JEE Physics Dual Nature Of Radiation And Matter MCQs Set 17

Answer: (a) \( 6.6\times10^{-19}\ \text{J} \)

Explanation:
1. Use \( E=h\nu \).
2. \( E=6.6\times10^{-34}\times10^{15}=6.6\times10^{-19}\ \text{J} \).

Teacher's Note:
1. Photon energy depends only on frequency. Multiply the exponents carefully: -34 + 15 = -19.

Answer: (a) \( 3\times10^{16} \)

Explanation:
1. Power on 1 cm² = \( 100\times10^{-4}=10^{-2}\ \text{W} \).
2. Energy of one photon = \( hc/\lambda=1.98\times10^{-25}/(6\times10^{-7})=3.3\times10^{-19}\ \text{J} \).
3. Number per second = \( 10^{-2}/(3.3\times10^{-19})\approx3\times10^{16} \).

Teacher's Note:
1. Number of photons per second equals power divided by energy of one photon. Remember to convert cm² to m².

Answer: (b) 2 : 1

Explanation:
1. For 'just liberated' electrons, the photon energy equals the work function: \( \phi=hc/\lambda_0 \).
2. So \( \phi_1:\phi_2=\lambda_2:\lambda_1=600:300=2:1 \).

Teacher's Note:
1. Threshold wavelength and work function are inversely proportional. The emitter that needs shorter wavelength has the larger work function.

Answer: (b) \( \nu_1\lt \nu_2\lt \nu_3 \)

Explanation:
1. From \( eV_0=h\nu-\phi \), a larger stopping potential means a larger frequency.
2. Hence \( \nu_1\lt \nu_2\lt \nu_3 \).

Teacher's Note:
1. Stopping potential grows linearly with frequency. Saturation current, in contrast, depends on intensity.

Answer: (d) the intensity of the incident light increases

Explanation:
1. The number of electrons emitted per second is proportional to the number of photons arriving per second.
2. That number is set by the intensity of the light (above the threshold frequency).

Teacher's Note:
1. Intensity controls how many electrons come out; frequency controls how energetic each one is.

Answer: (d) all of these

Explanation:
1. Wave theory predicts that enough intensity would eventually eject electrons at any frequency, so the threshold frequency is unexplained.
2. It also predicts maximum kinetic energy increasing with intensity, and a time lag to build up energy. Observation contradicts all three.

Teacher's Note:
1. Einstein's photon picture explains all three observations. Learn the three failures of the wave theory as a set.

Answer: (c) between 0 and 1 eV

Explanation:
1. Maximum kinetic energy = 3.8 - 2.8 = 1 eV.
2. Electrons that lose some energy inside the metal come out with less, so the kinetic energies lie between 0 and 1 eV.

Teacher's Note:
1. Einstein's equation gives only the maximum kinetic energy. The emitted electrons show a spread from zero up to that value.

Answer: (c) 4n and E

Explanation:
1. Intensity of a point source varies as \( 1/d^2 \), so halving the distance makes intensity 4 times.
2. The rate of emission is proportional to intensity, so it becomes 4n.
3. Maximum kinetic energy depends only on frequency, so it stays E.

Teacher's Note:
1. Intensity changes the number of photoelectrons, not their energy. Use the inverse square law for a point source.

Answer: (c) 4

Explanation:
1. \( eV_0=\tfrac12 mv^2\Rightarrow V_0=\dfrac{v^2}{2(e/m)} \).
2. \( V_0=\dfrac{(1.2\times10^{6})^2}{2\times1.8\times10^{11}}=\dfrac{1.44\times10^{12}}{3.6\times10^{11}}=4\ \text{V} \).

Teacher's Note:
1. Stopping potential measures the maximum kinetic energy per unit charge. Using e/m directly saves two steps.

Answer: (b) the stopping potential increases

Explanation:
1. A shorter wavelength means a higher frequency and higher photon energy.
2. Maximum kinetic energy \( =hc/\lambda-\phi \) rises, so the stopping potential rises.
3. The work function is a property of the metal and does not change.

Teacher's Note:
1. Work function belongs to the metal, not to the light. Everything else follows from the photon energy.

Answer: (b) 1.10 eV

Explanation:
1. Energy is inversely proportional to wavelength, so a 5000 Å photon has 2 × 1.23 = 2.46 eV.
2. Work function = 2.46 - 1.36 = 1.10 eV.

Teacher's Note:
1. The stopping potential in volt equals the maximum kinetic energy in eV. The saturation current is not needed here.

Answer: (b) \( K_A\lt K_B/2 \)

Explanation:
1. \( K_B=\dfrac{hc}{\lambda_B}-\phi \) and \( K_A=\dfrac{hc}{2\lambda_B}-\phi \).
2. \( \dfrac{K_B}{2}=\dfrac{hc}{2\lambda_B}-\dfrac{\phi}{2} \), which is larger than \( K_A \) by \( \phi/2 \).
3. So \( K_A\lt K_B/2 \).

Teacher's Note:
1. Kinetic energy is not proportional to frequency because of the work function. Compare with half of the other value instead of doubling.

Answer: (b) \( p\propto 1/\lambda \)

Explanation:
1. De Broglie relation: \( \lambda=h/p \).
2. Hence \( p=h/\lambda \), a hyperbola on a p versus \( \lambda \) graph.

Teacher's Note:
1. Momentum and wavelength are inversely related. A p versus lambda graph is a rectangular hyperbola.