CBSE Class 12 Physics Dual Nature of Radiation and Matter MCQs Set 07

Answer: (a) 620 nm

Explanation:
1. λ₀ = \( \frac{hc}{\varphi_0} \), and hc ≈ 1240 eV nm.
2. λ₀ = \( \frac{1240}{2.0} \) = 620 nm

Teacher's Note:
1. Light of wavelength longer than 620 nm will not eject electrons from this metal.
2. Using hc = 1240 eV nm saves converting eV to joule.

Answer: (c) 3.1 eV

Explanation:
1. E = \( \frac{hc}{\lambda} \) = \( \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9}} \) ≈ 4.97 × 10⁻¹⁹ J
2. In eV: \( \frac{4.97 \times 10^{-19}}{1.6 \times 10^{-19}} \) ≈ 3.1 eV

Teacher's Note:
1. 4.97 × 10⁻¹⁹ is the value in joule, not in eV; watch the unit.
2. Violet light (400 nm) has more energy per photon than red light (700 nm).

Answer: (b) 2 V

Explanation:
1. K_max = hν − φ₀ = 5 − 3 = 2 eV
2. eV₀ = K_max, so V₀ = 2 V

Teacher's Note:
1. A kinetic energy of x eV corresponds to a stopping potential of x volt.
2. Adding the energies (8 V) is a common slip.

Answer: (d) the saturation current doubles and the stopping potential is unchanged

Explanation:
1. Doubling the intensity doubles the number of photons per second, so twice as many electrons are emitted per second and the saturation current doubles.
2. Each photon still has energy hν, so K_max and hence V₀ do not change.
3. Answer: saturation current doubles, stopping potential unchanged

Teacher's Note:
1. Intensity controls the number of photoelectrons; frequency controls their maximum energy.
2. This is one of the results the wave theory could not explain.

Answer: (b) A and B have the same intensity and ν_A > ν_B

Explanation:
1. Both curves reach the same saturation current, so the two beams eject the same number of electrons per second: same intensity.
2. The stopping potential of A is larger in magnitude, so the photoelectrons of A have more kinetic energy; A has the higher frequency.
3. Answer: same intensity, ν_A > ν_B

Teacher's Note:
1. Read the height of the flat part for intensity and the point on the negative V axis for frequency.
2. Larger stopping potential always means higher frequency for the same metal.

Answer: (a) h/e

Explanation:
1. From Einstein's equation, V₀ = \( \frac{h}{e} \)ν − \( \frac{\varphi_0}{e} \)
2. This is a straight line y = mx + c with slope m = h/e.

Teacher's Note:
1. The slope is the same for every metal; only the intercept changes.
2. If the y axis is K_max (in J) instead of V₀, the slope is h.

Answer: (c) 6.7 × 10¹⁵

Explanation:
1. Energy of one photon, E = \( \frac{hc}{\lambda} \) = \( \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{663 \times 10^{-9}} \) = 3 × 10⁻¹⁹ J
2. n = \( \frac{P}{E} \) = \( \frac{2 \times 10^{-3}}{3 \times 10^{-19}} \) ≈ 6.7 × 10¹⁵ per second

Teacher's Note:
1. Convert mW to W before dividing.
2. Even a weak source emits an enormous number of photons each second.

Answer: (d) 1.33 × 10⁻²⁷ kg m s⁻¹

Explanation:
1. p = \( \frac{h}{\lambda} \) = \( \frac{6.63 \times 10^{-34}}{500 \times 10^{-9}} \)
2. p = 1.33 × 10⁻²⁷ kg m s⁻¹

Teacher's Note:
1. 3.98 × 10⁻¹⁹ is the photon's energy in joule, not its momentum.
2. A photon has momentum even though its rest mass is zero.

Answer: (b) 0.123 nm

Explanation:
1. For an electron, λ = \( \frac{1.227}{\sqrt{V}} \) nm
2. λ = \( \frac{1.227}{\sqrt{100}} \) = \( \frac{1.227}{10} \) ≈ 0.123 nm

Teacher's Note:
1. This wavelength is close to atomic spacings in crystals, so electrons can be diffracted.
2. Divide by √V, not by V.

Answer: (a) The electron

Explanation:
1. λ = \( \frac{h}{\sqrt{2mK}} \); for the same K, λ is proportional to \( \frac{1}{\sqrt{m}} \).
2. The electron has the much smaller mass, so the electron has the longer wavelength.

Teacher's Note:
1. Lighter particles have longer de Broglie wavelengths for the same kinetic energy.
2. Charge enters only when the particle is accelerated through a potential difference.

Answer: (d) momentum

Explanation:
1. For both, λ = \( \frac{h}{p} \), so equal λ means equal momentum.
2. Their energies differ: the photon has E = pc while the electron has K = p²/2m. So the answer is momentum.

Teacher's Note:
1. The de Broglie relation is the same for light and matter.
2. Speeds also differ: the photon moves at c, the electron much slower.

Answer: (c) The existence of a threshold frequency

Explanation:
1. In the wave picture, any frequency should eventually eject electrons if the light is intense enough or falls long enough.
2. Experiment shows no emission at all below ν₀, however intense the light. This is the threshold frequency, which wave theory cannot explain.

Teacher's Note:
1. Instantaneous emission and K_max independent of intensity are the other two failures.
2. The rise of current with intensity is consistent with both pictures.

Answer: (b) 4 eV

Explanation:
1. Work function φ₀ = 3 − 1 = 2 eV
2. With 6 eV photons: K_max = 6 − 2 = 4 eV

Teacher's Note:
1. K_max is not proportional to photon energy, because the work function is subtracted.
2. Find φ₀ first; it is fixed for the metal.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Emission is observed within about 10⁻⁹ s, so the Assertion is true.
2. In Einstein's picture, one photon gives all its energy hν to one electron at once, so no time is needed to collect energy. The Reason is true.
3. The Reason explains why there is no time lag.

Teacher's Note:
1. The wave theory predicted a delay of hours for dim light.
2. This holds even for very low intensity, provided ν ≥ ν₀.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. eV₀ = K_max = hν − φ₀, which contains no intensity term.
2. Both statements are true, and since V₀ is fixed by K_max, the Reason explains the Assertion.

Teacher's Note:
1. Intensity changes the number of electrons, not their maximum energy.
2. This is clear from the common stopping point of I–V curves for different intensities.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Electrons from deeper inside the metal lose energy before escaping, so their kinetic energies range from zero to K_max. The Assertion is true.
2. Monochromatic light has a single frequency, so every photon has energy hν. The Reason is true.
3. Equal photon energies would suggest equal electron energies; the spread is due to energy lost inside the metal, so the Reason does not explain the Assertion.

Teacher's Note:
1. K_max belongs to electrons from the surface, which lose only the work function.
2. When the Reason is true but does not account for the Assertion, choose (b).

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. For m ≈ 0.15 kg and v ≈ 30 m s⁻¹, λ = h/mv ≈ 10⁻³⁴ m, far smaller than any slit or atom.
2. So no diffraction or interference can be seen. Both statements are true and the Reason explains the Assertion.

Teacher's Note:
1. Matter waves are important only for very light particles such as electrons.
2. λ = h/p, so a large p means a tiny λ.

Answer: (d) A is false, but R is true.

Explanation:
1. Experiment shows that for the same intensity the saturation current is the same for all frequencies above ν₀. The Assertion is false.
2. E = hν, so photon energy does increase with frequency. The Reason is true.

Teacher's Note:
1. Higher frequency increases the stopping potential, not the saturation current.
2. Saturation current is set by the number of photons per second.

Answer: (c) A is true, but R is false.

Explanation:
1. A photon's momentum is p = h/λ = E/c, although its rest mass is zero. The Assertion is true.
2. Photons are electrically neutral, so electric and magnetic fields do not deflect them. The Reason is false.

Teacher's Note:
1. Light pressure on a surface is evidence of photon momentum.
2. Only charged particles are bent by electric and magnetic fields.

Answer: (c) 3.1 eV

Explanation:
1. E = \( \frac{1240}{\lambda \text{ (nm)}} \) eV = \( \frac{1240}{400} \) = 3.1 eV

Teacher's Note:
1. 2.07 eV is the photon energy for 600 nm.
2. Shorter wavelength means more energy per photon.

Answer: (a) 0.85 V

Explanation:
1. K_max = hν − φ₀ = 3.1 − 2.25 = 0.85 eV
2. V₀ = K_max/e = 0.85 V

Teacher's Note:
1. The voltage noted by the students is this stopping potential.
2. Subtract the work function; do not add it.

Answer: (d) 600 nm is longer than the threshold wavelength of about 550 nm

Explanation:
1. Threshold wavelength λ₀ = \( \frac{1240}{2.25} \) ≈ 551 nm
2. 600 nm > λ₀, so each photon (2.07 eV) has less energy than the work function and no electron escapes.
3. Answer: 600 nm is longer than the threshold wavelength (≈ 550 nm)

Teacher's Note:
1. Making the light brighter will not help below threshold.
2. Compare photon energy with φ₀: 2.07 eV < 2.25 eV.

Answer: (b) remains 0.85 V

Explanation:
1. Moving the lamp closer increases intensity, i.e. more photons per second.
2. Photon energy is unchanged, so K_max and V₀ stay the same: remains 0.85 V. Only the saturation current increases.

Teacher's Note:
1. Intensity affects the current, not the stopping potential.
2. Students often expect a brighter lamp to give faster electrons.

Answer: (a) 5.5 × 10⁵ m s⁻¹

Explanation:
1. K_max = 0.85 eV = 0.85 × 1.6 × 10⁻¹⁹ = 1.36 × 10⁻¹⁹ J
2. v_max = \( \sqrt{\frac{2K}{m}} \) = \( \sqrt{\frac{2 \times 1.36 \times 10^{-19}}{9.1 \times 10^{-31}}} \) ≈ 5.5 × 10⁵ m s⁻¹

Teacher's Note:
1. Convert eV to joule before using ½mv².
2. Remember the square root in the last step.

Answer: (b) 2.4 × 10⁻¹⁷ J

Explanation:
1. K = eV = 150 eV
2. K = 150 × 1.6 × 10⁻¹⁹ = 2.4 × 10⁻¹⁷ J

Teacher's Note:
1. An electron gains 1 eV for every volt it is accelerated through.
2. 150 J would be the energy of 1 C of charge, not of one electron.

Answer: (d) 6.6 × 10⁻²⁴ kg m s⁻¹

Explanation:
1. p = \( \sqrt{2mK} \) = \( \sqrt{2 \times 9.1 \times 10^{-31} \times 2.4 \times 10^{-17}} \)
2. p = \( \sqrt{4.37 \times 10^{-47}} \) ≈ 6.6 × 10⁻²⁴ kg m s⁻¹

Teacher's Note:
1. 4.4 × 10⁻⁴⁷ is p², before taking the square root.
2. K = p²/2m is a quick way to switch between K and p.

Answer: (c) 0.1 nm

Explanation:
1. λ = \( \frac{h}{p} \) = \( \frac{6.63 \times 10^{-34}}{6.6 \times 10^{-24}} \) ≈ 1.0 × 10⁻¹⁰ m
2. Check: λ = \( \frac{1.227}{\sqrt{150}} \) ≈ 0.100 nm. So λ ≈ 0.1 nm

Teacher's Note:
1. This is about 5000 times shorter than visible light, hence the finer detail.
2. The 1.227/√V formula applies only to electrons.

Answer: (a) half

Explanation:
1. λ = \( \frac{h}{\sqrt{2meV}} \), so λ is proportional to \( \frac{1}{\sqrt{V}} \).
2. V → 4V gives λ → \( \frac{\lambda}{\sqrt{4}} \) = half

Teacher's Note:
1. The square root halves the effect: 4 times V gives only 2 times p.
2. Higher voltage gives a sharper (higher resolution) image.

Answer: (d) about 43 times smaller than that of the electrons

Explanation:
1. Same charge magnitude and same V, so the same K. Then λ ∝ \( \frac{1}{\sqrt{m}} \).
2. \( \frac{\lambda_e}{\lambda_p} \) = \( \sqrt{\frac{m_p}{m_e}} \) = \( \sqrt{\frac{1.67 \times 10^{-27}}{9.1 \times 10^{-31}}} \) ≈ 43
3. So the proton wavelength is about 43 times smaller.

Teacher's Note:
1. 1836 is the mass ratio; the wavelength ratio is its square root.
2. The heavier particle always has the shorter wavelength for the same K.