Dual Nature of Radiation and Matter — Class 12 Physics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 12 Physics chapter "Dual Nature of Radiation and Matter" — 6 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 12 Physics chapter "Dual Nature of Radiation and Matter" — 6 important questions with detailed answers for…
Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated
NCERT Solutions for Class 12 Physics — Dual Nature of Radiation and Matter. Step-by-step answers to all 6 textbook questions from this chapter, written for CBSE board preparation and free to use.
Dual Nature of Radiation and Matter — All 6 Questions Solved
Q1. Define photoelectric effect. State Einstein's photoelectric equation and explain the concept of work function.
Photoelectric Effect: Emission of electrons from a material when light of sufficient frequency is incident on it.
Einstein's Photoelectric Equation:
hν = Φ + K.E._max
Where:
hν = energy of incident photon (h = 6.63 × 10⁻³⁴ J·s, ν = frequency)
Φ = work function (minimum energy to remove electron)
K.E._max = maximum kinetic energy of emitted electron
K.E._max = ½ m_e v²_max
Work Function:
Minimum energy required to remove electron from surface of metal
Φ = hν₀ where ν₀ = threshold frequency
Φ = hc/λ₀ where λ₀ = threshold wavelength
Typical values:
Aluminum: Φ = 4.28 eV
Sodium: Φ = 2.75 eV
Zinc: Φ = 3.63 eV
Key Features:
1. No emission if ν < ν₀ (below threshold), regardless of intensity
2. Emission occurs instantaneously when ν ≥ ν₀
3. K.E._max depends on frequency, not intensity
4. Number of photoelectrons depends on intensity
5. Maximum kinetic energy: K.E._max = hν - Φ = h(ν - ν₀)
Stopping Potential:
Voltage needed to stop fastest electrons:
eV_s = K.E._max = hν - Φ
V_s = (h/e)ν - (Φ/e)
Q2. Light of wavelength 300 nm falls on sodium surface (work function 2.75 eV). Find the maximum kinetic energy of photoelectrons and stopping potential.
Given:
Wavelength: λ = 300 nm = 300 × 10⁻⁹ m
Work function: Φ = 2.75 eV
Planck's constant: h = 6.63 × 10⁻³⁴ J·s
Speed of light: c = 3 × 10⁸ m/s
Electron charge: e = 1.6 × 10⁻¹⁹ C
Photon Energy:
E = hc/λ = (6.63 × 10⁻³⁴ × 3 × 10⁸)/(300 × 10⁻⁹)
E = (19.89 × 10⁻²⁶)/(3 × 10⁻⁷) = 6.63 × 10⁻¹⁹ J
Converting to eV:
E = (6.63 × 10⁻¹⁹)/(1.6 × 10⁻¹⁹) = 4.14 eV
Maximum Kinetic Energy:
K.E._max = hν - Φ = 4.14 - 2.75 = 1.39 eV
In Joules:
K.E._max = 1.39 × 1.6 × 10⁻¹⁹ = 2.22 × 10⁻¹⁹ J
Stopping Potential:
V_s = K.E._max/e = (2.22 × 10⁻¹⁹)/(1.6 × 10⁻¹⁹)
V_s = 1.39 V
Or directly: V_s = K.E._max(in eV)/e = 1.39 V
Answers:
- Maximum kinetic energy: 1.39 eV or 2.22 × 10⁻¹⁹ J
- Stopping potential: 1.39 V
- Threshold wavelength: λ₀ = hc/Φ = 450 nm (light will cause emission)
Q3. What is de Broglie wavelength? Derive its expression and explain wave-particle duality.
de Broglie Wavelength: Wavelength associated with a moving particle, demonstrating wave-particle duality.
Expression Derivation:
From Einstein's energy relation: E = hν
For photon: E = pc (where p = momentum)
Combining: hν = pc
h/λ = p (since c = νλ)
λ = h/p
for any particle with momentum p:
λ = h/p = h/(mv)
Where:
h = 6.63 × 10⁻³⁴ J·s (Planck's constant)
p = mv = momentum of particle
m = mass of particle
v = velocity of particle
For relativistic particles:
λ = h/√(2mK.E.) where K.E. = kinetic energy
Wave-Particle Duality:
Matter exhibits both wave and particle properties:
Particle Properties:
- Definite position
- Definite momentum
- Energy = ½mv²
- Localized
Wave Properties:
- Wavelength λ = h/p
- Frequency ν = E/h
- Diffraction and interference
- Delocalized
Examples:
1. Electron: λ = 1.2 nm at 1000 V acceleration
2. Neutron: comparable wavelength to X-rays
3. Macroscopic object: λ ≈ 10⁻³³ m (undetectable)
Significance:
- Explains electron orbitals in atoms
- Standing waves in atoms
- Quantum mechanics foundation
Q4. An electron is accelerated through 1000 V. Calculate its de Broglie wavelength.
Given:
Accelerating voltage: V = 1000 V
Electron mass: m = 9.11 × 10⁻³¹ kg
Electron charge: e = 1.6 × 10⁻¹⁹ C
Planck's constant: h = 6.63 × 10⁻³⁴ J·s
Kinetic Energy Gained:
K.E. = eV = 1.6 × 10⁻¹⁹ × 1000 = 1.6 × 10⁻¹⁶ J
For non-relativistic case (valid here):
K.E. = ½mv²
v² = 2K.E./m = (2 × 1.6 × 10⁻¹⁶)/(9.11 × 10⁻³¹)
v² = (3.2 × 10⁻¹⁶)/(9.11 × 10⁻³¹) = 3.51 × 10¹⁴
v = 1.87 × 10⁷ m/s
Momentum:
p = mv = 9.11 × 10⁻³¹ × 1.87 × 10⁷ = 1.70 × 10⁻²³ kg·m/s
de Broglie Wavelength:
λ = h/p = (6.63 × 10⁻³⁴)/(1.70 × 10⁻²³)
λ = 3.9 × 10⁻¹¹ m = 0.039 nm = 39 pm
Alternative using energy formula:
λ = h/√(2mK.E.) = (6.63 × 10⁻³⁴)/√(2 × 9.11 × 10⁻³¹ × 1.6 × 10⁻¹⁶)
λ = (6.63 × 10⁻³⁴)/√(2.92 × 10⁻⁴⁶)
λ = (6.63 × 10⁻³⁴)/(1.71 × 10⁻²³) ≈ 3.9 × 10⁻¹¹ m
Answer:
- de Broglie wavelength: 3.9 × 10⁻¹¹ m or 0.039 nm or 39 pm
- Comparable to X-ray wavelengths
- Shows wave-like behavior of electrons
Q5. Explain Davisson-Germer experiment. How does it prove wave nature of electrons?
Davisson-Germer Experiment (1927):
Demonstrated diffraction of electrons, proving their wave nature.
Setup:
- Electrons accelerated through known potential difference
- Directed at nickel crystal surface
- Intensity of scattered electrons measured at various angles
- Using Bragg's law for crystal diffraction
Procedure:
1. Electrons accelerated by voltage V
2. Electrons hit nickel crystal at angle θ to surface
3. Crystal acts as diffraction grating (atomic spacing ≈ 0.2 nm)
4. Intensity of reflected electrons measured by detector
5. Temperature raised to anneal crystal and align atoms
Key Observation:
At certain angles, constructive interference produced sharp diffraction maxima
At other angles, destructive interference produced minima
Bragg's Diffraction Condition:
2d sin θ = nλ
Where d = interatomic spacing = 0.215 nm (nickel)
θ = Bragg angle
n = order of diffraction
λ = de Broglie wavelength
Results:
For V = 54 V at θ = 50°:
Path difference = 2 × 0.215 × sin(50°) = 0.33 nm
Match found with λ calculated from de Broglie: λ = h/√(2meV) = 0.167 nm
Second order: 2λ = 0.334 nm ✓ (perfect match)
Proof of Wave Nature:
1. Clear diffraction pattern like light waves
2. Wavelength matches de Broglie prediction
3. Constructive and destructive interference observed
4. Confirms electron-crystal interaction via waves
5. Validates quantum mechanics predictions
Significance:
- Experimental confirmation of de Broglie hypothesis
- Foundation of electron microscopy
- Demonstrated quantum mechanics validity
Q6. Define Compton scattering. Derive the Compton wavelength shift formula.
Compton Scattering: Collision between high-energy photon and free electron, where photon transfers energy and momentum to electron.
Observation:
Wavelength of scattered photon > wavelength of incident photon
Δλ = λ' - λ (wavelength shift)
Derivation using Conservation Laws:
Energy Conservation:
hν + m_e c² = hν' + E_e
Where E_e = √[(p_e c)² + (m_e c²)²]
Momentum Conservation (along direction of photon):
hν/c = hν'/c cos θ + p_e cos φ
Momentum Conservation (perpendicular):
0 = hν'/c sin θ - p_e sin φ
For photon scattered at angle θ:
After algebraic manipulation (eliminating electron momentum and angle):
Compton Wavelength Shift:
Δλ = λ' - λ = (h/m_e c)(1 - cos θ)
Or: Δλ = λ_c(1 - cos θ)
Where λ_c = h/(m_e c) = Compton wavelength
λ_c = 6.63 × 10⁻³⁴/(9.11 × 10⁻³¹ × 3 × 10⁸)
λ_c = 2.43 × 10⁻¹² m = 2.43 pm (picometers)
Key Features:
1. Shift independent of initial wavelength
2. Maximum shift at θ = 180°: Δλ_max = 2λ_c = 4.86 pm
3. Zero shift at θ = 0° (forward scattering, no collision)
4. Recoil electron gains kinetic energy
5. Energy of scattered photon: hν' = hν/(1 + (hν/m_e c²)(1 - cos θ))
Significance:
- Proves photons are particles with momentum
- Provides evidence for photon-electron interaction
- Validates quantum mechanics and relativity together
Dual Nature of Radiation and Matter — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. What is the minimum energy required to eject an electron from the surface of a metal?
- A. Kinetic energy
- B. Potential energy
- C. Work function
- D. Threshold energy
Answer: C. Work function — The work function is defined as the minimum energy an electron needs to overcome the attractive forces holding it in the metal and escape from the surface.
MCQ 2. In the photoelectric effect, if the intensity of incident radiation is increased, what happens to the photoelectric current?
- A. It decreases
- B. It increases
- C. It remains constant
- D. It first increases then decreases
Answer: B. It increases — The photoelectric current is directly proportional to the number of photoelectrons emitted per second, which in turn depends on the intensity of the incident radiation. Higher intensity means more photons per second, thus more emitted electrons.
MCQ 3. According to Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons is given by:
- A. $KE_{max} = hf + \phi_0$
- B. $KE_{max} = hf - \phi_0$
- C. $KE_{max} = \phi_0 - hf$
- D. $KE_{max} = hf$
Answer: B. $KE_{max} = hf - \phi_0$ — Einstein's photoelectric equation states that the energy of the incident photon ($hf$) is used partly to overcome the work function ($\phi_0$) and the remaining part is converted into the maximum kinetic energy of the emitted electron.
MCQ 4. The energy of a photon with frequency $f$ is given by:
- A. $hf$
- B. $hc/\lambda$
- C. $h/\lambda$
- D. Both A and B
Answer: D. Both A and B — The energy of a photon is given by $E = hf$, where $h$ is Planck's constant and $f$ is the frequency. Since $c = f\lambda$, we can also write $f = c/\lambda$, so $E = hc/\lambda$.
MCQ 5. If the work function of a metal is $\phi_0$, what is its threshold frequency $f_0$?
- A. $f_0 = h/\phi_0$
- B. $f_0 = \phi_0/h$
- C. $f_0 = \phi_0 h$
- D. $f_0 = \sqrt{\phi_0/h}$
Answer: B. $f_0 = \phi_0/h$ — The work function is the minimum energy required to eject an electron, which corresponds to the photon energy at the threshold frequency: $\phi_0 = hf_0$. Therefore, $f_0 = \phi_0/h$.
MCQ 6. Light of frequency $10^{15}$ Hz is incident on a photocell. The work function of the cathode material is 2.1 eV. What is the maximum kinetic energy of the emitted photoelectrons? (Given $h = 4.14 \times 10^{-15}$ eV s)
- A. 1.5 eV
- B. 2.04 eV
- C. 4.14 eV
- D. 0.5 eV
Answer: B. 2.04 eV — Energy of incident photon $E = hf = (4.14 \times 10^{-15} \text{ eV s}) \times (10^{15} \text{ Hz}) = 4.14 \text{ eV}$. Maximum kinetic energy $KE_{max} = E - \phi_0 = 4.14 \text{ eV} - 2.1 \text{ eV} = 2.04 \text{ eV}$.
MCQ 7. The stopping potential in the photoelectric effect is the minimum negative anode potential required to:
- A. Increase the photoelectric current
- B. Decrease the work function
- C. Stop the most energetic photoelectrons from reaching the anode
- D. Increase the frequency of incident light
Answer: C. Stop the most energetic photoelectrons from reaching the anode — Stopping potential is the retarding potential that is just sufficient to stop the fastest moving photoelectrons from reaching the collector, thereby making the photoelectric current zero.
MCQ 8. A graph plotting stopping potential ($V_s$) versus frequency ($f$) of incident radiation in a photoelectric experiment is a straight line. What does the slope of this line represent?
- A. Planck's constant
- B. Work function
- C. $h/e$
- D. $e/h$
Answer: C. $h/e$ — From Einstein's equation, $KE_{max} = hf - \phi_0$. Also, $KE_{max} = eV_s$. So, $eV_s = hf - \phi_0$, which gives $V_s = (h/e)f - \phi_0/e$. Comparing with $y = mx + c$, the slope $m = h/e$.
MCQ 9. The de Broglie wavelength of a particle is given by the relation:
- A. $\lambda = mv/h$
- B. $\lambda = h/mv$
- C. $\lambda = hv/m$
- D. $\lambda = m/hv$
Answer: B. $\lambda = h/mv$ — According to de Broglie hypothesis, the wavelength associated with a particle is inversely proportional to its momentum. $\lambda = h/p = h/mv$.
MCQ 10. The Davisson-Germer experiment demonstrated the:
- A. Particle nature of light
- B. Wave nature of electrons
- C. Photoelectric effect
- D. Quantization of energy levels in atoms
Answer: B. Wave nature of electrons — The Davisson-Germer experiment observed the diffraction of electrons by a nickel crystal, providing experimental evidence for the wave nature of matter, specifically electrons, as proposed by de Broglie.
MCQ 11. An electron is accelerated through a potential difference V. Its de Broglie wavelength is proportional to:
- A. $V$
- B. $V^{-1}$
- C. $V^{1/2}$
- D. $V^{-1/2}$
Answer: D. $V^{-1/2}$ — The kinetic energy gained by an electron is $KE = eV$. Also, $KE = p^2/(2m)$. So $p = \sqrt{2mKE} = \sqrt{2meV}$. De Broglie wavelength $\lambda = h/p = h/\sqrt{2meV}$. Thus, $\lambda \propto V^{-1/2}$.
MCQ 12. The momentum of a photon of wavelength $\lambda$ is:
- A. $h\lambda/c$
- B. $h/\lambda$
- C. $c/\lambda h$
- D. $\lambda/h$
Answer: B. $h/\lambda$ — The energy of a photon is $E = pc$ and also $E = hc/\lambda$. Equating these, $pc = hc/\lambda$, which gives $p = h/\lambda$.
MCQ 13. The phenomenon of emission of electrons from the surface of a metal when light of suitable frequency falls on it is called:
- A. Compton effect
- B. Photoelectric effect
- C. Zeeman effect
- D. Hall effect
Answer: B. Photoelectric effect — The photoelectric effect is the emission of electrons when electromagnetic radiation, such as light, hits a material.
MCQ 14. In the photoelectric effect, if the intensity of incident light is doubled while keeping its frequency constant, what happens to the maximum kinetic energy of the emitted photoelectrons?
- A. It doubles
- B. It halves
- C. It remains unchanged
- D. It increases by a factor of four
Answer: C. It remains unchanged — The maximum kinetic energy of photoelectrons depends only on the frequency of incident light and the work function of the material, not on the intensity of light. Intensity only affects the number of emitted electrons (photocurrent).
MCQ 15. An electron and a proton are accelerated through the same potential difference. The ratio of their de Broglie wavelengths ($\lambda_e / \lambda_p$) is:
- A. $\sqrt{m_p/m_e}$
- B. $\sqrt{m_e/m_p}$
- C. $m_p/m_e$
- D. $m_e/m_p$
Answer: A. $\sqrt{m_p/m_e}$ — For a particle accelerated through potential V, $\lambda = h/\sqrt{2meV}$. So, $\lambda_e = h/\sqrt{2m_e eV}$ and $\lambda_p = h/\sqrt{2m_p eV}$. The ratio $\lambda_e / \lambda_p = \sqrt{m_p/m_e}$.
MCQ 16. Which of the following statements best describes wave-particle duality?
- A. All particles behave like waves, and all waves behave like particles.
- B. Light sometimes behaves as a wave and sometimes as a particle, depending on the phenomenon observed.
- C. Only photons exhibit wave-particle duality.
- D. Particles can only exhibit wave-like properties, and waves can only exhibit particle-like properties.
Answer: B. Light sometimes behaves as a wave and sometimes as a particle, depending on the phenomenon observed. — Wave-particle duality states that light and matter exhibit properties of both waves and particles. For instance, light shows wave nature in interference/diffraction and particle nature in the photoelectric effect.
MCQ 17. Which of the following is true for a photon?
- A. It has rest mass.
- B. It is a charged particle.
- C. Its energy is independent of its frequency.
- D. It travels at the speed of light in vacuum.
Answer: D. It travels at the speed of light in vacuum. — Photons are elementary particles of light, have zero rest mass, no electric charge, and always travel at the speed of light in a vacuum. Their energy is directly proportional to their frequency ($E=hf$).
MCQ 18. If the work function of a metal is $\phi_0$, what is its cut-off wavelength $\lambda_0$?
- A. $\lambda_0 = hc/\phi_0$
- B. $\lambda_0 = \phi_0/hc$
- C. $\lambda_0 = h\phi_0/c$
- D. $\lambda_0 = h/c\phi_0$
Answer: A. $\lambda_0 = hc/\phi_0$ — The work function is the minimum energy required to eject an electron, which corresponds to the photon energy at the threshold wavelength: $\phi_0 = hc/\lambda_0$. Therefore, $\lambda_0 = hc/\phi_0$.
MCQ 19. An electron has a de Broglie wavelength of $10^{-10}$ m. What is its kinetic energy? (Given $h = 6.6 \times 10^{-34}$ J s, $m_e = 9.1 \times 10^{-31}$ kg)
- A. $1.2 \times 10^{-17}$ J
- B. $2.4 \times 10^{-17}$ J
- C. $6.6 \times 10^{-20}$ J
- D. $9.1 \times 10^{-31}$ J
Answer: B. $2.4 \times 10^{-17}$ J — Using $\lambda = h/p$, we get $p = h/\lambda = (6.6 \times 10^{-34}) / (10^{-10}) = 6.6 \times 10^{-24}$ kg m/s. Kinetic energy $KE = p^2/(2m_e) = (6.6 \times 10^{-24})^2 / (2 \times 9.1 \times 10^{-31}) = 2.409 \times 10^{-17}$ J. Option B is the closest value.
MCQ 20. If the wavelength of light is increased, what happens to the energy of its photons?
- A. It increases
- B. It decreases
- C. It remains constant
- D. It depends on the intensity
Answer: B. It decreases — The energy of a photon is given by $E = hc/\lambda$. Since $h$ and $c$ are constants, energy is inversely proportional to wavelength. As $\lambda$ increases, $E$ decreases.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.