Atoms: NEET and JEE questions with solutions

Answer: (C) see figure

Concept. Rutherford's scattering formula gives the number of $\alpha$-particles scattered at angle $\theta$:
$$N(\theta) \propto \frac{1}{\sin^4(\theta/2)}$$

Step 1. Small angles. Most $\alpha$-particles pass almost undeflected, so $N(\theta)$ is very large near $\theta = 0$.

Step 2. Large angles. As $\theta$ grows, $\sin(\theta/2)$ grows and $N$ falls very steeply; only a few particles come back near $\theta = 180°$.

Step 3. Choose the graph. The correct plot falls steadily and sharply with $\theta$, which is graph (3).

Answer. Graph (3) (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $1$

Concept. In a circular orbit the Coulomb attraction provides the centripetal force:
$$\frac{mv^2}{r} = \frac{1}{4\pi\epsilon_0}\frac{e^2}{r^2} \;\Rightarrow\; r = \frac{1}{4\pi\epsilon_0}\frac{e^2}{mv^2}$$

Step 1. Substitute. $v^2 = 25.6 \times 10^{10}\ \text{m}^2\text{s}^{-2}$:
$$r = \frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{9 \times 10^{-31} \times 25.6 \times 10^{10}} = \frac{9 \times 2.56 \times 10^{-29}}{9 \times 2.56 \times 10^{-20}} = 10^{-9}\ \text{m}$$

Step 2. Compare. $r = 1 \times 10^{-9}$ m, so $x = 1$.

Answer. $1$ (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $r \propto n^{2/3};\ v \propto n^{1/3}$

Concept. Combine the force condition for a circular orbit with Bohr's quantisation $mvr = \dfrac{nh}{2\pi}$.

Step 1. Force condition. The constant force $F$ is the centripetal force: $\dfrac{mv^2}{r} = F$, so $v^2 \propto r$.

Step 2. Quantisation. $mvr = \dfrac{nh}{2\pi}$, so $v \propto \dfrac{n}{r}$.

Step 3. Eliminate $v$.
$$\frac{n^2}{r^2} \propto r \;\Rightarrow\; r^3 \propto n^2 \;\Rightarrow\; r \propto n^{2/3}$$

Step 4. Speed. $v \propto \dfrac{n}{r} \propto \dfrac{n}{n^{2/3}} = n^{1/3}$.

Answer. $r \propto n^{2/3}$, $v \propto n^{1/3}$ (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $0.67$ nm

Concept. Bohr's condition means a whole number of de Broglie wavelengths fits around the orbit: $2\pi r_n = n\lambda$.

Step 1. Radius of the $n = 2$ orbit. $r_2 = n^2a_0 = 4 \times 0.052 = 0.208$ nm.

Step 2. Wavelength.
$$\lambda = \frac{2\pi r_2}{2} = \pi \times 0.208 \approx 0.65\ \text{nm}$$
The closest option is $0.67$ nm.

Answer. $0.67$ nm (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) A-III, B-IV, C-II, D-I

Concept. All four are Balmer lines: $\dfrac{1}{\lambda} = R\left(\dfrac{1}{4} - \dfrac{1}{n_2^2}\right)$. A larger $n_2$ gives a larger energy gap and a shorter wavelength.

Step 1. Order the wavelengths. $\lambda_{3\to2} > \lambda_{4\to2} > \lambda_{5\to2} > \lambda_{6\to2}$.

Step 2. Match.

• A ($3 \to 2$, H$_\alpha$): $656.3$ nm → III.
• B ($4 \to 2$, H$_\beta$): $486.1$ nm → IV.
• C ($5 \to 2$, H$_\gamma$): $434.1$ nm → II.
• D ($6 \to 2$, H$_\delta$): $410.2$ nm → I.

Answer. A-III, B-IV, C-II, D-I (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) Statement I is correct but Statement II is incorrect.

Concept. Judge each statement against the structure of the atom and how spectra arise.

Step 1. Statement I. An atom has as many electrons as protons, so it is electrically neutral → correct.

Step 2. Statement II. Atoms emit their characteristic spectrum only when excited (by heating or an electric discharge) and falling back to lower levels, not on their own. Also, radioactive atoms are not stable → incorrect.

Answer. Statement I is correct but Statement II is incorrect (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $4\lambda$

Concept. The shortest wavelength of a series is its series limit ($n_2 \to \infty$): $\dfrac{1}{\lambda_{\min}} = \dfrac{R}{n_1^2}$, so $\lambda_{\min} \propto n_1^2$.

Step 1. Lower levels. Balmer: $n_1 = 2$. Brackett: $n_1 = 4$.

Step 2. Ratio.
$$\frac{\lambda_{\text{Brackett}}}{\lambda_{\text{Balmer}}} = \frac{4^2}{2^2} = 4 \;\Rightarrow\; \lambda_{\text{Brackett}} = 4\lambda$$

Answer. $4\lambda$ (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $4.77$ Å

Concept. In hydrogen the Bohr radius grows as $r_n = n^2 r_1$.

Step 1. Substitute. $r_1 = 5.3 \times 10^{-11}$ m $= 0.53$ Å:
$$r_3 = 9 \times 0.53 = 4.77\ \text{Å}$$

Answer. $4.77$ Å (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) Singly ionised neon atom (Ne$^+$)

Concept. Bohr's model treats a single electron in the field of a nucleus, so it holds only for one-electron (hydrogen-like) systems.

Step 1. Count electrons. H, deuterium and He$^+$ each have one electron. Ne$^+$ ($Z = 10$) still has $9$ electrons, so the model does not apply.

Answer. Singly ionised neon (Ne$^+$) (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $3$

Concept. A single atom emits one photon per jump; the most photons come from stepping down one level at a time.

Step 1. Path. $4\to3\to2\to1$: three photons.

Answer. Option D.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $\dfrac{1}{m}$

Concept. At closest approach in a head-on collision all kinetic energy has become electric potential energy: $\tfrac12mv^2=\dfrac{k(2e)(Ze)}{d}$.

Step 1. Distance. $d=\dfrac{4kZe^2}{mv^2}\propto\dfrac{1}{m}$, $\dfrac1{v^2}$ and $Z$.

Answer. $\dfrac1m$ (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $-1.51$ eV

Concept. Energy levels of hydrogen: $E_n = -\dfrac{13.6}{n^2}$ eV.

Step 1. Substitute $n = 3$.
$$E_3 = -\frac{13.6}{9} \approx -1.51\ \text{eV}$$

Answer. $-1.51$ eV (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $-1$

Concept. In every Bohr orbit $U = -2K$, so $E = K + U = -K$.

Step 1. Ratio.
$$\frac{K}{E} = \frac{K}{-K} = -1$$

Answer. $-1$ (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $2.2 \times 10^6$ m/s

Concept. In the Bohr model $v_n = v_1\dfrac{Z}{n}$, where $v_1 = 2.2 \times 10^6$ m/s is hydrogen's first-orbit speed.

Step 1. Substitute. For Li$^{2+}$, $Z = 3$, $n = 3$:
$$v = 2.2 \times 10^6 \times \frac{3}{3} = 2.2 \times 10^6\ \text{m/s}$$

Answer. $2.2 \times 10^6$ m/s (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $1.06$ Å

Concept. For a hydrogen-like ion $r_n = r_1\dfrac{n^2}{Z}$ with $r_1 = 0.53$ Å.

Step 1. Substitute. He$^+$: $Z = 2$, $n = 2$:
$$r = 0.53 \times \frac{4}{2} = 1.06\ \text{Å}$$

Answer. $1.06$ Å (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $54.4$ eV

Concept. Ground-state energy of a hydrogen-like ion: $E_1 = -13.6Z^2$ eV. The ionisation energy is $-E_1$.

Step 1. Substitute $Z = 2$. $E_1 = -13.6 \times 4 = -54.4$ eV, so $54.4$ eV removes the electron.

Answer. $54.4$ eV (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $6566$ Å

Concept. Use $\dfrac{1}{\lambda} = R\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)$ for both lines and take the ratio, so $R$ cancels.

Step 1. Lyman first line ($2 \to 1$). $\dfrac{1}{\lambda_L} = R\cdot\dfrac{3}{4}$.

Step 2. Balmer first line ($3 \to 2$). $\dfrac{1}{\lambda_B} = R\left(\dfrac14 - \dfrac19\right) = R\cdot\dfrac{5}{36}$.

Step 3. Ratio.
$$\lambda_B = \lambda_L \times \frac{3/4}{5/36} = 1216 \times 5.4 \approx 6566\ \text{Å}$$

Answer. $6566$ Å (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (C) Lyman series

Concept. The region of a series depends on its lower level: transitions to $n = 1$ have the largest energy gaps.

Step 1. Place each series. Lyman (to $n = 1$): ultraviolet. Balmer (to $n = 2$): visible. Paschen, Brackett, Pfund: infrared.

Answer. Lyman series (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $10$

Concept. In a sample, atoms in level $n$ can fall by every possible route, giving $\dfrac{n(n - 1)}{2}$ different lines.

Step 1. Substitute $n = 5$.
$$\frac{5 \times 4}{2} = 10$$

Answer. $10$ (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $102$ nm

Concept. Rydberg formula: $\dfrac{1}{\lambda} = R\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)$.

Step 1. Substitute. $n_1 = 1$, $n_2 = 3$:
$$\frac{1}{\lambda} = R\left(1 - \frac{1}{9}\right) = \frac{8 \times 1.1 \times 10^7}{9} \approx 9.78 \times 10^6\ \text{m}^{-1}$$

Step 2. Wavelength. $\lambda \approx 1.02 \times 10^{-7}$ m $= 102$ nm.

Answer. $102$ nm (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $\dfrac{3h}{2\pi}$

Concept. Bohr's quantisation condition: $L = \dfrac{nh}{2\pi}$.

Step 1. Substitute $n = 3$. $L = \dfrac{3h}{2\pi}$.

Answer. $\dfrac{3h}{2\pi}$ (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (C) most of the atom is empty space

Concept. An $\alpha$-particle is deflected only when it passes close to something massive and charged.

Step 1. Interpret. Most $\alpha$-particles passed almost undeflected, so they met nothing in their way: most of the atom is empty space. (The rare large-angle deflections showed that the positive charge and mass sit in a tiny nucleus.)

Answer. Most of the atom is empty space (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $12.09$ eV

Concept. The second excited state is $n = 3$; the energy needed is $E_3 - E_1$ with $E_n = -\dfrac{13.6}{n^2}$ eV.

Step 1. Energy of $n = 3$. $E_3 = -\dfrac{13.6}{9} = -1.51$ eV.

Step 2. Excitation energy. $E_3 - E_1 = -1.51 + 13.6 = 12.09$ eV.

Answer. $12.09$ eV (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $\dfrac{1}{n^3}$

Concept. The frequency of revolution is $f = \dfrac{v}{2\pi r}$, with $v \propto \dfrac{1}{n}$ and $r \propto n^2$ in the Bohr model.

Step 1. Combine.
$$f \propto \frac{1/n}{n^2} = \frac{1}{n^3}$$

Answer. $\dfrac{1}{n^3}$ (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: 2.55

Concept. The photon carries the energy difference: $E = 13.6\left(\dfrac{1}{n_f^2} - \dfrac{1}{n_i^2}\right)$ eV.

Step 1. Substitute. $n_i = 4$, $n_f = 2$:
$$E = 13.6\left(\frac14 - \frac{1}{16}\right) = 13.6 \times \frac{3}{16} = 2.55\ \text{eV}$$

Answer. $2.55$.

Solution by Sreeraj P, M.Sc Physics

Answer: 91

Concept. The shortest Lyman wavelength is the series limit, $n = \infty \to 1$: $\dfrac{1}{\lambda} = R$.

Step 1. Substitute.
$$\lambda = \frac{1}{1.097 \times 10^7} = 9.12 \times 10^{-8}\ \text{m} \approx 91\ \text{nm}$$

Answer. $91$.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $20$ and $27$

Concept. Photon momentum $p = \dfrac{h}{\lambda} \propto \dfrac{1}{\lambda} \propto \left(\dfrac{1}{2^2} - \dfrac{1}{n^2}\right)$ for Balmer lines.

Step 1. First line ($n = 3$). $\dfrac14 - \dfrac19 = \dfrac{5}{36}$.

Step 2. Second line ($n = 4$). $\dfrac14 - \dfrac{1}{16} = \dfrac{3}{16}$.

Step 3. Ratio.
$$\frac{p_1}{p_2} = \frac{5/36}{3/16} = \frac{80}{108} = \frac{20}{27}$$

Answer. $\alpha = 20$, $\beta = 27$ (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $486$

Concept. $r\propto n^2$, so the radius ratio $16:4$ gives $n_i=4$, $n_f=2$. Then $\dfrac1\lambda=R\left(\dfrac1{n_f^2}-\dfrac1{n_i^2}\right)$.

Step 1. Wave number. $\dfrac1\lambda=R\left(\dfrac14-\dfrac1{16}\right)=\dfrac{3R}{16}$.

Step 2. Wavelength. $\lambda=\dfrac{16}{3\times1.0973\times10^7}=4.86\times10^{-7}$ m $=486$ nm.

Answer. Option C.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) A, C, E Only

Concept. Large-angle scattering needs a nearly head-on approach (small impact parameter) to a tiny, heavy nucleus.

Step 1. Check. A: true, the nucleus is tiny. B: false; charges are $2e$ and $79e$. C: true. D: all particles had about the same energy, false. E: true.

Answer. A, C, E only (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $-0.38$

Concept. $L=\dfrac{nh}{2\pi}$ gives $n$; then $E_n=-\dfrac{13.6}{n^2}$ eV.

Step 1. Orbit. $\dfrac{3h}{\pi}=\dfrac{6h}{2\pi}$, so $n=6$.

Step 2. Energy. $-\dfrac{13.6}{36}=-0.38$ eV.

Answer. Option C.

Solution by Sreeraj P, M.Sc Physics

Answer: 32

Concept. The orbiting electron is a current loop: $I=ef\propto\dfrac{1}{n^3}$, and $B=\dfrac{\mu_0I}{2r}$ with $r\propto n^2$, so $B\propto n^{-5}$.

Step 1. Ratio. $\dfrac{B_2}{B_4}=\left(\dfrac42\right)^5=32$.

Answer. $32$.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) 6 and 5

Concept. $r\propto\dfrac{n^2}{Z}$ and $v\propto\dfrac Zn$. Equal radii mean $\dfrac{n_1^2}{Z_1}=\dfrac{n_2^2}{Z_2}=k$, so $Z=\dfrac{n^2}k$ and $v\propto\dfrac{Z}{n}=\dfrac nk$: the speeds are in the ratio of the $n$'s.

Step 1. Ratio of $n$. $\dfrac{n_1}{n_2}=\dfrac{3\times10^5}{2.5\times10^5}=\dfrac65$.

Step 2. Pick. $6$ and $5$.

Answer. Option D.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $4$

Concept. Bohr's condition $2\pi r_n = n\lambda_n$ gives $\lambda_n = \dfrac{2\pi r_n}{n}$.

Step 1. Identify the state. The third excited state is $n = 4$. Check: $\dfrac{8.48 \times 10^{-10}}{5.3 \times 10^{-11}} = 16 = 4^2$.

Step 2. Ratio.
$$\frac{\lambda_4}{\lambda_1} = \frac{r_4/4}{r_1/1} = \frac{16}{4} = 4$$

Answer. $4$ (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $\sqrt{\dfrac{h}{\pi eB}}$

Concept. Combine the magnetic circle $r=\dfrac{mv}{eB}$ with $mvr=\dfrac{nh}{2\pi}$.

Step 1. Substitute. $mv=eBr$, so $eBr^2=\dfrac{nh}{2\pi}$, $r=\sqrt{\dfrac{nh}{2\pi eB}}$.

Step 2. First excited ($n=2$). $r=\sqrt{\dfrac{h}{\pi eB}}$.

Answer. Option A.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $3000\ \text{Å}$

Concept. Energies add: $E_{AC}=E_{AB}+E_{BC}$, so $\dfrac1{\lambda_{AC}}=\dfrac1{\lambda_{AB}}+\dfrac1{\lambda_{BC}}$.

Step 1. Solve. $\dfrac1{\lambda_{AB}}=\dfrac1{2000}-\dfrac1{6000}=\dfrac{1}{3000}$.

Answer. $3000$ Å (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $\dfrac{1}{n^3}$

Concept. The frequency of revolution is $f = \dfrac{v}{2\pi r}$, with $v \propto \dfrac{1}{n}$ and $r \propto n^2$.

Step 1. Combine.
$$f \propto \frac{1/n}{n^2} = \frac{1}{n^3}$$

Answer. $\dfrac{1}{n^3}$ (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $6$

Concept. In a sample excited to level $n$, atoms fall by every possible route, giving $\dfrac{n(n - 1)}{2}$ distinct lines.

Step 1. Substitute $n = 4$.
$$\frac{4 \times 3}{2} = 6$$
($4\to3$, $4\to2$, $4\to1$, $3\to2$, $3\to1$, $2\to1$.)

Answer. $6$ (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $12.1\ \text{eV}$

Concept. Balmer lines end on $n=2$, so the atom must first reach at least $n=3$.

Step 1. Energy. $13.6\left(1-\dfrac19\right)=12.1$ eV.

Answer. Option D.

Solution by Sreeraj P, M.Sc Physics

Answer: 32

Concept. For one ion $\nu\propto Z^2\left(\dfrac1{n_f^2}-\dfrac1{n_i^2}\right)$; $Z^2$ cancels in a ratio.

Step 1. Ratio. $\dfrac{\nu_{3\to1}}{\nu_{2\to1}}=\dfrac{8/9}{3/4}=\dfrac{32}{27}$.

Step 2. Frequency. $\nu_{3\to1}=3\times10^{15}\times\dfrac{32}{27}=\dfrac{32}{9}\times10^{15}$ Hz.

Answer. $x=32$.

Solution by Sreeraj P, M.Sc Physics

Answer: 7

Concept. The photon's momentum $p = \dfrac{E}{c}$ goes to the atom as recoil. The recoil energy $\dfrac{p^2}{2M} = \dfrac{E^2}{2Mc^2}$ is taken from the photon, so $\dfrac{\Delta\lambda}{\lambda} \approx \dfrac{\Delta E}{E} = \dfrac{E}{2Mc^2}$.

Step 1. Transition energy.
$$E = 13.6\left(\frac14 - \frac19\right) = 1.89\ \text{eV}$$

Step 2. Rest energy of the atom.
$$Mc^2 = 1.6 \times 10^{-27} \times 9 \times 10^{16} = 1.44 \times 10^{-10}\ \text{J} = 9 \times 10^8\ \text{eV}$$

Step 3. Percentage change.
$$\frac{\Delta\lambda}{\lambda} \times 100 = \frac{1.89}{1.8 \times 10^9} \times 100 \approx 1 \times 10^{-7}\ \%$$

Answer. $7$.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $2\dfrac{h}{\pi}$

Concept. Bohr's quantisation: $L=\dfrac{nh}{2\pi}$.

Step 1. $n=4$. $L=\dfrac{4h}{2\pi}=\dfrac{2h}{\pi}$.

Answer. Option D.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $2E = U$

Concept. For an inverse-square attraction, $K=\dfrac{|U|}{2}$, so $E=K+U=\dfrac U2$.

Answer. $2E=U$ (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $\sqrt r$

Concept. $L\propto n$ and $r\propto n^2$, so $L\propto\sqrt r$.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: 6588

Concept. Lyman limit $=\dfrac1R=915$ Å. Longest Balmer ($3\to2$) $=\dfrac{36}{5R}$.

Step 1. Value. $\dfrac{36}{5}\times915=6588$ Å.

Answer. $6588$.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $4:1$

Concept. Shortest wavelength = series limit: $\dfrac1\lambda=\dfrac{R}{n_f^2}$.

Step 1. Ratio. $\dfrac{\lambda_B}{\lambda_L}=\dfrac{4/R}{1/R}=4$.

Answer. $4:1$ (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: 16

Concept. $r_n=n^2a_0$, $E_n=\dfrac{E}{n^2}$.

Step 1. Orbit. $\dfrac{8.48}{0.529}=16=n^2$, so $n=4$.

Step 2. Energy. $\dfrac{E}{16}$.

Answer. $x=16$.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $1.876\times10^{-6}\ \text{m}$

Concept. The longest wavelength of a series comes from the smallest energy gap, here $n = 4 \to 3$.

Step 1. Rydberg formula.
$$\frac{1}{\lambda} = R_H\left(\frac19 - \frac{1}{16}\right) = \frac{7R_H}{144}$$

Step 2. Wavelength.
$$\lambda = \frac{144}{7 \times 1.097 \times 10^7} \approx 1.876 \times 10^{-6}\ \text{m}$$

Answer. $1.876 \times 10^{-6}$ m (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: 122

Concept. The first dip at $10.2$ V means the first excitation energy is $10.2$ eV; the atom emits a photon of this energy on returning, $\lambda = \dfrac{hc}{E}$.

Step 1. Wavelength.
$$\lambda = \frac{1245}{10.2} \approx 122\ \text{nm}$$

Answer. $122$.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $\dfrac{16}{9}R$

Concept. $r_n\propto n^2$ for a given atom.

Step 1. Ratio. $r_4=R\dfrac{16}{9}$.

Answer. $\dfrac{16}9R$ (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: 144

Concept. The longest wavelength of a series is its first line; for Paschen, $4\to3$.

Step 1. Wave number. $\dfrac1\lambda=R\left(\dfrac19-\dfrac1{16}\right)=\dfrac{7R}{144}$.

Step 2. Wavelength. $\lambda=\dfrac{144}{7R}$.

Answer. $\alpha=144$.

Solution by Sreeraj P, M.Sc Physics

Answer: 17

Concept. Momentum is conserved: the atom recoils with $p=\dfrac Ec$ of the photon.

Step 1. Photon momentum. $\dfrac{10.2\times1.6\times10^{-19}}{3\times10^8}=5.44\times10^{-27}$ kg m/s.

Step 2. Speed. $\dfrac{5.44\times10^{-27}}{1.6\times10^{-27}}=3.4=\dfrac{17}{5}$ m/s.

Answer. $x=17$.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) Statement I is true but Statement II is false

Concept. Statement II describes Thomson's plum-pudding model, which Rutherford's experiment replaced; it is not a special case of it.

Answer. Statement I true, II false (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: 121

Concept. Balmer lines end on $n=2$, so the atom must first reach at least $n=3$. An electron accelerated through $V$ volts brings $eV$ of energy.

Step 1. Energy to $n=3$. $13.6\left(1-\dfrac19\right)=12.09$ eV.

Step 2. Voltage. $12.1$ V $=\dfrac{121}{10}$ V.

Answer. $\alpha=121$.

Solution by Sreeraj P, M.Sc Physics

Answer: 156

Concept. Same energy balance: $\tfrac12mv^2=\dfrac{k(2e)(Ze)}{d}$.

Step 1. Energy. $\dfrac{9\times10^9\times2\times80\times(1.6\times10^{-19})^2}{4.5\times10^{-14}}=8.19\times10^{-13}$ J.

Step 2. Speed. $v=\sqrt{\dfrac{2\times8.19\times10^{-13}}{6.72\times10^{-27}}}=1.56\times10^7$ m/s.

Answer. $156$.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) 1

Concept. Find which level the absorbed energy reaches, then count the possible downward transitions.

Step 1. Level reached. $E_2 - E_1 = -3.4 - (-13.6) = 10.2$ eV, so the atom goes to $n = 2$.

Step 2. Count lines. From $n = 2$ the only transition is $2 \to 1$: one line.

Answer. $1$ (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $\dfrac12$

Concept. In every Bohr orbit $K=-\dfrac U2$, so $\dfrac{K}{|U|}=\dfrac12$, whatever $n$.

Answer. Option C.

Solution by Sreeraj P, M.Sc Physics

Answer: 6

Concept. Find $n$ from $E_n=-\dfrac{13.6}{n^2}$ eV; the number of possible downward transitions is $\dfrac{n(n-1)}{2}$.

Step 1. Level. $\dfrac{13.6}{n^2}=0.85$, so $n=4$.

Step 2. Transitions. $\dfrac{4\times3}{2}=6$.

Answer. $6$.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $1$

Concept. Magnetic moment $\mu=IA=ef\cdot\pi r^2$, with $f\propto n^{-3}$ and $r^2\propto n^4$.

Step 1. Combine. $\mu\propto n^{-3}\cdot n^4=n$. (Directly: $\mu=\dfrac{e}{2m}L\propto n$.)

Answer. $x=1$ (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $\dfrac{27}{32}\lambda$

Concept. Lyman lines: $\dfrac{1}{\lambda} = R\left(1 - \dfrac{1}{n^2}\right)$. Take the ratio of the first ($n = 2$) and second ($n = 3$) members.

Step 1. First member. $\dfrac{1}{\lambda} = R\left(1 - \dfrac14\right) = \dfrac{3R}{4}$.

Step 2. Second member. $\dfrac{1}{\lambda'} = R\left(1 - \dfrac19\right) = \dfrac{8R}{9}$.

Step 3. Ratio.
$$\frac{\lambda'}{\lambda} = \frac{3/4}{8/9} = \frac{27}{32}$$

Answer. $\dfrac{27}{32}\lambda$ (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $8.4\times10^6$

Concept. $v_n\propto\dfrac Zn$.

Step 1. Value. $v_3=3.6\times10^6\times\dfrac73=8.4\times10^6$ m/s.

Answer. Option D.

Solution by Sreeraj P, M.Sc Physics

Answer: 27

Concept. $\dfrac{1}{\lambda} \propto \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}$, so a wavelength ratio is the inverse ratio of these brackets.

Step 1. Transition 1 ($3 \to 1$). $1 - \dfrac19 = \dfrac89$.

Step 2. Transition 2 ($2 \to 1$). $1 - \dfrac14 = \dfrac34$.

Step 3. Ratio.
$$\frac{\lambda_1}{\lambda_2} = \frac{3/4}{8/9} = \frac{27}{32}$$

Answer. $27$.

Solution by Sreeraj P, M.Sc Physics

Answer: 136

Concept. Hydrogen-like ion: $E_n=-\dfrac{13.6Z^2}{n^2}$ eV. Second excited state means $n=3$.

Step 1. Li$^{2+}$. $\dfrac{13.6\times9}{9}=13.6$ eV $=136\times10^{-1}$ eV.

Answer. $136$.

Solution by Sreeraj P, M.Sc Physics

Answer: 27

Concept. Second excited to first excited is $3\to2$; third excited to second orbit is $4\to2$.

Step 1. Brackets. $\dfrac5{36}$ and $\dfrac3{16}$.

Step 2. Ratio. $\lambda=\dfrac{5/36}{3/16}\lambda_0=\dfrac{20}{27}\lambda_0$.

Answer. $x=27$.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) D

Concept. Photon energy $E=\dfrac{1240}{124.1}\approx10.0$ eV. Find the arrow whose energy gap is $10$ eV.

Step 1. Gaps. A: $2.2$ eV; B: $5.2$ eV; C: $3.0$ eV; D: $10.0$ eV.

Answer. Transition D (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: 4125

Concept. Energy kept by the atom $=$ energy absorbed $-$ energy emitted $=hc\left(\dfrac1{\lambda_1}-\dfrac1{\lambda_2}\right)$.

Step 1. $hc$ in eV nm. $\dfrac{6.6\times10^{-34}\times3\times10^8}{1.6\times10^{-19}}=1237.5$ eV nm.

Step 2. Net energy. $1237.5\left(\dfrac1{500}-\dfrac1{600}\right)=\dfrac{1237.5}{3000}=0.4125$ eV $=4125\times10^{-4}$ eV.

Answer. $4125$.

Solution by Sreeraj P, M.Sc Physics

Answer: 5

Concept. The first, second and third excited levels are $n = 2, 3, 4$. Use $\dfrac{1}{\lambda} \propto \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}$ for each transition.

Step 1. $\lambda_1$ ($3 \to 2$). $\dfrac14 - \dfrac19 = \dfrac{5}{36}$.

Step 2. $\lambda_2$ ($4 \to 3$). $\dfrac19 - \dfrac{1}{16} = \dfrac{7}{144}$.

Step 3. Ratio.
$$\frac{\lambda_1}{\lambda_2} = \frac{7/144}{5/36} = \frac{7}{20} = \frac{7}{4 \times 5}$$

Answer. $5$.

Solution by Sreeraj P, M.Sc Physics

Answer: 3668

Concept. The lowest wavelength of a series is its series limit: $\dfrac{1}{\lambda} = \dfrac{R}{n_1^2}$, so $\lambda_{\min} \propto n_1^2$.

Step 1. Ratio. Lyman $n_1 = 1$, Balmer $n_1 = 2$:
$$\lambda_{\text{Balmer}} = 4 \times 917 = 3668\ \text{Å}$$

Answer. $3668$.

Solution by Sreeraj P, M.Sc Physics

Answer: 3

Concept. Six emitted wavelengths tell us the highest level reached; the incident photon supplied exactly the energy to reach it from $n = 1$.

Step 1. Highest level.
$$\frac{n(n - 1)}{2} = 6 \;\Rightarrow\; n = 4$$

Step 2. Photon energy.
$$E = 13.6\left(1 - \frac{1}{16}\right) = 12.75\ \text{eV}$$

Step 3. Frequency.
$$\nu = \frac{E}{h} = \frac{12.75}{4.25 \times 10^{-15}} = 3 \times 10^{15}\ \text{Hz}$$

Answer. $3$.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $-13.6\ \text{eV}$

Concept. For a hydrogen-like ion $E_n = -13.6\dfrac{Z^2}{n^2}$ eV. Following the official key, "first state" is read as the first excited state, $n = 2$.

Step 1. Substitute. He$^+$: $Z = 2$, $n = 2$:
$$E = -13.6 \times \frac{4}{4} = -13.6\ \text{eV}$$
(The ground state of He$^+$ would be $-54.4$ eV.)

Answer. $-13.6$ eV (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) 3

Concept. The electron beam can lift atoms only to levels within $12.5$ eV of the ground state.

Step 1. Levels. $E_3-E_1=12.09$ eV $<12.5$; $E_4-E_1=12.75$ eV $>12.5$. So up to $n=3$.

Step 2. Lines. $\dfrac{3\times2}{2}=3$.

Answer. Option D.

Solution by Sreeraj P, M.Sc Physics

Answer: 2

Concept. $r\propto\dfrac{n^2}{Z}$.

Step 1. He$^+$, $n=2$. $\dfrac42=2$. Be$^{3+}$, $n=4$. $\dfrac{16}{4}=4$.

Step 2. Ratio. $\dfrac{r_2}{r_1}=2$.

Answer. $2$.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) D

Concept. $\lambda = \dfrac{hc}{\Delta E}$, so the shortest wavelength comes from the largest energy gap.

Step 1. Compare the gaps. D ($3 \to 1$) releases $12.09$ eV; A ($0.66$ eV), B ($2.55$ eV) and C ($1.89$ eV) are much smaller.

Answer. Transition D (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: 425

Concept. $r=\dfrac{n^2}{Z}a_0$.

Step 1. Value. $\dfrac{25}{3}\times0.51=4.25$ Å $=425\times10^{-12}$ m.

Answer. $425$.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $\sqrt n$

Concept. The force from $U = \dfrac12m\omega^2r^2$ is $F = m\omega^2r$ towards the origin; combine it with Bohr's quantisation $mvr = \dfrac{nh}{2\pi}$.

Step 1. Circular orbit.
$$\frac{mv^2}{r} = m\omega^2r \;\Rightarrow\; v = \omega r$$

Step 2. Quantisation.
$$m\omega r^2 = \frac{nh}{2\pi} \;\Rightarrow\; r \propto \sqrt{n}$$

Answer. $\sqrt{n}$ (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: 16

Concept. Binding energy $=|E|=\dfrac{ke^2}{2r}$.

Step 1. Radius. $r=\dfrac{9\times10^9\times(1.6\times10^{-19})^2}{2\times12.8\times1.6\times10^{-19}}=5.625\times10^{-11}$ m $=\dfrac{9}{16}\times10^{-10}$ m.

Answer. $x=16$.

Solution by Sreeraj P, M.Sc Physics

Answer: 27

Concept. $\dfrac1\lambda=R\left(\dfrac1{n_f^2}-\dfrac1{n_i^2}\right)$. H$_\alpha$: $3\to2$; H$_\beta$: $4\to2$.

Step 1. Brackets. $\dfrac5{36}$ and $\dfrac3{16}$.

Step 2. Ratio. $\dfrac{\lambda_\alpha}{\lambda_\beta}=\dfrac{3/16}{5/36}=\dfrac{27}{20}$.

Answer. $x=27$.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $L$

Concept. Bohr's quantisation: $L = \dfrac{nh}{2\pi}$, so $L_n = nL_1$.

Step 1. Change. First orbit: $L$. Second orbit: $2L$. Change $= 2L - L = L$.

Answer. $L$ (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: 828

Concept. Find the final level from the energy, then $L=\dfrac{nh}{2\pi}$.

Step 1. Level. $-13.6+12.75=-0.85$ eV $=-\dfrac{13.6}{16}$, so $n=4$.

Step 2. Angular momentum. $\dfrac{4h}{2\pi}=\dfrac{2\times4.14\times10^{-15}}{\pi}=\dfrac{828}{\pi}\times10^{-17}$ eV s.

Answer. $828$.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $40.8\ \text{eV}$

Concept. $\Delta E=13.6Z^2\left(\dfrac1{n_f^2}-\dfrac1{n_i^2}\right)$.

Step 1. Value. $13.6\times16\times\left(\dfrac14-\dfrac1{16}\right)=217.6\times\dfrac{3}{16}=40.8$ eV.

Answer. Option D.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) A classical atom based on Rutherford's model is doomed to collapse.

Concept. In Rutherford's model electrons orbit a tiny heavy nucleus. Classically an accelerating charge radiates, loses energy and spirals in, so the atom should collapse.

Step 1. Other options. Electrons in orbit are not in static equilibrium (A false); mass is concentrated, not continuous, in Rutherford's model (B false); in Thomson's model too the positive part carries most mass (D false).

Answer. Option C.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $K$ decreases, $P$ and $E$ increase.

Concept. $K=\dfrac{13.6}{n^2}$ eV, $U=-2K$, $E=-K$. Going to a higher $n$ makes $K$ smaller, so $U$ and $E$ (both negative) rise towards zero.

Step 1. Changes. $K$ decreases; $P$ and $E$ increase.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $2 : 1$

Concept. $v\propto\dfrac Zn$; same $n$.

Step 1. Ratio. $\dfrac{Z_{He}}{Z_H}=2$.

Answer. $2:1$ (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $\dfrac{nh}{2\pi r}$

Concept. Bohr's condition: $L=mvr=\dfrac{nh}{2\pi}$, and the momentum is $p=mv$.

Step 1. Momentum. $p=\dfrac{L}{r}=\dfrac{nh}{2\pi r}$.

Answer. Option A.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $\sqrt{\dfrac{\lambda R}{\lambda R - 1}}$

Concept. To the ground state: $\dfrac1\lambda=R\left(1-\dfrac1{n^2}\right)$. Solve for $n$.

Step 1. Rearrange. $\dfrac1{n^2}=1-\dfrac1{\lambda R}=\dfrac{\lambda R-1}{\lambda R}$.

Step 2. Level. $n=\sqrt{\dfrac{\lambda R}{\lambda R-1}}$.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: 5

Concept. Energy $\propto\dfrac14-\dfrac1{n^2}$.

Step 1. Two cases. $3\to2$: $\dfrac5{36}$. $\infty\to2$: $\dfrac14=\dfrac9{36}$.

Step 2. Ratio. $\dfrac59=\dfrac{x}{x+4}\Rightarrow x=5$.

Answer. $5$.

Solution by Sreeraj P, M.Sc Physics

Answer: 5

Concept. Write each wavelength as a multiple of $\dfrac1R$: $\dfrac1\lambda=R\left(\dfrac1{n_f^2}-\dfrac1{n_i^2}\right)$.

Step 1. Lyman, first line ($2\to1$). $\lambda=\dfrac{4}{3R}$.

Step 2. Paschen, third line ($6\to3$). $\dfrac1{\lambda_P}=R\left(\dfrac19-\dfrac1{36}\right)=\dfrac R{12}$, so $\lambda_P=\dfrac{12}R$.

Step 3. Balmer, second line ($4\to2$). $\dfrac1{\lambda_B}=\dfrac{3R}{16}$, so $\lambda_B=\dfrac{16}{3R}$.

Step 4. Difference. $\dfrac{12}R-\dfrac{16}{3R}=\dfrac{20}{3R}=5\lambda$.

Answer. $a=5$.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $1.05\times10^{-34}$ Js

Concept. $10.2$ eV takes the electron from $n=1$ to $n=2$; $L$ rises by $\dfrac{h}{2\pi}$.

Step 1. Change. $\dfrac{6.6\times10^{-34}}{2\pi}=1.05\times10^{-34}$ J s.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: 114

Concept. Li$^{++}$ ($Z=3$): $\Delta E=13.6\times9\left(1-\dfrac19\right)=108.8$ eV.

Step 1. Wavelength. $\dfrac{1242}{108.8}=11.4$ nm $=114$ Å.

Answer. $x=114$.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) Statement I is incorrect but Statement II is true.

Concept. Bohr's frequency condition: a jump from the higher level $E_2$ to the lower $E_1$ emits $hf=E_2-E_1$. Going up needs energy to be absorbed.

Step 1. Statement I. It gives $hf=E_1-E_2$, a negative value, for a jump upward: incorrect.

Step 2. Statement II. Correct: that is the frequency condition.

Answer. Option D.

Solution by Sreeraj P, M.Sc Physics

Answer: 112

Concept. The electron passes any point $f=\dfrac{v}{2\pi r}$ times a second, so the current is $I=ef$.

Step 1. Frequency. $f=\dfrac{2.2\times10^6}{2\times\frac{22}{7}\times0.5\times10^{-10}}=7.0\times10^{15}$ s$^{-1}$.

Step 2. Current. $I=1.6\times10^{-19}\times7.0\times10^{15}=1.12\times10^{-3}$ A $=112\times10^{-2}$ mA.

Answer. $112$.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $1.41$ eV

Concept. The photon carries the electron's energy plus the binding energy of the level it falls into: second excited state, $n=3$, $-1.51$ eV.

Step 1. Photon. $3+1.51=4.51$ eV.

Step 2. Work function. $\dfrac{12400}{4000}=3.1$ eV.

Step 3. Photoelectron. $4.51-3.1=1.41$ eV.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: 10

Concept. $K_\alpha$ is emitted when the vacancy moves from the $K$ shell to the $L$ shell, so $E_{K_\alpha} = E_K - E_L$, where $E_K$ and $E_L$ are the energies of the atom with a $K$ or $L$ vacancy.

Step 1. Photon energy.
$$E_{K_\alpha} = \frac{hc}{\lambda} = \frac{4.14 \times 10^{-15} \times 3 \times 10^8}{0.071 \times 10^{-9}} \approx 17.5\ \text{keV}$$

Step 2. Energy with an $L$ vacancy.
$$E_L = E_K - E_{K_\alpha} = 27.5 - 17.5 = 10\ \text{keV}$$

Answer. $10$.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $1.45\times10^9$ MHz

Concept. On capture, the photon carries away the electron's kinetic energy plus the binding energy of the level formed.

Step 1. Energy. First excited state: $3.4$ eV; photon $2.6+3.4=6.0$ eV.

Step 2. Frequency. $\nu=\dfrac{6.0\times1.6\times10^{-19}}{6.6\times10^{-34}}=1.45\times10^{15}$ Hz $=1.45\times10^9$ MHz.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) The series limit of Lyman series, third member of Balmer series and second member of Paschen series.

Concept. A series is named by its lower level. Members are counted from the nearest upper level: for Balmer, $3\to2$ is the first, $4\to2$ the second, $5\to2$ the third.

Step 1. A. From the continuum ($n=\infty$) to $n=1$: Lyman series limit.

Step 2. B. $5\to2$: third member of the Balmer series.

Step 3. C. $5\to3$: second member of the Paschen series ($4\to3$ is the first).

Answer. Option C.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $n = 2$ to $n = 1$

Concept. The levels crowd together as $n$ rises ($E_n\propto-\dfrac1{n^2}$), so the gap between adjacent levels is largest at the bottom.

Step 1. Gaps. $2\to1$: $10.2$ eV; $3\to2$: $1.89$ eV; $4\to3$: $0.66$ eV; $5\to4$: $0.31$ eV.

Answer. $n=2$ to $n=1$ (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $7 : 135$

Concept. Third Lyman line: $4\to1$; first Paschen: $4\to3$.

Step 1. Brackets. $1-\dfrac1{16}=\dfrac{15}{16}$; $\dfrac19-\dfrac1{16}=\dfrac7{144}$.

Step 2. Ratio. $\dfrac{\lambda_1}{\lambda_2}=\dfrac{7/144}{15/16}=\dfrac{7}{135}$.

Answer. Option A.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $4.17$ m s$^{-1}$

Concept. Momentum is conserved: the atom recoils with momentum equal to the photon's, $Mv = \dfrac{E}{c}$.

Step 1. Photon energy.
$$E = 13.6\left(1 - \frac{1}{25}\right) = 13.06\ \text{eV} = 2.09 \times 10^{-18}\ \text{J}$$

Step 2. Recoil speed.
$$v = \frac{E}{Mc} = \frac{2.09 \times 10^{-18}}{1.67 \times 10^{-27} \times 3 \times 10^8} \approx 4.17\ \text{m s}^{-1}$$

Answer. $4.17$ m s$^{-1}$ (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $2.46\times10^{15}$

Concept. For a given ion $f \propto \left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)$; $Z$ cancels in a ratio.

Step 1. Ratio.
$$\frac{f_{2\to1}}{f_{3\to1}} = \frac{1 - \frac14}{1 - \frac19} = \frac{3/4}{8/9} = \frac{27}{32}$$

Step 2. Frequency.
$$f_{2\to1} = 2.92 \times 10^{15} \times \frac{27}{32} \approx 2.46 \times 10^{15}\ \text{Hz}$$

Answer. $2.46 \times 10^{15}$ Hz (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: 15

Concept. A sample (many atoms) from level $n$ can give all $\dfrac{n(n-1)}{2}$ lines.

Step 1. $n=6$. $\dfrac{6\times5}{2}=15$.

Answer. $15$.

Solution by Sreeraj P, M.Sc Physics

Answer: 486

Concept. Balmer lines: $\dfrac{1}{\lambda} = R\left(\dfrac{1}{2^2} - \dfrac{1}{n^2}\right)$. Take the ratio of the first ($n = 3$) and second ($n = 4$) members.

Step 1. Brackets. First: $\dfrac14 - \dfrac19 = \dfrac{5}{36}$. Second: $\dfrac14 - \dfrac{1}{16} = \dfrac{3}{16}$.

Step 2. Wavelength.
$$\lambda_2 = \lambda_1 \times \frac{5/36}{3/16} = 6561 \times \frac{20}{27} = 4860\ \text{Å} = 486\ \text{nm}$$

Answer. $486$.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $11.4$ nm

Concept. Ionisation from the ground state is $13.6Z^2$ eV $=Z^2$ Rydbergs, which gives $Z$. Then the $3\to1$ photon.

Step 1. Atomic number. $Z^2=9$, $Z=3$.

Step 2. Photon energy. $13.6\times9\left(1-\dfrac19\right)=108.8$ eV.

Step 3. Wavelength. $\dfrac{1240}{108.8}=11.4$ nm.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $7.8\times10^{14}$

Concept. $f=\dfrac{v}{2\pi r}\propto\dfrac{1/n}{n^2}=\dfrac{1}{n^3}$.

Step 1. Ground state. $f_1=\dfrac{1}{1.6\times10^{-16}}=6.25\times10^{15}$ Hz.

Step 2. $n=2$. $\dfrac{6.25\times10^{15}}{8}=7.8\times10^{14}$ Hz.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) see figure

Concept. Rutherford's formula gives the number scattered at angle $\theta$ as $N\propto\dfrac{1}{\sin^4(\theta/2)}$: most $\alpha$ particles pass almost straight through; very few are turned through large angles.

Step 1. Shape. Very large at small $\theta$, falling steeply, and small but not zero towards $\pi$.

Answer. Option D.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $\dfrac{1}{n^{3}}$

Concept. $\nu\propto\dfrac1{n^2}-\dfrac1{(n+1)^2}=\dfrac{2n+1}{n^2(n+1)^2}\approx\dfrac{2}{n^3}$ for large $n$.

Answer. $\dfrac1{n^3}$ (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: 51

Concept. Momentum is conserved; the kinetic energy that disappears excites the atom ($10.2$ eV for $1\to2$).

Step 1. Momentum. $mu=Mv$ with $M=5m$, so $v=\dfrac u5$.

Step 2. Energy lost. $\tfrac12mu^2-\tfrac12(5m)\dfrac{u^2}{25}=\tfrac12mu^2\cdot\dfrac45=10.2$ eV.

Step 3. Initial KE. $\tfrac12mu^2=12.75$ eV $=\dfrac{51}4$ eV.

Answer. $N=51$.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $10.8\ \text{nm}$

Concept. Six lines means the ion was raised to $n=4$ ($\dfrac{4\times3}{2}=6$). The absorbed photon gives the $1\to4$ energy.

Step 1. Energy. $13.6\times9\times\left(1-\dfrac1{16}\right)=114.75$ eV.

Step 2. Wavelength. $\dfrac{1240}{114.75}=10.8$ nm.

Answer. Option A.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $n = 2 \to n = 4$

Concept. The H photon has $10.2$ eV. He$^+$ levels are $-\dfrac{54.4}{n^2}$; find a jump of exactly $10.2$ eV.

Step 1. Try $2\to4$. $54.4\left(\dfrac14-\dfrac1{16}\right)=10.2$ eV. (From $n=1$ the smallest jump is $40.8$ eV.)

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $488.9\ \text{nm}$

Concept. Balmer lines: $\dfrac{1}{\lambda} = R\left(\dfrac14 - \dfrac{1}{n^2}\right)$. Take the ratio, so $R$ cancels.

Step 1. Brackets.
$$\frac{1}{\lambda_1} = \frac{5R}{36},\qquad \frac{1}{\lambda_2} = \frac{3R}{16}$$

Step 2. Wavelength.
$$\lambda_2 = \lambda_1 \times \frac{5/36}{3/16} = 660 \times \frac{80}{108} \approx 488.9\ \text{nm}$$

Answer. $488.9$ nm (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $13.60\ \text{eV}$

Concept. Ionisation energy from level $n$ $=\dfrac{13.6Z^2}{n^2}$ eV. First excited: $n=2$.

Step 1. He$^+$. $\dfrac{13.6\times4}{4}=13.6$ eV.

Answer. Option A.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $16a_0$

Concept. The photon must supply the gap to a level exactly: $13.6\left(1-\dfrac1{n^2}\right)$ eV. Then $r_n=n^2a_0$.

Step 1. Photon energy. $\dfrac{1240}{98}=12.65$ eV, nearest the $1\to4$ gap ($12.75$ eV).

Step 2. Radius. $r_4=16a_0$.

Answer. $16a_0$ (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $\dfrac{20}{27}\lambda$

Concept. M, L, N shells are $n = 3, 2, 4$. Use $\dfrac{1}{\lambda} \propto \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}$ and take the ratio.

Step 1. M → L. $\dfrac14 - \dfrac19 = \dfrac{5}{36}$.

Step 2. N → L. $\dfrac14 - \dfrac{1}{16} = \dfrac{3}{16}$.

Step 3. Ratio.
$$\lambda' = \lambda \times \frac{5/36}{3/16} = \frac{20}{27}\lambda$$

Answer. $\dfrac{20}{27}\lambda$ (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $r_n \propto \sqrt n$, $E_n \propto n$

Concept. Apply $mvr=\dfrac{nh}{2\pi}$ to the spring-like force $F=kr$.

Step 1. Speed. $\dfrac{mv^2}{r}=kr\Rightarrow v=r\sqrt{k/m}$.

Step 2. Radius. $mr^2\sqrt{k/m}=\dfrac{nh}{2\pi}\Rightarrow r^2\propto n$, $r\propto\sqrt n$.

Step 3. Energy. $E=\tfrac12mv^2+\tfrac12kr^2=kr^2\propto n$.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $250$ nm

Concept. The energy the electron loses goes into exciting a mercury atom; the atom then emits a photon of that energy.

Step 1. Energy. $5.6-0.7=4.9$ eV.

Step 2. Wavelength. $\dfrac{1240}{4.9}=253$ nm, close to $250$ nm.

Answer. Option A.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $n = 5$

Concept. Two photons in succession carry the full drop from $n$ to the ground state of He$^+$ ($Z=2$, $E_1=-54.4$ eV).

Step 1. Photon energies. $\dfrac{1240}{108.5}=11.43$ eV; $\dfrac{1240}{30.4}=40.8$ eV; total $52.2$ eV.

Step 2. Level. $54.4\left(1-\dfrac1{n^2}\right)=52.2\Rightarrow\dfrac1{n^2}=0.04$, $n=5$.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $9.7$ Å

Concept. Bohr's condition is the same as fitting $n$ de Broglie waves on the orbit: $2\pi r=n\lambda$. Second excited state: $n=3$.

Step 1. Wavelength. $\lambda=\dfrac{2\pi\times4.65}{3}=9.7$ Å.

Answer. Option C.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $\dfrac{20}{7}$

Concept. Third excited $\to$ second excited is $4\to3$; then $3\to2$.

Step 1. Brackets. $\dfrac19-\dfrac1{16}=\dfrac7{144}$; $\dfrac14-\dfrac19=\dfrac{5}{36}=\dfrac{20}{144}$.

Step 2. Ratio. $\dfrac{\lambda_1}{\lambda_2}=\dfrac{20}{7}$.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $79\ \text{eV}$

Concept. Ionising helium completely takes two steps: remove the first electron (energy $E_1$), then the electron of He$^+$, which needs $13.6\times2^2=54.4$ eV.

Step 1. First electron. $54.4=2.2E_1$, so $E_1=24.7$ eV.

Step 2. Total. $24.7+54.4=79.1$ eV.

Answer. About $79$ eV (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) (A), (C), (D)

Concept. In Bohr's results $r\propto\dfrac1m$, $v$ does not depend on $m$, $E\propto m$, and $p=mv\propto m$.

Step 1. Check. (A) true, (B) false, (C) true, (D) true.

Answer. (A), (C), (D) (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $r = \dfrac13$

Concept. $\lambda=\dfrac{hc}{\Delta E}$, so the ratio of wavelengths is the inverse ratio of the energy gaps.

Step 1. Gaps. $\lambda_1$: $-E\to-2E$, gap $E$. $\lambda_2$: $-E\to-\dfrac{4E}3$, gap $\dfrac E3$.

Step 2. Ratio. $r=\dfrac{\lambda_1}{\lambda_2}=\dfrac{E/3}{E}=\dfrac13$.

Answer. Option A.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $n^{-5}$

Concept. The orbiting electron is a current loop $I = \dfrac{ev}{2\pi r}$, whose field at the centre is $B = \dfrac{\mu_0I}{2r}$.

Step 1. Field at the nucleus.
$$B = \frac{\mu_0}{2r} \cdot \frac{ev}{2\pi r} = \frac{\mu_0ev}{4\pi r^2}$$

Step 2. Dependence on $n$. $v \propto \dfrac{1}{n}$ and $r \propto n^2$:
$$B \propto \frac{n^{-1}}{n^4} = n^{-5}$$

Answer. $n^{-5}$ (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $\dfrac{h^2}{4\pi^2m^2r^3}$

Concept. Centripetal acceleration $a=\dfrac{v^2}{r}$, with $v$ from Bohr's condition $mvr=\dfrac{h}{2\pi}$ for $n=1$.