $\dfrac{v_1^2}{v_2^2}=\dfrac{E_1-\phi}{E_2-\phi}$.
Step 1. Equation. $4=\dfrac{4-\phi}{2.5-\phi}$, so $10-4\phi=4-\phi$.
Step 2. Work function. $\phi=2$ eV.
Answer. $2$.
Solution by Sreeraj P, M.Sc Physics
An electron, a doubly ionized helium ion ($\text{He}^{++}$) and a proton are having the same kinetic energy. The relation between their respective de Broglie wavelengths $\lambda_e$, $\lambda_{\text{He}^{++}}$ and $\lambda_p$ is:
Answer: (C) $\lambda_e > \lambda_p > \lambda_{\text{He}^{++}}$
Concept. Equal KE: $\lambda \propto \dfrac{1}{\sqrt m}$.
Step 1. Order. $m_e < m_p < m_{\text{He}}$, so $\lambda_e > \lambda_p > \lambda_{\text{He}^{++}}$.
Answer. $\lambda_e > \lambda_p > \lambda_{\text{He}^{++}}$ (Option C).
Solution by Sreeraj P, M.Sc Physics
Assuming the nitrogen molecule is moving with r.m.s. velocity at $400$ K, the de Broglie wavelength of the nitrogen molecule is close to: (Given: nitrogen molecule weight: $4.64\times10^{-26}$ kg, Boltzmann constant: $1.38\times10^{-23}\ \text{J K}^{-1}$, Planck constant: $6.63\times10^{-34}$ J s)
Answer: (A) $0.24$ Å
Concept. At the rms speed, $p=mv_{rms}=\sqrt{3mkT}$ and $\lambda=\dfrac hp$.
Step 1. Momentum. $p=\sqrt{3\times4.64\times10^{-26}\times1.38\times10^{-23}\times400}=2.77\times10^{-23}$ kg m s$^{-1}$.
Step 2. Wavelength. $\lambda=\dfrac{6.63\times10^{-34}}{2.77\times10^{-23}}=2.4\times10^{-11}$ m $=0.24$ Å.
Answer. Option A.
Solution by Sreeraj P, M.Sc Physics
A beam of electromagnetic radiation of intensity $6.4\times10^{-5}\ \text{W/cm}^2$ is comprised of wavelength $\lambda = 310$ nm. It falls normally on a metal (work function $\varphi = 2$ eV) of surface area $1\ \text{cm}^2$. If one in $10^3$ photons ejects an electron, the total number of electrons ejected in $1$ s is $10^x$. ($hc = 1240$ eV nm, $1$ eV $= 1.6\times10^{-19}$ J), then $x$ is ______.
Numerical value type. Enter your answer.
Answer: 11
Concept. Photons per second $= \dfrac{P}{E_{\text{photon}}}$; one in $10^3$ ejects an electron.
Step 1. Power. $6.4 \times 10^{-5}$ W on $1\ \text{cm}^2$.
Step 2. Photon energy. $\dfrac{1240}{310} = 4$ eV $= 6.4 \times 10^{-19}$ J (above the work function).
Step 3. Counts. $10^{14}$ photons/s, so $10^{11}$ electrons/s: $x = 11$.
Answer. $11$.
Solution by Sreeraj P, M.Sc Physics
An electron (of mass $m$) and a photon have the same energy $E$ in the range of a few eV. The ratio of the de Broglie wavelength associated with the electron and the wavelength of the photon is ($c$ = speed of light in vacuum):
Answer: (C) $\dfrac1c\left(\dfrac{E}{2m}\right)^{1/2}$
Concept. Electron $\lambda_e = \dfrac{h}{\sqrt{2mE}}$ (non-relativistic); photon $\lambda_p = \dfrac{hc}{E}$.
Step 1. Ratio.
$$\frac{\lambda_e}{\lambda_p} = \frac{E}{c\sqrt{2mE}} = \frac1c\left(\frac{E}{2m}\right)^{1/2}$$
(The official answer key lists $\left(\frac{E}{2m}\right)^{1/2}$, which has the dimension of speed and cannot be a ratio of wavelengths.)
Answer. $\dfrac1c\left(\dfrac{E}{2m}\right)^{1/2}$ (Option C).
Solution by Sreeraj P, M.Sc Physics
When a photon of energy $4.0$ eV strikes the surface of a metal A, the ejected photoelectrons have maximum kinetic energy $T_A$ eV and de Broglie wavelength $\lambda_A$. The maximum kinetic energy of photoelectrons liberated from another metal B by a photon of energy $4.50$ eV is $T_B = (T_A - 1.5)$ eV. If the de Broglie wavelength of these photoelectrons is $\lambda_B = 2\lambda_A$, then the work function of metal B is:
Answer: (A) $4$ eV
Concept. For electrons $\lambda=\dfrac{h}{\sqrt{2mT}}$, so $\lambda_B=2\lambda_A$ means $T_B=\dfrac{T_A}{4}$.
Step 1. Energies. $\dfrac{T_A}{4}=T_A-1.5$, so $T_A=2$ eV and $T_B=0.5$ eV.
Step 2. Work function of B. $4.5-0.5=4$ eV.
Answer. $4$ eV (Option A).
Solution by Sreeraj P, M.Sc Physics
When radiation of wavelength $\lambda$ is used to illuminate a metallic surface, the stopping potential is $V$. When the same surface is illuminated with radiation of wavelength $3\lambda$, the stopping potential is $\dfrac{V}{4}$. If the threshold wavelength for the metallic surface is $n\lambda$ then value of $n$ will be ______.
Numerical value type. Enter your answer.
Answer: 9
Concept. Einstein's equation at both wavelengths.
Step 1. Equations. $\dfrac{hc}{\lambda} - \phi = eV$, $\dfrac{hc}{3\lambda} - \phi = \dfrac{eV}{4}$.
Step 2. Eliminate $V$. $\dfrac{hc}{\lambda} - \phi = \dfrac{4hc}{3\lambda} - 4\phi \Rightarrow \phi = \dfrac{hc}{9\lambda}$.
Step 3. Threshold. $\lambda_0 = 9\lambda$, $n = 9$.
Answer. $9$.
Solution by Sreeraj P, M.Sc Physics
A particle is moving $5$ times as fast as an electron. The ratio of the de-Broglie wavelength of the particle to that of the electron is $1.878 \times 10^{-4}$. The mass of the particle is close to:
Answer: (D) $9.7\times10^{-28}\ \text{kg}$
Concept. $\lambda=\dfrac{h}{mv}$, so $\dfrac{\lambda_p}{\lambda_e}=\dfrac{m_ev_e}{m_pv_p}$.
Step 1. Equation. $1.878\times10^{-4}=\dfrac{m_e}{5m_p}$.
Step 2. Mass. $m_p=\dfrac{9.1\times10^{-31}}{5\times1.878\times10^{-4}}=9.7\times10^{-28}$ kg.
Answer. Option D.
Solution by Sreeraj P, M.Sc Physics
When the wavelength of radiation falling on a metal is changed from $500\ \text{nm}$ to $200\ \text{nm}$, the maximum kinetic energy of the photoelectrons becomes three times larger. The work function of the metal is close to:
Answer: (D) $0.61\ \text{eV}$
Concept. $K_{\max} = E - \phi$ at both wavelengths.
Step 1. Photon energies. $2.48$ eV and $6.2$ eV.
Step 2. Equation. $6.2 - \phi = 3(2.48 - \phi) \Rightarrow \phi \approx 0.62$ eV; the closest option is $0.61$ eV.
Answer. $0.61$ eV (Option D).
Solution by Sreeraj P, M.Sc Physics
Two sources of light emit X-rays of wavelength $1\ \text{nm}$ and visible light of wavelength $500\ \text{nm}$, respectively. Both the sources emit light of the same power $200\ \text{W}$. The ratio of the number density of photons of X-rays to the number density of photons of the visible light of the given wavelengths is:
Answer: (A) $\dfrac{1}{500}$
Concept. Same power: photon number $\propto \lambda$.
Step 1. Ratio. $\dfrac{1}{500}$.
Answer. $\dfrac{1}{500}$ (Option A).
Solution by Sreeraj P, M.Sc Physics
Particle $A$ of mass $m_A = \dfrac m2$ moving along the $x$-axis with velocity $v_0$ collides elastically with another particle $B$ at rest having mass $m_B = \dfrac m3$. If both the particles move along the $x$-axis after the collision, the change $\Delta\lambda$ in the wavelength of the particle $A$, in terms of its de-Broglie wavelength $(\lambda_0)$ before the collision is:
Answer: (D) $\Delta\lambda = 4\lambda_0$
Concept. Head-on elastic collision with B at rest: $v_A'=\dfrac{m_A-m_B}{m_A+m_B}v_0$. Then $\lambda\propto\dfrac1v$ for A.
Step 1. Speed after. $v_A'=\dfrac{\frac12-\frac13}{\frac12+\frac13}v_0=\dfrac{v_0}5$.
Step 2. Wavelength. $\lambda'=5\lambda_0$, so $\Delta\lambda=4\lambda_0$.
Answer. Option D.
Solution by Sreeraj P, M.Sc Physics
The given figure shows a few data points in a photoelectric effect experiment for a certain metal, plotted as stopping potential against frequency. The minimum energy for ejection of electrons from its surface is: (Planck's constant $h = 6.62\times10^{-34}\ \text{J s}$)
Answer: (A) $2.27\ \text{eV}$
Concept. The stopping potential is zero at the threshold frequency; $\phi = hf_0$.
Step 1. Threshold. From the graph, $f_0 = 5.5 \times 10^{14}$ Hz.
Step 2. Work function.
$$\phi = \frac{6.62 \times 10^{-34} \times 5.5 \times 10^{14}}{1.6 \times 10^{-19}} \approx 2.27\ \text{eV}$$
Answer. $2.27$ eV (Option A).
Solution by Sreeraj P, M.Sc Physics
In a photoelectric effect experiment, the graph of stopping potential $V_s$ versus reciprocal of wavelength obtained is a straight line as shown in the figure. As the intensity of incident radiation is increased:
Answer: (D) Graph does not change
Concept. $V_s=\dfrac{hc}{e}\cdot\dfrac1\lambda-\dfrac\phi e$: the slope ($hc/e$) and intercept ($\phi$) do not involve intensity.
Step 1. Effect. A brighter beam gives more electrons but the same $V_s$ at each $\lambda$: the graph is unchanged.
Answer. Option D.
Solution by Sreeraj P, M.Sc Physics
A beam of electrons of energy $E$ scatters from a target having atomic spacing of $1\ \text{Å}$. The first maximum intensity occurs at $\theta = 60^\circ$. Then $E$ (in eV) is ______. (Planck's constant $h = 6.64\times10^{-34}$ J s, $1\ \text{eV} = 1.6\times10^{-19}$ J, electron mass $m = 9.1\times10^{-31}$ kg)
Numerical value type. Enter your answer.
Answer: 50
Concept. Bragg condition for the first maximum: $2d\sin\theta = \lambda$; then $E = \dfrac{h^2}{2m\lambda^2}$.
Step 1. Wavelength. $\lambda = 2 \times 1 \times \sin60° = \sqrt3$ Å.
Step 2. Energy.
$$E = \frac{(6.64 \times 10^{-34})^2}{2 \times 9.1 \times 10^{-31} \times 3 \times 10^{-20}} \approx 8.1 \times 10^{-18}\ \text{J} \approx 50\ \text{eV}$$
Answer. $50$.
Solution by Sreeraj P, M.Sc Physics
The magnetic field associated with a light wave is given, at the origin, by $B = B_0\left[\sin(3.14\times10^7)ct + \sin(6.28\times10^7)ct\right]$. If this light falls on a silver plate having a work function of $4.7\ \text{eV}$, what will be the maximum kinetic energy of the photoelectrons? $(c = 3\times10^8\ \text{m s}^{-1},\ h = 6.6\times10^{-34}\ \text{J s})$
Answer: (B) $7.72\ \text{eV}$
Concept. The light contains two frequencies, $\omega=kc$ with $k=3.14\times10^7$ and $6.28\times10^7$ m$^{-1}$. The higher frequency gives the maximum kinetic energy.
Step 1. Higher frequency. $\omega=6.28\times10^7\times3\times10^8$, so $\nu=\dfrac{\omega}{2\pi}=3\times10^{15}$ Hz.
Step 2. Photon energy. $h\nu=\dfrac{6.6\times10^{-34}\times3\times10^{15}}{1.6\times10^{-19}}=12.4$ eV.
Step 3. Kinetic energy. $12.4-4.7=7.7$ eV.
Answer. $7.72$ eV (Option B).
Solution by Sreeraj P, M.Sc Physics
In an electron microscope, the resolution that can be achieved is of the order of the wavelength of electrons used. To resolve a width of $7.5\times10^{-12}\ \text{m}$, the minimum electron energy required is close to:
Answer: (B) $25\ \text{keV}$
Concept. Take $\lambda$ equal to the width to resolve; then $K=\dfrac{p^2}{2m}=\dfrac{h^2}{2m\lambda^2}$.
Step 1. Energy. $K=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(7.5\times10^{-12})^2}=4.3\times10^{-15}$ J.
Step 2. In eV. $\dfrac{4.3\times10^{-15}}{1.6\times10^{-19}}=2.7\times10^4$ eV $\approx25$ keV.
Answer. Option B.
Solution by Sreeraj P, M.Sc Physics
A metal plate of area $1\times10^{-4}\ \text{m}^2$ is illuminated by a radiation of intensity $16\ \text{mW/m}^2$. The work function of the metal is $5\ \text{eV}$. The energy of the incident photons is $10\ \text{eV}$ and only $10\%$ of it produces photo electrons. The number of emitted photo electrons per second and their maximum energy, respectively, will be: $[1\ \text{eV} = 1.6\times10^{-19}\ \text{J}]$
Answer: (C) $10^{11}$ and $5\ \text{eV}$
Concept. Photons per second $=\dfrac{IA}{E}$; electrons per second $=$ efficiency $\times$ photons; $K_{max}=E-\phi$.
Step 1. Power on the plate. $16\times10^{-3}\times10^{-4}=1.6\times10^{-6}$ W.
Step 2. Photons. $\dfrac{1.6\times10^{-6}}{10\times1.6\times10^{-19}}=10^{12}$ per second; $10\%$ gives $10^{11}$ electrons per second.
Step 3. Energy. $10-5=5$ eV.
Answer. $10^{11}$ and $5$ eV (Option C).
Solution by Sreeraj P, M.Sc Physics
In a photoelectric effect experiment, the threshold wavelength of light is $380\ \text{nm}$. If the wavelength of incident light is $260\ \text{nm}$, the maximum kinetic energy of emitted electrons will be (Given $E\,(\text{in eV}) = \dfrac{1237}{\lambda\,(\text{in nm})}$)
Answer: (C) $1.5\ \text{eV}$
Concept. $K_{max}=\dfrac{hc}{\lambda}-\dfrac{hc}{\lambda_0}$, with $hc=1237$ eV nm.
Step 1. Energies. $\dfrac{1237}{260}=4.76$ eV; $\dfrac{1237}{380}=3.26$ eV.
Step 2. Difference. $K_{max}=1.50$ eV.
Answer. $1.5$ eV (Option C).
Solution by Sreeraj P, M.Sc Physics
A $2\ \text{mW}$ laser operates at a wavelength of $500\ \text{nm}$. The number of photons that will be emitted per second is: [Given Planck's constant $h = 6.6\times10^{-34}\ \text{J s}$, speed of light $c = 3.0\times10^8\ \text{m/s}$]
Answer: (B) $5\times10^{15}$
Concept. Photons per second $= \dfrac{P}{hc/\lambda}$.
Step 1. Photon energy. $3.96 \times 10^{-19}$ J.
Step 2. Number. $\dfrac{2 \times 10^{-3}}{3.96 \times 10^{-19}} \approx 5 \times 10^{15}$.
Answer. $5 \times 10^{15}$ (Option B).
Solution by Sreeraj P, M.Sc Physics
The surface of certain metal is first illuminated with light of wavelength $\lambda_1 = 350\ \text{nm}$ and then, by a light of wavelength $\lambda_2 = 540\ \text{nm}$. It is found that the maximum speed of the photoelectrons in the two cases differ by a factor of 2. The work function of the metal (in eV) is close to (Energy of photon $= \dfrac{1240}{\lambda\ \text{in nm}}\ \text{eV}$)
Answer: (B) $1.8$
Concept. Speeds differ by 2, so maximum KEs are in ratio $4 : 1$.
Step 1. Photon energies. $E_1 = 3.54$ eV, $E_2 = 2.30$ eV.
Step 2. Equation. $E_1 - \phi = 4(E_2 - \phi) \Rightarrow 3\phi = 9.19 - 3.54 = 5.65$.
Step 3. Result. $\phi \approx 1.9$ eV, closest option $1.8$.
Answer. $1.8$ (Option B).
Solution by Sreeraj P, M.Sc Physics
Two particles move at right angle to each other. Their de Broglie wavelengths are $\lambda_1$ and $\lambda_2$ respectively. The particles suffer perfectly inelastic collision. The de Broglie wavelength $\lambda$ of the final particle, is given by
Answer: (D) $\dfrac1{\lambda^2} = \dfrac1{\lambda_1^2} + \dfrac1{\lambda_2^2}$
Concept. In a perfectly inelastic collision the joined particle carries the vector sum of the momenta. At right angles: $p^2=p_1^2+p_2^2$, with $p=\dfrac h\lambda$.
Step 1. Substitute. $\dfrac{h^2}{\lambda^2}=\dfrac{h^2}{\lambda_1^2}+\dfrac{h^2}{\lambda_2^2}$.
Answer. $\dfrac1{\lambda^2}=\dfrac1{\lambda_1^2}+\dfrac1{\lambda_2^2}$ (Option D).
Solution by Sreeraj P, M.Sc Physics
The electric field of light wave is given as $\vec E = 10^{-3}\cos\left(\dfrac{2\pi x}{5\times10^{-7}} - 2\pi\times6\times10^{14}t\right)\hat x\ \dfrac{\text{N}}{\text{C}}$. This light falls on a metal plate of work function $2\ \text{eV}$. The stopping potential of the photo-electrons is (Given, $E\ (\text{in eV}) = \dfrac{12375}{\lambda\ (\text{in Å})}$)
Answer: (D) $0.48\ \text{V}$
Concept. Read the wavelength from the field: $k=\dfrac{2\pi}{\lambda}$ gives $\lambda=5\times10^{-7}$ m $=5000$ Å (and $c/\nu$ agrees). Then $eV_0=E-\phi$.
Step 1. Photon energy. $E=\dfrac{12375}{5000}=2.475$ eV.
Step 2. Stopping potential. $V_0=2.475-2=0.475\approx0.48$ V.
Answer. Option D.
Solution by Sreeraj P, M.Sc Physics
A particle P is formed due to a completely inelastic collision of particles x and y having de Broglie wavelengths $\lambda_x$ and $\lambda_y$ respectively. If x and y were moving in opposite directions, then the de Broglie wavelength of P is
Answer: (C) $\dfrac{\lambda_x\lambda_y}{|\lambda_x - \lambda_y|}$
Concept. Moving in opposite directions, the combined momentum is the difference: $p=|p_x-p_y|$.
Step 1. Substitute. $\dfrac h\lambda=\left|\dfrac h{\lambda_x}-\dfrac h{\lambda_y}\right|$.
Step 2. Solve. $\lambda=\dfrac{\lambda_x\lambda_y}{|\lambda_x-\lambda_y|}$.
Answer. Option C.
Solution by Sreeraj P, M.Sc Physics
If the de Broglie wavelength of an electron is equal to $10^{-3}$ times the wavelength of a photon of frequency $6\times10^{14}$ Hz, then the speed of electron is equal to: (Speed of light $= 3\times10^8$ m/s, Planck's constant $= 6.63\times10^{-34}$ J s, mass of electron $= 9.1\times10^{-31}$ kg)
Answer: (D) $1.45\times10^{6}$ m/s
Concept. Photon $\lambda=\dfrac c\nu$; electron $v=\dfrac{h}{m\lambda_e}$.
Step 1. Wavelengths. $\lambda_{ph}=\dfrac{3\times10^8}{6\times10^{14}}=5\times10^{-7}$ m; $\lambda_e=5\times10^{-10}$ m.
Step 2. Speed. $v=\dfrac{6.63\times10^{-34}}{9.1\times10^{-31}\times5\times10^{-10}}=1.45\times10^6$ m/s.
Answer. Option D.
Solution by Sreeraj P, M.Sc Physics
In a photoelectric experiment, the wavelength of the light incident on a metal is changed from $300$ nm to $400$ nm. The decrease in the stopping potential is close to: $\left(\dfrac{hc}{e} = 1240\ \text{nm V}\right)$
Answer: (C) $1.0$ V
Concept. Subtract Einstein's equation for the two wavelengths; $\phi$ cancels: $\Delta V_0=\dfrac{hc}{e}\left(\dfrac1{\lambda_1}-\dfrac1{\lambda_2}\right)$.
Step 1. Change. $1240\left(\dfrac1{300}-\dfrac1{400}\right)=\dfrac{1240}{1200}=1.03$ V.
Answer. About $1.0$ V (Option C).
Solution by Sreeraj P, M.Sc Physics
A particle A of mass $m$ and charge $q$ is accelerated by a potential difference of $50$ V. Another particle B of mass $4m$ and charge $q$ is accelerated by a potential difference of $2500$ V. The ratio of de Broglie wavelengths $\dfrac{\lambda_A}{\lambda_B}$ is close to:
Answer: (D) $14.14$
Concept. $\lambda = \dfrac{h}{\sqrt{2mqV}}$.
Step 1. Ratio.
$$\frac{\lambda_A}{\lambda_B} = \sqrt{\frac{4m \times 2500}{m \times 50}} = \sqrt{200} \approx 14.14$$
Answer. $14.14$ (Option D).
Solution by Sreeraj P, M.Sc Physics
The stopping potential $V_0$ (in volt) as a function of frequency $(\nu)$ for a sodium emitter, is shown in the figure. The work function of sodium, from the data plotted in the figure, will be: (Given: Planck's constant $h = 6.63\times10^{-34}$ J s, electron charge $e = 1.6\times10^{-19}$ C)
Answer: (D) $1.66$ eV
Concept. The $V_0$–$\nu$ line meets the frequency axis at the threshold $\nu_0$, and $\phi=h\nu_0$.
Step 1. Threshold. $V_0=0$ at $\nu_0=4\times10^{14}$ Hz.
Step 2. Work function. $\phi=\dfrac{6.63\times10^{-34}\times4\times10^{14}}{1.6\times10^{-19}}=1.66$ eV.
Answer. Option D.
Solution by Sreeraj P, M.Sc Physics
Two electrons are moving with non-relativistic speeds perpendicular to each other. If corresponding de Broglie wavelengths are $\lambda_1$ and $\lambda_2$, their de Broglie wavelength in the frame of reference attached to their centre of mass is:
Answer: (B) $\lambda_{CM} = \dfrac{2\lambda_1\lambda_2}{\sqrt{\lambda_1^2 + \lambda_2^2}}$
Concept. For two equal masses, in the centre-of-mass frame each has momentum $\dfrac12|\vec p_1-\vec p_2|$. The momenta are perpendicular, so $|\vec p_1-\vec p_2|=\sqrt{p_1^2+p_2^2}$.
Step 1. Momentum in the CM frame. $p=\dfrac h2\sqrt{\dfrac1{\lambda_1^2}+\dfrac1{\lambda_2^2}}$.
Step 2. Wavelength. $\lambda_{CM}=\dfrac hp=\dfrac{2\lambda_1\lambda_2}{\sqrt{\lambda_1^2+\lambda_2^2}}$.
Answer. Option B.
Solution by Sreeraj P, M.Sc Physics
If the de Broglie wavelengths associated with a proton and an $\alpha$-particle are equal, then the ratio of velocities of the proton and the $\alpha$-particle will be:
Answer: (C) $4 : 1$
Concept. Equal wavelengths mean equal momenta.
Step 1. Ratio. $\dfrac{v_p}{v_\alpha} = \dfrac{m_\alpha}{m_p} = 4$.
Answer. $4 : 1$ (Option C).
Solution by Sreeraj P, M.Sc Physics
The de-Broglie wavelength $(\lambda_B)$ associated with the electron orbiting in the second excited state of hydrogen atom is related to that in the ground state $(\lambda_G)$ by:
Answer: (A) $\lambda_B = 3\lambda_G$
Concept. In Bohr orbit $n$ the circumference holds $n$ wavelengths: $2\pi r_n=n\lambda$. With $r_n\propto n^2$, $\lambda\propto n$.
Step 1. States. Ground state $n=1$; second excited state $n=3$.
Step 2. Ratio. $\lambda_B=3\lambda_G$.
Answer. Option A.
Solution by Sreeraj P, M.Sc Physics
An electron beam is accelerated by a potential difference $V$ to hit a metallic target to produce X-rays. It produces continuous as well as characteristic X-rays. If $\lambda_{\min}$ is the smallest possible wavelength of X-ray in the spectrum, the variation of $\log\lambda_{\min}$ with $\log V$ is correctly represented in:
Answer: (B) see figure
Concept. $\lambda_{\min} = \dfrac{hc}{eV}$.
Step 1. Logs. $\log\lambda_{\min} = \log\dfrac{hc}{e} - \log V$: a falling straight line of slope $-1$, graph (2).
Answer. Graph (2) (Option B).
Solution by Sreeraj P, M.Sc Physics
A particle $A$ of mass $m$ and initial velocity $v$ collides with a particle $B$ of mass $\dfrac m2$ which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths $\lambda_A$ to $\lambda_B$ after the collision is:
Answer: (C) $\dfrac{\lambda_A}{\lambda_B} = 2$
Concept. Elastic head-on collision with B at rest: $v_A'=\dfrac{m_A-m_B}{m_A+m_B}v$, $v_B'=\dfrac{2m_A}{m_A+m_B}v$. Compare momenta.
Step 1. Speeds. $v_A'=\dfrac{m/2}{3m/2}v=\dfrac v3$; $v_B'=\dfrac{2m}{3m/2}v=\dfrac{4v}3$.
Step 2. Momenta. $p_A=\dfrac{mv}3$, $p_B=\dfrac m2\cdot\dfrac{4v}3=\dfrac{2mv}3$.
Step 3. Ratio. $\dfrac{\lambda_A}{\lambda_B}=\dfrac{p_B}{p_A}=2$.
Answer. Option C.
Solution by Sreeraj P, M.Sc Physics
The maximum velocity of the photoelectrons emitted from the surface is $v$ when light of frequency $n$ falls on a metal surface. If the incident frequency is increased to $3n$, the maximum velocity of the ejected photoelectrons will be:
Answer: (A) more than $\sqrt3v$
Concept. $\tfrac12mv^2=hn-\phi$. With $3n$: $\tfrac12mv'^2=3hn-\phi=3(hn-\phi)+2\phi$.
Step 1. Compare. $\tfrac12mv'^2>3\times\tfrac12mv^2$, so $v'>\sqrt3v$.
Answer. More than $\sqrt3v$ (Option A).
Solution by Sreeraj P, M.Sc Physics
A laser light of wavelength $660$ nm is used to weld Retina detachment. If a laser pulse of width $60$ ms and power $0.5$ kW is used, the approximate number of photons in the pulse are: (Take Planck's constant, $h = 6.62\times10^{-34}$ J s)
Answer: (C) $10^{20}$
Concept. A pulse of power $P$ lasting $t$ carries energy $Pt$; each photon carries $\dfrac{hc}{\lambda}$.
Step 1. Energy. $Pt=500\times0.06=30$ J.
Step 2. Photon energy. $\dfrac{hc}\lambda=\dfrac{6.62\times10^{-34}\times3\times10^8}{660\times10^{-9}}=3.0\times10^{-19}$ J.
Step 3. Number. $N=\dfrac{30}{3.0\times10^{-19}}=1.0\times10^{20}$.
Answer. $10^{20}$ (Option C).
Solution by Sreeraj P, M.Sc Physics
When two monochromatic lights of frequency, $\nu$ and $\dfrac{\nu}{2}$ are incident on a photoelectric metal, their stopping potential becomes $\dfrac{V_s}{2}$ and $V_s$ respectively. The threshold frequency for this metal is
Answer: (A) $\dfrac{3}{2}\nu$
Concept. Einstein's equation $eV = h\nu - h\nu_0$ for both cases.
Step 1. Equations. $\dfrac{eV_s}{2} = h\nu - h\nu_0$ and $eV_s = \dfrac{h\nu}{2} - h\nu_0$.
Step 2. Eliminate. $2h\nu - 2h\nu_0 = \dfrac{h\nu}{2} - h\nu_0 \Rightarrow \nu_0 = \dfrac32\nu$.
(This threshold is above both incident frequencies, so the data are not physically consistent; solving the equations as given leads to $\dfrac32\nu$, the expected answer.)
Answer. $\dfrac32\nu$ (Option A).
Solution by Sreeraj P, M.Sc Physics
The graph which shows the variation of the de Broglie wavelength ($\lambda$) of a particle and its associated momentum ($p$) is
Answer: (A) see figure
Concept. $\lambda = \dfrac hp$, so $\lambda p = $ constant.
Step 1. Shape. A rectangular hyperbola: graph (1).
Answer. Graph (1) (Option A).
Solution by Sreeraj P, M.Sc Physics
A monochromatic light is incident on a metallic plate having work function $\phi$. An electron, emitted normally to the plate from a point A with maximum kinetic energy, enters a constant magnetic field perpendicular to the initial velocity of the electron. The electron passes through a curve and hits back the plate at a point B. The distance between A and B is:
(Given: the magnitude of charge of an electron is $e$ and mass is $m$, $h$ is Planck's constant and $c$ is velocity of light. Take the magnetic field to exist throughout the path of the electron.)
Answer: (C) $\dfrac{\sqrt{8m\left(\frac{hc}{\lambda} - \phi\right)}}{eB}$
Concept. The electron leaves normally, the field is perpendicular to its velocity, so it moves on a semicircle and lands a diameter $2r$ away; $r=\dfrac{mv}{eB}=\dfrac{\sqrt{2mK}}{eB}$.
Step 1. Kinetic energy. $K=\dfrac{hc}\lambda-\phi$.
Step 2. Distance. $AB=2r=\dfrac{2\sqrt{2mK}}{eB}=\dfrac{\sqrt{8m\left(\frac{hc}\lambda-\phi\right)}}{eB}$.
Answer. Option C.
Solution by Sreeraj P, M.Sc Physics
An electron with mass $m$ with an initial velocity ($t = 0$) $\vec v = v_0\hat i$ ($v_0 > 0$) enters a magnetic field $\vec B = B_0\hat j$. If the initial de-Broglie wavelength at $t = 0$ is $\lambda_0$ then its value after time $t$ would be:
Answer: (D) $\lambda_0$
Concept. A magnetic force is always perpendicular to the velocity, so it does no work: the speed, momentum magnitude and hence $\lambda$ stay constant.
Answer. $\lambda_0$ (Option D).
Solution by Sreeraj P, M.Sc Physics
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: In the photoelectric effect, on increasing the intensity of incident light the stopping potential increases.
Reason R: Increase in intensity of light increases the rate of photoelectrons emitted, provided the frequency of incident light is greater than the threshold frequency.
In the light of the above statements, choose the correct answer from the options given below:
Answer: (B) A is false but R is true
Concept. Intensity sets the number of photons per second, so it changes the photocurrent, not $K_{max}$ or $V_0$.
Step 1. Assertion. False: $V_0$ does not change with intensity.
Step 2. Reason. True (above the threshold frequency).
Answer. A is false but R is true (Option B).
Solution by Sreeraj P, M.Sc Physics
A photo-emissive substance is illuminated with radiation of wavelength $\lambda_i$ so that it releases electrons with de Broglie wavelength $\lambda_e$. The longest wavelength of radiation that can emit photoelectrons is $\lambda_0$. The expression for the de Broglie wavelength is: ($m$: mass of the electron, $h$: Planck's constant and $c$: speed of light)
Answer: (A) $\lambda_e = \sqrt{\dfrac{h}{2mc\left(\frac1{\lambda_i} - \frac1{\lambda_0}\right)}}$
Concept. $K_{max}=hc\left(\dfrac1{\lambda_i}-\dfrac1{\lambda_0}\right)$, and the electron's wavelength is $\lambda_e=\dfrac{h}{\sqrt{2mK_{max}}}$.
Step 1. Substitute. $\lambda_e=\dfrac{h}{\sqrt{2mhc\left(\frac1{\lambda_i}-\frac1{\lambda_0}\right)}}=\sqrt{\dfrac{h}{2mc\left(\frac1{\lambda_i}-\frac1{\lambda_0}\right)}}$.
Answer. Option A.
Solution by Sreeraj P, M.Sc Physics
An electron is released from rest near an infinite non-conducting sheet of uniform charge density $-\sigma$. The rate of change of the de Broglie wavelength associated with the electron varies inversely as the $n^{\text{th}}$ power of time. The numerical value of $n$ is ______.
Numerical value type. Enter your answer.
Answer: 2
Concept. The sheet's field $\dfrac{\sigma}{2\varepsilon_0}$ is uniform, so from rest $p = mat$; $\lambda = \dfrac hp$.
Step 1. Wavelength. $\lambda = \dfrac{h}{mat}$.
Step 2. Rate. $\dfrac{d\lambda}{dt} = -\dfrac{h}{ma} \cdot \dfrac{1}{t^2}$, so $n = 2$.
Answer. $2$.
Solution by Sreeraj P, M.Sc Physics
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