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NCERT Exercises and Solutions: Dual Nature of Radiation and Matter

🎓 Class 12 Physics CBSE Theory Ch 11 – Dual Nature of Radiation and Matter ⏱ ~8 min
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NCERT Exercises and Solutions: Dual Nature of Radiation and Matter

Chapter 11 — Summary & Key Formulae

Big ideas in one minute
  • Free electrons are bound to a metal by the work function φ₀; four mechanisms (thermionic, field, photoelectric, secondary) liberate them.
  • Hertz, Hallwachs and Lenard discovered that light above a threshold frequency ejects photoelectrons; intensity sets the count, frequency sets the energy.
  • Wave theory could not explain a threshold or instantaneous emission. Einstein (1905) introduced the photon: Kmax = hν − φ₀.
  • Millikan (1916) verified Einstein's law and used the slope of the V₀–ν line to remeasure Planck's constant.
  • A photon has E = hν, p = h/λ, charge 0, rest mass 0; Compton scattering proved its momentum.
  • de Broglie (1924) symmetrised the duality: every particle has a wave λ = h/p.
  • Davisson and Germer (1927) confirmed it: 54 V electrons diffract from a Ni crystal at 50°, giving λ = 0.165 nm.
QuantitySymbolEquation / value
Photon energyEhν = hc/λ
Photon momentumphν/c = h/λ
Threshold frequencyν₀φ₀/h
Einstein's PE equationKmaxhν − φ₀
Stopping potentialV₀(hν − φ₀)/e
Slope of V₀ vs ν linemh/e
de Broglie wavelengthλh/p = h/(mv)
e⁻ accelerated by V voltsλ1.227/√V nm
Planck constanth6.626 × 10⁻³⁴ J·s
Electron massme9.11 × 10⁻³¹ kg
Electron chargee1.602 × 10⁻¹⁹ C

NCERT Exercises — Worked Solutions

Exercise 11.1 — Photons from a sodium lamp

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

The shortest-wavelength X-ray photon arises when an electron loses all its KE in one go.
\(eV = h\nu_\text{max}\) ⟹ \(\nu_\text{max} = eV/h = (1.6\times10^{-19})(30\times10^3)/(6.626\times10^{-34}) = 7.24\times10^{18}\) Hz.
\(\lambda_\text{min} = c/\nu_\text{max} = (3\times10^8)/(7.24\times10^{18}) = 4.14\times10^{-11}\) m = 0.0414 nm.
Exercise 11.2 — Caesium photoemitter

The work function of caesium metal is 2.14 eV. When light of frequency \(6\times10^{14}\) Hz is incident on the metal surface, find: (a) the maximum kinetic energy of emitted electrons, (b) the stopping potential, and (c) the maximum speed of the photoelectrons.

(a) Photon energy E = hν = (4.136×10⁻¹⁵ eV·s)(6×10¹⁴ Hz) = 2.48 eV.
\(K_\text{max} = E - \phi_0 = 2.48 - 2.14 = 0.34\) eV.
(b) \(V_0 = K_\text{max}/e = 0.34\) V.
(c) \(K_\text{max} = 0.34 \times 1.6\times10^{-19} = 5.44\times10^{-20}\) J.
\(v_\text{max} = \sqrt{2K/m_e} = \sqrt{2(5.44\times10^{-20})/(9.11\times10^{-31})} = 3.46\times10^{5}\) m/s.
Exercise 11.3 — Stopping potential

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

\(K_\text{max} = eV_0 = (1.6\times10^{-19})(1.5) = 2.4\times10^{-19}\) J = 1.5 eV.
Exercise 11.4 — Number of photons from a He-Ne laser

Monochromatic light of wavelength 632.8 nm is produced by a He-Ne laser at a power of 9.42 mW. Find: (a) the energy and momentum of each photon, (b) the number of photons reaching a target per second on average, and (c) how fast a hydrogen atom would have to travel to have the same momentum as a photon.

(a) E = hc/λ = (6.626×10⁻³⁴)(3×10⁸)/(632.8×10⁻⁹) = 3.14×10⁻¹⁹ J ≈ 1.96 eV.
p = h/λ = 6.626×10⁻³⁴/632.8×10⁻⁹ = 1.05×10⁻²⁷ kg·m/s.
(b) N = P/E = 9.42×10⁻³ / 3.14×10⁻¹⁹ = 3.0×10¹⁶ photons/s.
(c) Mass of H atom ≈ 1.67×10⁻²⁷ kg. \(v = p/m = 1.05\times10^{-27}/1.67\times10^{-27} = 0.63\) m/s — slower than walking pace!
Exercise 11.5 — Energy flux from the Sun

The energy flux of sunlight reaching the Earth is 1.388 × 10³ W/m². Estimate the number of photons (per square metre per second) striking the Earth, assuming an average wavelength of 550 nm.

Photon energy at 550 nm = hc/λ = 1240/550 eV = 2.25 eV = 3.61×10⁻¹⁹ J.
N = (1.388×10³)/(3.61×10⁻¹⁹) = 3.84×10²¹ photons m⁻² s⁻¹.
Exercise 11.6 — Slope of V₀ vs ν

In an experiment on photoelectric effect, the slope of the cut-off voltage vs frequency graph is found to be 4.12×10⁻¹⁵ V·s. Calculate the value of Planck's constant.

slope = h/e ⟹ h = slope × e = (4.12×10⁻¹⁵)(1.6×10⁻¹⁹) = 6.59×10⁻³⁴ J·s — within 1% of the textbook value 6.626×10⁻³⁴.
Exercise 11.7 — Sodium lamp photons

A 100 W sodium lamp radiates energy uniformly in all directions. Wavelength of sodium light = 589 nm. (a) Energy per photon associated with the sodium light. (b) Rate of photons emitted from the lamp.

(a) E = hc/λ = 1240/589 eV = 2.11 eV = 3.38×10⁻¹⁹ J.
(b) N = P/E = 100/3.38×10⁻¹⁹ = 2.96×10²⁰ photons/s.
Exercise 11.8 — Threshold frequency of photoemission

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.

\(eV_0 = h(\nu - \nu_0) = (6.626\times10^{-34})(8.2 - 3.3)\times10^{14} = (6.626\times10^{-34})(4.9\times10^{14}) = 3.25\times10^{-19}\) J.
\(V_0 = 3.25\times10^{-19}/1.6\times10^{-19} \approx \) 2.03 V.
Exercise 11.9 — Aluminium photoemitter

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

Photon energy E = 1240/330 = 3.76 eV. Since E (3.76 eV) < φ₀ (4.2 eV), no photoelectric emission occurs at this wavelength.
Exercise 11.10 — Electron and photon at same wavelength

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

\(K_\text{max} = \tfrac12 m_e v^2 = \tfrac12(9.11\times10^{-31})(6.0\times10^{5})^2 = 1.64\times10^{-19}\) J.
\(h\nu_0 = h\nu - K_\text{max} = (6.626\times10^{-34})(7.21\times10^{14}) - 1.64\times10^{-19} = 4.78\times10^{-19} - 1.64\times10^{-19} = 3.14\times10^{-19}\) J.
\(\nu_0 = 3.14\times10^{-19}/6.626\times10^{-34} = \) 4.74×10¹⁴ Hz.
Exercise 11.11 — Mo wavelength range

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

Photon energy E = 1240/488 = 2.54 eV.
\(\phi_0 = E - eV_0 = 2.54 - 0.38 = \) 2.16 eV — close to caesium.
Exercise 11.12 — de Broglie wavelength of an electron

Calculate the (a) momentum, and (b) de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.

KE = 56 eV = 8.96×10⁻¹⁸ J.
(a) \(p = \sqrt{2m_eK} = \sqrt{2(9.11\times10^{-31})(8.96\times10^{-18})} = \) 4.04×10⁻²⁴ kg·m/s.
(b) \(\lambda = h/p = 6.626\times10^{-34}/4.04\times10^{-24} = 1.64\times10^{-10}\) m = 0.164 nm.
Exercise 11.13 — de Broglie wavelength of a fast electron

What is the (a) momentum, (b) speed, and (c) de Broglie wavelength of an electron with kinetic energy of 120 eV?

K = 120 eV = 1.92×10⁻¹⁷ J.
(a) \(p = \sqrt{2m_eK} = \sqrt{2(9.11\times10^{-31})(1.92\times10^{-17})} = \) 5.92×10⁻²⁴ kg·m/s.
(b) \(v = p/m_e = 5.92\times10^{-24}/9.11\times10^{-31} = \) 6.50×10⁶ m/s (about 2% of c).
(c) \(\lambda = h/p = 6.626\times10^{-34}/5.92\times10^{-24} = \) 1.12×10⁻¹⁰ m = 0.112 nm.
Exercise 11.14 — de Broglie wavelength of light photons of equal energy

The wavelength of light from the spectral emission line of sodium is 589 nm. Find the kinetic energy at which (a) an electron, and (b) a neutron, would have the same de Broglie wavelength.

\(p = h/\lambda = 6.626\times10^{-34}/589\times10^{-9} = 1.125\times10^{-27}\) kg·m/s.
(a) Electron: \(K_e = p^2/(2m_e) = (1.125\times10^{-27})^2/(2\times9.11\times10^{-31}) = 6.95\times10^{-25}\) J ≈ 4.34×10⁻⁶ eV.
(b) Neutron: \(K_n = p^2/(2m_n) = 6.95\times10^{-25}\times(m_e/m_n) = 6.95\times10^{-25}/1839 = \) 3.78×10⁻²⁸ J ≈ 2.36×10⁻⁹ eV.
Exercise 11.15 — de Broglie wavelength of an electron and a bullet

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?

(a) λ = h/(mv) = 6.626×10⁻³⁴/(0.04×1000) = 1.66×10⁻³⁵ m.
(b) λ = 6.626×10⁻³⁴/(0.06×1) = 1.10×10⁻³² m.
(c) λ = 6.626×10⁻³⁴/(10⁻⁹×2.2) = 3.01×10⁻²⁵ m.
All are vastly smaller than any physical aperture; their wave nature is undetectable.
Exercise 11.16 — Photon energy of equal momentum

An electron and a photon each have a wavelength of 1.00 nm. Find (a) their momenta, (b) the energy of the photon, and (c) the kinetic energy of the electron.

(a) p = h/λ = 6.626×10⁻³⁴/10⁻⁹ = 6.626×10⁻²⁵ kg·m/s (same for both).
(b) Photon: E = pc = 6.626×10⁻²⁵×3×10⁸ = 1.99×10⁻¹⁶ J = 1240 eV.
(c) Electron: K = p²/(2mₑ) = (6.626×10⁻²⁵)²/(2×9.11×10⁻³¹) = 2.41×10⁻¹⁹ J = 1.51 eV.
The photon is far more energetic — this is why high-energy X-rays use shorter λ to probe smaller distances.
Exercise 11.17 — Thermal neutron wavelength

(a) For what kinetic energy of a neutron will the associated de Broglie wavelength be 1.40×10⁻¹⁰ m? (b) Also find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of (3/2)kT at 300 K.

(a) p = h/λ = 6.626×10⁻³⁴/1.4×10⁻¹⁰ = 4.73×10⁻²⁴ kg·m/s.
K = p²/(2mₙ) = (4.73×10⁻²⁴)²/(2×1.675×10⁻²⁷) = 6.69×10⁻²¹ J ≈ 0.0418 eV.
(b) Average K = (3/2)kT = (1.5)(1.38×10⁻²³)(300) = 6.21×10⁻²¹ J.
p = √(2mₙK) = √(2×1.675×10⁻²⁷×6.21×10⁻²¹) = 4.56×10⁻²⁴ kg·m/s.
λ = h/p = 6.626×10⁻³⁴/4.56×10⁻²⁴ = 1.45×10⁻¹⁰ m ≈ 0.145 nm.
Activity — Self-test in 60 seconds

Without scrolling back, write down (a) the four types of electron emission, (b) Einstein's photoelectric equation, (c) the value of h/e in SI units, and (d) the formula for the de Broglie wavelength of a 100 V electron.

(a) Thermionic, field, photoelectric, secondary.
(b) hν = φ₀ + ½mv²max.
(c) h/e ≈ 4.14 × 10⁻¹⁵ V·s.
(d) λ = 1.227/√100 = 0.1227 nm.

Interactive Revision — Drag the Slider, Match the Phenomenon

Slide the photon energy and watch which physical effect is dominant in that range. The colour bar maps each phenomenon to its typical energy.

1.0 eV
Photoelectric effect (visible light on alkali metals)
10⁻⁶10⁻²110²10⁷ eV

Competency-Based Questions — Mixed Revision

Q1. The photoelectric current depends linearly on:

  • (a) Frequency of light only
  • (b) Intensity of light only
  • (c) Both intensity and frequency
  • (d) Wavelength only
(b) Above threshold, the saturation photocurrent depends linearly on intensity (number of photons per second).

Q2. de Broglie wavelength of an electron accelerated through 100 V is approximately:

  • (a) 12.27 nm
  • (b) 1.227 nm
  • (c) 0.1227 nm
  • (d) 0.0123 nm
(c) λ = 1.227/√100 = 0.1227 nm.

Q3. (Short Answer) Why do red and blue photons produce different stopping potentials in a photocell?

Blue photons have higher frequency, hence larger energy hν. Once the same φ₀ is paid, more KE is left over, so V₀ is larger for blue.

Q4. (Fill in the blank) The slope of the V₀-versus-ν line equals ______, the same for every metal.

h/e ≈ 4.14 × 10⁻¹⁵ V·s.

Q5. (HOT) An electron and a proton have equal de Broglie wavelengths. Compare (i) their momenta, (ii) their kinetic energies, (iii) their speeds.

(i) Equal momenta (since p = h/λ).
(ii) K = p²/(2m), so K_e/K_p = m_p/m_e ≈ 1836. The electron has 1836× the KE.
(iii) v = p/m, so v_e/v_p = m_p/m_e ≈ 1836. Electron moves 1836× faster.

Assertion–Reason — Mixed Revision

Options: (A) Both true, R correct explanation of A. (B) Both true, R not the correct explanation. (C) A true, R false. (D) A false, R true.

Assertion: A photon and an electron with the same momentum have the same de Broglie wavelength.

Reason: The de Broglie relation λ = h/p applies equally to massless and massive particles.

(A) Both correct and the reason explains the assertion. λ depends only on p, not on m.

Assertion: The photoelectric effect supports the particle nature of light, while electron diffraction supports the wave nature of matter.

Reason: Both are predicted by the unified equations E = hν and λ = h/p.

(A) Both correct and the reason explains the assertion — wave-particle duality with one Planck constant.

Assertion: Increasing the wavelength of light below the threshold value increases the photocurrent.

Reason: Longer wavelength means lower frequency.

(D) Assertion is false (longer λ below threshold actually means smaller hν, so no current). Reason is true.

Frequently Asked Questions - NCERT Exercises and Solutions: Dual Nature of Radiation and Matter

What are the key NCERT exercise types in Chapter 11 Dual Nature of Radiation and Matter?
NCERT Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter exercises cover conceptual MCQs, short numerical problems (2-3 marks), long numerical derivations (5 marks), and assertion-reason questions. The MyAiSchool solution set provides step-by-step worked solutions for every NCERT exercise question, aligned with the CBSE board exam pattern. Students should focus on dimensional analysis, formula application, and unit consistency to score full marks.
How should students approach numerical problems in Dual Nature of Radiation and Matter?
For numerical problems in NCERT Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter: (1) list all given data with units, (2) write the relevant formula(e), (3) substitute values carefully, (4) calculate with proper significant figures, (5) state the final answer with the correct unit. Always draw a free-body or schematic diagram where relevant. The MyAiSchool solutions follow this 5-step CBSE-aligned format consistently.
What are the most-asked CBSE board questions from Chapter 11?
From NCERT Class 12 Physics Chapter 11 (Dual Nature of Radiation and Matter), the most-asked CBSE board questions test conceptual understanding of fundamental principles, application of derived formulas to standard scenarios, and proof-style derivations. 5-mark questions usually combine derivation + application. The MyAiSchool exercise set tags each question by board frequency so students can prioritize high-yield problems before exams.
How do I check the dimensional correctness of my answer?
Dimensional correctness in NCERT Class 12 Physics is verified by ensuring the LHS and RHS of any equation have the same dimensional formula. For Chapter 11 Dual Nature of Radiation and Matter problems, write the dimensions of each quantity, substitute, and simplify. If the dimensions match, the equation is dimensionally valid (necessary but not sufficient). The MyAiSchool solutions include dimensional checks in every numerical answer.
What are common mistakes students make in Chapter 11 exercises?
Common mistakes in NCERT Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter exercises include: (1) unit conversion errors (CGS vs SI), (2) sign convention mistakes for charges/currents/vectors, (3) forgetting to consider all field/force contributions, (4) algebra errors during derivations, (5) misreading the problem. The MyAiSchool solutions highlight these traps with red-flag annotations so students learn to avoid them.
How does the MyAiSchool solution differ from other NCERT solution sets?
MyAiSchool Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter solutions use NEP 2024-aligned pedagogy with Bloom Taxonomy tagged questions (L1 Remember to L6 Create), step-by-step working with reasoning, dimensional checks, interactive simulations, and Competency-Based Questions (CBQs) for board exam practice. Each solution is verified against NCERT and CBSE marking schemes for accuracy.
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