PiTheory

PiTheory JEE Physics Practice & Revision Pack · Class 12 · Chapter 11 · JEE Main

Dual Nature of Radiation and Matter

Chapter-wise practice and revision for JEE Main: concise notes on photons, the photoelectric effect and Einstein's equation, and de Broglie matter waves, concept figures, and 47 selected previous-year JEE Main questions covering the major concepts and question patterns, each with a step-by-step solution.

47 selected questions3 topics3 concept figuresPreview · first 3 of 47 questions

How to study this chapter

  1. Look at the topic's figures, then read its notes (5 to 10 minutes). They give the major equations and methods the questions use.
  2. Solve the topic's questions with a timer: 2 minutes for Easy, 3 to 4 for Medium and Hard.
  3. Check the solution. Read the Point to note line even if you were right: it is the idea you will reuse.
  4. Mark each question Got it or Revise. Come back to the Revise list after two days.

Where the marks come from

Main idea of each of the 153 previous-year JEE questions in the PiTheory bank

Photoelectric70
de Broglie68
Photons15
01

Topic 1 · Q1–Q9

Photons: energy, momentum and photon flux

15 of the 153 JEE PYQs in the bank test this as their main idea

Quick notes

$$E=h\nu=\frac{hc}{\lambda},\qquad p=\frac{h}{\lambda}=\frac Ec,\qquad n=\frac{P}{h\nu}=\frac{P\lambda}{hc}$$

Handy: $hc=1240$ eV nm, so $E(\text{eV})=\dfrac{1240}{\lambda(\text{nm})}$. $1$ eV $=1.6\times10^{-19}$ J.

  • Same power: number of photons per second $\propto\lambda$.
  • Through area $A$: $n=\dfrac{IA}{h\nu}$.
  • X-ray tube at $V$: shortest wavelength $\dfrac{hc}{eV}$ (Duane–Hunt).
  • Relativistic energy: $E^2=p^2c^2+m^2c^4$; a photon has $m=0$.
  • Photoelectric and Compton effects need photons; interference, diffraction and polarisation are wave effects.
Trap. Shorter wavelength means larger energy and momentum per photon.

Practice: first 3 of 9 questions

Q1JEE Main 2026 (28 Jan, Shift 2)Easy

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)

  1. A$10\times10^{15}$
  2. B$5\times10^{16}$
  3. C$5\times10^{15}$
  4. D$2\times10^{16}$
Answer and solution

Answer: Option 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.

Point to note. $n=P\lambda/hc$.

Q2JEE Main 2024 (1 Feb, Shift 2)Easy

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 $h = 6.63\times10^{-34}\ \text{J s}$)

  1. A$9\times10^{18}$
  2. B$6\times10^{15}$
  3. C$5\times10^{15}$
  4. D$7\times10^{16}$
Answer and solution

Answer: Option 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.

Point to note. $n=P/h\nu$.

Q3JEE Main 2024 (29 Jan, Shift 2)Easy

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:

  1. A$1:5$
  2. B$1:3$
  3. C$5:3$
  4. D$3:5$
Answer and solution

Answer: Option 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).

Point to note. $n\propto\lambda$.

44 more questions and 2 more topics in the full chapter

  1. 2Photoelectric effect and Einstein's equation20 questions with solutions
  2. 3Matter waves: de Broglie wavelength18 questions with solutions
  3. ✓Last-day checklist

The complete Dual Nature of Radiation and Matter chapter is in the Class 12 Pack and the Complete Pack, as a PDF on WhatsApp. Want to see a whole chapter first? Laws of Motion and Current Electricity are free in full.

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