NCERT Solutions for Class 12 Physics Chapter 11: Dual Nature of Radiation and Matter (NCERT 2026–27)

These Class 12 Physics Chapter 11 solutions cover Dual Nature of Radiation and Matter with complete, step-by-step answers to every NCERT exercise (11.1–11.11). All numericals are solved with full working and verified units, matching the official NCERT answer key. The chapter explains how light behaves both as a wave and as a stream of particles (photons), introduces the photoelectric effect and Einstein’s photoelectric equation, and extends the idea of wave–particle duality to matter through the de Broglie relation.

Class: 12 Subject: Physics Chapter: 11 Topic: Dual Nature of Radiation and Matter Exercises: 11.1–11.11 Session: 2026–27
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Class 12 Physics Chapter 11 Solutions – Overview

Chapter 11, Dual Nature of Radiation and Matter, brings together two of the most important ideas in modern physics. First, it shows that electromagnetic radiation — long known to be a wave (interference, diffraction, polarisation) — also behaves as a stream of energy packets called photons when it interacts with matter. The strongest evidence for this is the photoelectric effect: the emission of electrons from a metal surface when light of suitable frequency falls on it. The wave theory could not explain the existence of a threshold frequency, the independence of maximum kinetic energy from intensity, or the instantaneous nature of emission. Einstein’s photoelectric equation, Kmax = hν − φ0, explained all of these elegantly. Second, the chapter completes the symmetry of nature: if waves can act as particles, then particles can act as waves. The de Broglie relation, λ = h/p, gives the wavelength of matter waves, which is measurable only for sub-atomic particles such as electrons.

Key Concepts & Formulas

Work function (φ0): the minimum energy required to eject an electron from a metal surface. Measured in eV; 1 eV = 1.602 × 10−19 J.

Photoelectric effect: emission of electrons from a metal when illuminated by light whose frequency exceeds the threshold frequency ν0.

Photon: a quantum of radiation with energy E = hν = hc/λ and momentum p = hν/c = h/λ, travelling at speed c; electrically neutral.

de Broglie wave: every moving particle has an associated wavelength λ = h/p = h/(mv).

Photon energy: E = hν = hc/λ

Photon momentum: p = hν/c = h/λ

Einstein’s photoelectric equation: Kmax = ½mv2max = hν − φ0 = h(ν − ν0)

Stopping potential: eV0 = Kmax  ⇒  V0 = (h/e)ν − (φ0/e)

Threshold frequency: ν0 = φ0/h

de Broglie wavelength: λ = h/p = h/(mv)

Constants used: h = 6.63 × 10−34 J·s; c = 3 × 108 m/s; e = 1.6 × 10−19 C; me = 9.11 × 10−31 kg.

NCERT Exercises (11.1–11.11) – Solutions

Questions are reproduced verbatim from the NCERT textbook; all working is original and numerical answers are cross-checked with the official NCERT answer key.

11.1 Find the (a) maximum frequency, and (b) minimum wavelength of X-rays produced by 30 kV electrons.

SOLUTION Energy gained by each electron: E = eV = (1.6 × 10−19 C)(30 × 103 V) = 4.8 × 10−15 J. (a) The maximum X-ray frequency corresponds to the whole energy converted to one photon: νmax = E/h = (4.8 × 10−15)/(6.63 × 10−34) = 7.24 × 1018 Hz. (b) Minimum wavelength: λmin = c/νmax = (3 × 108)/(7.24 × 1018) = 4.14 × 10−11 m = 0.041 nm.

11.2 The work function of caesium metal is 2.14 eV. When light of frequency 6 ×1014 Hz is incident on the metal surface, photoemission of electrons occurs. What is the (a) maximum kinetic energy of the emitted electrons, (b) Stopping potential, and (c) maximum speed of the emitted photoelectrons?

SOLUTION (a) Photon energy: E = hν = (6.63 × 10−34)(6 × 1014) = 3.98 × 10−19 J = 2.49 eV. Kmax = hν − φ0 = 2.49 − 2.14 = 0.34 eV (= 0.54 × 10−19 J). (b) Stopping potential: V0 = Kmax/e = 0.34 eV / e = 0.34 V. (c) vmax = √(2Kmax/m) = √[(2 × 0.54 × 10−19)/(9.11 × 10−31)] = √(1.19 × 1011) ≈ 3.44 × 105 m/s = 344 km/s.

11.3 The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?

SOLUTION Kmax = eV0 = (1.6 × 10−19 C)(1.5 V) = 2.4 × 10−19 J = 1.5 eV (= 2.4 × 10−19 J).

11.4 Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 mW. (a) Find the energy and momentum of each photon in the light beam, (b) How many photons per second, on the average, arrive at a target irradiated by this beam? (Assume the beam to have uniform cross-section which is less than the target area), and (c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon?

SOLUTION (a) Energy of each photon: E = hc/λ = (6.63 × 10−34)(3 × 108)/(632.8 × 10−9) = 3.14 × 10−19 J (≈ 1.96 eV). Momentum: p = E/c = (3.14 × 10−19)/(3 × 108) = 1.05 × 10−27 kg·m/s. (b) Number of photons per second: N = P/E = (9.42 × 10−3)/(3.14 × 10−19) = 3 × 1016 photons/s. (c) For equal momentum, mHv = p, so v = p/mH = (1.05 × 10−27)/(1.67 × 10−27) = 0.63 m/s (mass of H atom ≈ 1.67 × 10−27 kg).

11.5 In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12 × 10−15 V s. Calculate the value of Planck’s constant.

SOLUTION From V0 = (h/e)ν − (φ0/e), the slope of the V0–ν graph is h/e. ∴ h = slope × e = (4.12 × 10−15 V s)(1.6 × 10−19 C) = 6.59 × 10−34 J·s.

11.6 The threshold frequency for a certain metal is 3.3 × 1014 Hz. If light of frequency 8.2 × 1014 Hz is incident on the metal, predict the cut-off voltage for the photoelectric emission.

SOLUTION Cut-off (stopping) voltage: V0 = h(ν − ν0)/e. V0 = (6.63 × 10−34)(8.2 × 1014 − 3.3 × 1014)/(1.6 × 10−19) = (6.63 × 10−34)(4.9 × 1014)/(1.6 × 10−19) = (3.25 × 10−19)/(1.6 × 10−19) = 2.0 V.

11.7 The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm?

SOLUTION Energy of incident photon: E = hc/λ = (6.63 × 10−34)(3 × 108)/(330 × 10−9) = 6.03 × 10−19 J. Convert to eV: E = (6.03 × 10−19)/(1.6 × 10−19) = 3.77 eV. Since E (3.77 eV) < φ0 (4.2 eV), i.e. ν < ν0, the photon energy is below the work function. No, the metal will not give photoelectric emission.

11.8 Light of frequency 7.21 × 1014 Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 × 105 m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons?

SOLUTION Maximum kinetic energy: Kmax = ½mv2max = ½(9.11 × 10−31)(6.0 × 105)2 = 1.64 × 10−19 J. From Kmax = h(ν − ν0): ν0 = ν − Kmax/h = 7.21 × 1014 − (1.64 × 10−19)/(6.63 × 10−34). = 7.21 × 1014 − 2.47 × 1014 = 4.73 × 1014 Hz.

11.9 Light of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photoelectrons is 0.38 V. Find the work function of the material from which the emitter is made.

SOLUTION Energy of incident photon: E = hc/λ = (6.63 × 10−34)(3 × 108)/(488 × 10−9) = 4.08 × 10−19 J = 2.55 eV. Maximum kinetic energy: Kmax = eV0 = 0.38 eV. Work function: φ0 = E − Kmax = 2.55 − 0.38 = 2.16 eV (= 3.46 × 10−19 J).

11.10 What is the de Broglie wavelength of (a) a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s, (b) a ball of mass 0.060 kg moving at a speed of 1.0 m/s, and (c) a dust particle of mass 1.0 × 10−9 kg drifting with a speed of 2.2 m/s?

SOLUTION de Broglie wavelength: λ = h/(mv). (a) λ = (6.63 × 10−34)/(0.040 × 1000) = (6.63 × 10−34)/(40) = 1.7 × 10−35 m. (b) λ = (6.63 × 10−34)/(0.060 × 1.0) = (6.63 × 10−34)/(0.060) = 1.1 × 10−32 m. (c) λ = (6.63 × 10−34)/(1.0 × 10−9 × 2.2) = (6.63 × 10−34)/(2.2 × 10−9) = 3.0 × 10−25 m. (The NCERT key prints this as 3.0 × 10−23 m; the unit-correct value from λ = h/mv is 3.0 × 10−25 m.) In every case the wavelength is far too small to measure, which is why macroscopic objects show no observable wave behaviour.

11.11 Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).

SOLUTION For a photon of frequency ν, the momentum is p = hν/c (since E = pc and E = hν). Its de Broglie wavelength is λ = h/p = h/(hν/c) = c/ν. But c/ν is precisely the wavelength of the electromagnetic radiation itself. Hence the de Broglie wavelength of a photon equals the wavelength of the radiation of which it is a quantum. Proved.

Extra Practice Questions

Short Answer Type Questions

Q1. Define work function and threshold frequency.

ANSWERThe work function (φ0) is the minimum energy needed to free an electron from a metal surface. The threshold frequency (ν0 = φ0/h) is the minimum frequency of incident light below which no photoelectric emission occurs, however intense the light.

Q2. Why does the maximum kinetic energy of photoelectrons not depend on the intensity of incident light?

ANSWERIn the photon picture, one electron absorbs exactly one photon of energy hν. Kmax = hν − φ0 depends only on frequency and the metal. Increasing intensity raises the number of photons (hence the number of electrons and the current), not the energy per photon, so Kmax is unchanged.

Q3. State two properties of a photon.

ANSWERA photon has energy E = hν and momentum p = hν/c, travels at the speed of light, and is electrically neutral (not deflected by electric or magnetic fields).

Q4. Calculate the energy (in eV) of a photon of wavelength 600 nm.

ANSWERE = hc/λ = (6.63 × 10−34)(3 × 108)/(600 × 10−9) = 3.32 × 10−19 J = 3.32 × 10−19/1.6 × 10−19 = 2.07 eV.

Q5. Why is the de Broglie wavelength of a cricket ball not observable?

ANSWERBecause λ = h/(mv) and the ball’s mass is very large, its momentum is enormous compared with h. The resulting wavelength (~10−34 m) is far smaller than any object or instrument can detect, so its wave nature is undetectable.

Long Answer Type Questions

Q1. State Einstein’s photoelectric equation and explain how it accounts for the experimental laws of the photoelectric effect.

ANSWEREinstein’s equation is Kmax = hν − φ0, where a single electron absorbs a single quantum hν. (i) Since Kmax depends linearly on ν but not on intensity, it explains why the maximum kinetic energy rises with frequency and is independent of intensity. (ii) Emission is possible only if hν > φ0, i.e. ν > ν0 = φ0/h, explaining the threshold frequency. (iii) Intensity = number of photons per unit area per second, so more intensity means more electrons and higher current, explaining current ∝ intensity. (iv) Absorption of one photon by one electron is a single instantaneous event, explaining the absence of any time lag. The wave theory failed on all four counts, which the photon picture resolves.

Q2. Describe how the variation of photocurrent depends on intensity, collector potential and frequency of incident light.

ANSWER(i) With intensity: at fixed frequency and accelerating potential, photocurrent increases linearly with intensity because the number of photoelectrons emitted per second is proportional to the number of incident photons. (ii) With collector potential: as the positive potential of the collector rises, current increases and then reaches a constant saturation value when all emitted electrons are collected; applying a negative (retarding) potential reduces the current until, at the stopping potential V0, it becomes zero. (iii) With frequency: higher frequency gives a larger (more negative) stopping potential, since Kmax = eV0 increases with frequency; the saturation current, however, stays the same if intensity is unchanged.

Q3. Explain de Broglie’s hypothesis of matter waves and discuss its significance.

ANSWERde Broglie argued that since radiation shows both wave and particle behaviour, matter must be symmetrical and also possess wave-like character. He proposed that a particle of momentum p has a wavelength λ = h/p = h/(mv). The left side is a wave attribute and the right side a particle attribute, linked by Planck’s constant h, so the relation expresses wave–particle duality of matter. The wavelength is independent of the particle’s charge. For macroscopic bodies it is immeasurably small, but for sub-atomic particles such as electrons it is comparable to atomic spacings and is therefore measurable — later confirmed by electron diffraction (Davisson–Germer experiment). This idea was the foundation of wave mechanics.

MCQs & Assertion–Reason

1. The maximum kinetic energy of photoelectrons depends on the:

(a) intensity of light    (b) frequency of light    (c) angle of incidence    (d) area of the metal

2. The stopping potential in a photoelectric experiment is independent of:

(a) frequency of light    (b) work function    (c) intensity of light    (d) nature of emitter

3. The momentum of a photon of wavelength λ is:

(a) hλ    (b) h/λ    (c) hc/λ    (d) λ/h

4. Below the threshold frequency, photoelectric emission:

(a) increases with intensity    (b) is delayed    (c) does not occur at all    (d) is instantaneous

5. One electron volt (1 eV) equals:

(a) 1.6 × 10−19 J    (b) 1.6 × 1019 J    (c) 6.63 × 10−34 J    (d) 3 × 108 J

6. The de Broglie wavelength of a particle is inversely proportional to its:

(a) charge    (b) momentum    (c) wavelength    (d) temperature

7. The slope of the stopping potential versus frequency graph equals:

(a) h    (b) e    (c) h/e    (d) e/h

8. Photoelectric emission is essentially:

(a) a delayed process    (b) an instantaneous process    (c) impossible in metals    (d) caused by heat only

9. Photons are:

(a) positively charged    (b) negatively charged    (c) electrically neutral    (d) deflected by magnetic fields

10. The wave nature of matter is significant only for:

(a) heavy macroscopic bodies    (b) sub-atomic particles    (c) slow-moving cars    (d) cricket balls

Answer key: 1-(b), 2-(c), 3-(b), 4-(c), 5-(a), 6-(b), 7-(c), 8-(b), 9-(c), 10-(b).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: Photoelectric emission does not occur below the threshold frequency.

Reason: Below the threshold frequency the photon energy is less than the work function.

A-R 2. Assertion: The maximum kinetic energy of photoelectrons increases with the intensity of light.

Reason: Intensity determines the number of photons incident per unit area per unit time.

A-R 3. Assertion: A photon carries momentum even though it has no rest mass.

Reason: A photon’s momentum is given by p = hν/c = h/λ.

A-R 4. Assertion: The de Broglie wavelength of a moving cricket ball is too small to measure.

Reason: The de Broglie wavelength is inversely proportional to the momentum of the particle.

A-R 5. Assertion: The classical wave theory of light fully explains the photoelectric effect.

Reason: The wave theory predicts a threshold frequency and instantaneous emission.

Answer key: 1-(A), 2-(D), 3-(A), 4-(A), 5-(D).

Common Mistakes & Exam Tips

Common mistakes to avoid

  • Confusing intensity with frequency — intensity changes the number of photoelectrons (and the current), while frequency changes their maximum kinetic energy.
  • Forgetting to convert eV to joules (1 eV = 1.6 × 10−19 J) before using vmax = √(2K/m).
  • Mixing up photon momentum p = h/λ with energy E = hc/λ.
  • Writing the de Broglie relation as λ = h·mv instead of λ = h/(mv).
  • Assuming emission is delayed at low intensity — photoelectric emission is always instantaneous (~10−9 s).

How to score full marks in this chapter

Always start numericals by listing the constants and converting units (nm → m, eV → J, kV → V). For photoelectric problems, use Kmax = hν − φ0 and eV0 = Kmax consistently. Quote the threshold condition ν > ν0 when checking whether emission occurs. For de Broglie problems, compute momentum first, then divide h by it, and comment on why macroscopic wavelengths are unmeasurable. Carry units through every step — markers award method marks for clear working.

Frequently Asked Questions

What is Class 12 Physics Chapter 11 about?

Chapter 11, Dual Nature of Radiation and Matter, explains that light behaves both as a wave and as particles (photons), covers the photoelectric effect and Einstein’s photoelectric equation, and introduces de Broglie’s idea of matter waves with the relation λ = h/p.

What is Einstein’s photoelectric equation?

It states Kmax = hν − φ0, where Kmax is the maximum kinetic energy of the emitted electrons, hν is the photon energy and φ0 is the work function. It explains the threshold frequency, the dependence of Kmax on frequency, and the instantaneous nature of emission.

Are these Class 12 Physics Chapter 11 solutions free?

Yes. All ClearStudy NCERT Solutions for Class 12 Physics are free and follow the official NCERT textbook for session 2026–27, with every numerical solved step by step and verified against the NCERT answer key.

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