JEE Physics Dual Nature Of Radiation and Matter MCQs Set 11

Answer: (c) B only

Explanation:
1. A is false: \( eV_s=h\nu-\phi \) shows the stopping potential depends on the frequency as well as the work function.
2. B is true: more intensity means more photons per second and hence more photoelectrons, so a larger saturation current.
3. C is false: the maximum kinetic energy depends on frequency, not on intensity.
4. D is false: wave theory cannot explain the threshold frequency or the instantaneous emission.

Teacher's Note:
1. Intensity controls the number of photoelectrons; frequency controls their maximum energy.
2. The photon picture is needed to explain the photoelectric effect.

Answer: (b) A-III, B-IV, C-I, D-II

Explanation:
1. h = E/f has dimensions of energy x time = [M L² T⁻¹] (III).
2. Stopping potential is a potential difference, energy per charge: [M L² T⁻² / (A T)] = [M L² T⁻³ A⁻¹] (IV).
3. Work function is an energy: [M L² T⁻²] (I).
4. Momentum is mass x velocity: [M L T⁻¹] (II).

Teacher's Note:
1. Reduce each quantity to a defining relation: E = hf, V = W/q, and p = mv.
2. Remember that charge has dimensions [A T].

Answer: (b) \( \lambda_\alpha<\lambda_p<\lambda_e \)

Explanation:
1. For a given kinetic energy K: \( \lambda=\frac{h}{\sqrt{2mK}} \), so \( \lambda\propto\frac{1}{\sqrt m} \).
2. Masses satisfy \( m_\alpha>m_p>m_e \), therefore \( \lambda_\alpha<\lambda_p<\lambda_e \).

Teacher's Note:
1. At equal kinetic energy the heavier particle has the larger momentum and hence the shorter wavelength.
2. Charge does not appear in this relation.

Answer: (d) \( \frac{\lambda_0}{\sqrt2} \)

Explanation:
1. For an electron accelerated through V: \( \lambda=\frac{h}{\sqrt{2meV}}\propto\frac{1}{\sqrt V} \).
2. \( \lambda'=\lambda_0\sqrt{\frac{20}{40}}=\frac{\lambda_0}{\sqrt2} \)

Teacher's Note:
1. The wavelength falls as the square root of the accelerating voltage, not directly with it.
2. Doubling the voltage divides the wavelength by root 2.

Answer: (d) Both statement I and statement II are correct

Explanation:
1. Statement I is correct: the stopping potential depends on frequency and work function only; power changes only the number of photoelectrons.
2. Statement II is correct: \( K_{max}=\frac{hc}{\lambda}-\phi \), so it depends on the wavelength.

Teacher's Note:
1. Power (intensity) and frequency play different roles in the photoelectric effect.
2. Shorter wavelength means higher photon energy and larger K_max.

Answer: (d) C and D only

Explanation:
1. Emission needs a wavelength shorter than the threshold wavelength, 5500 Å (visible green).
2. Infra-red wavelengths are longer than 5500 Å, so no photoelectrons are emitted whatever the power.
3. Ultra-violet wavelengths are shorter, so both the 75 W and the 10 W ultra-violet lamps cause emission.

Teacher's Note:
1. Emission depends on the frequency of the light, never on its power, once the threshold is crossed or not.
2. A higher-power source only emits more photons of the same energy.

Answer: (c) 8

Explanation:
1. For a particle of mass M and charge q accelerated through V: \( \lambda=\frac{h}{\sqrt{2MqV}} \).
2. \( \frac{\lambda_\alpha}{\lambda_p}=\sqrt{\frac{m_pq_p}{m_\alpha q_\alpha}}=\sqrt{\frac{1\times1}{4\times2}}=\frac{1}{\sqrt8} \), so m = 8.

Teacher's Note:
1. Both the mass and the charge of the particle matter when the accelerating potential is fixed.
2. An alpha particle has mass 4 u and charge 2e.

Answer: (b) \( 2\times10^{-17}\ \text{kg m/s} \)

Explanation:
1. Energy absorbed: \( E=Pt=20\times10^{-3}\times300\times10^{-9}=6\times10^{-9}\ \text{J} \).
2. For complete absorption the momentum gained equals the momentum of the light: \( p=\frac{E}{c}=\frac{6\times10^{-9}}{3\times10^{8}}=2\times10^{-17}\ \text{kg m/s} \).

Teacher's Note:
1. A photon of energy E carries momentum E/c.
2. If the light were reflected, the momentum transfer would be doubled.

Answer: (a) \( 4\times10^{-8}\ \text{N} \)

Explanation:
1. Half of the emitted power, P/2, falls on the hemisphere. A ray making angle \( \theta \) with the axis is reflected straight back along itself, so it transfers momentum twice its own.
2. Power in the angle range \( d\theta \) is \( \frac P2\sin\theta\,d\theta \) and its axial force component is \( \frac{2}{c}\cos\theta\cdot\frac P2\sin\theta\,d\theta \).
3. \( F=\frac{P}{c}\int_0^{\pi/2}\sin\theta\cos\theta\,d\theta=\frac{P}{2c}=\frac{24}{2\times3\times10^{8}}=4\times10^{-8}\ \text{N} \)

Teacher's Note:
1. By symmetry the sideways components cancel, so only the component along the axis remains.
2. The radius of the hemisphere does not affect the force.

Answer: (b) \( \frac94 \)

Explanation:
1. Since \( \lambda\propto\frac{1}{\sqrt V} \), \( \frac{\lambda}{1.5\lambda}=\sqrt{\frac{V_2}{V_1}}\Rightarrow\frac{V_2}{V_1}=\frac49 \).
2. \( \frac{V_1}{V_2}=\frac94 \)

Teacher's Note:
1. A longer wavelength means a lower accelerating voltage.
2. Square the ratio of wavelengths to get the ratio of voltages.

Answer: (d) Metal A will not emit photo-electrons

Explanation:
1. Photon energy: \( E=\frac{1240\ \text{eV nm}}{350\ \text{nm}}\approx3.54\ \text{eV} \).
2. Metal A needs 4.8 eV, which is more than 3.54 eV, so no emission.
3. Metal B needs only 2.2 eV, so it emits photoelectrons with kinetic energy about 1.3 eV.

Teacher's Note:
1. Compare the photon energy with each work function.
2. Use E (eV) = 1240 / wavelength (nm) as a quick conversion.

Answer: (b) 4 : 1

Explanation:
1. Equal wavelength means equal momentum p.
2. Kinetic energy in terms of momentum: \( K=\frac{p^2}{2m} \).
3. \( \frac{K_p}{K_\alpha}=\frac{m_\alpha}{m_p}=4 \)

Teacher's Note:
1. With the same momentum, the lighter particle has the larger kinetic energy.
2. The mass of an alpha particle is about four times that of a proton.

Answer: (a) \( \frac{v_1}{v_2}=\frac12 \)

Explanation:
1. Einstein's equation: \( \tfrac12mv^2=hf-hf_0 \).
2. \( K_1=2hf_0-hf_0=hf_0,\qquad K_2=5hf_0-hf_0=4hf_0 \).
3. \( \frac{v_1}{v_2}=\sqrt{\frac{K_1}{K_2}}=\sqrt{\frac14}=\frac12 \)

Teacher's Note:
1. Velocity depends on the square root of the kinetic energy.
2. The work function is hf0, so each kinetic energy is h times (f - f0).