Question

The work functions of two metals are 2.75 eV and 2 eV respectively. If these are irradiated by photons of energy 3 eV, the ratio of maximum momenta of the photoelectrons emitted respectively by them is

• A. 1:2
• B. 1:3
• C. 1:4
• D. 2:1
• E. 4:1

Hint:

When asked for a ratio of momenta, always calculate the kinetic energies first. The momentum ratio is simply the square root of the energy ratio.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The photoelectric effect is described by Einstein's equation, which dictates that the maximum kinetic energy of emitted photoelectrons equals the energy of the incident photons minus the metal's work function.
Step 2: Key Formula or Approach:
1. Einstein's Photoelectric Equation: \[ K_{\text{max}} = E - \Phi \] where \( K_{\text{max}} \) is maximum kinetic energy, \( E \) is photon energy, and \( \Phi \) is the work function. 2. The relation between kinetic energy and momentum (\( p \)): \[ p = \sqrt{2mK_{\text{max}}} \] Since the mass of the electron (\( m \)) is constant, the ratio of momenta for two different metals is: \[ \frac{p_1}{p_2} = \sqrt{\frac{K_1}{K_2}} \]
Step 3: Detailed Explanation:
Let's list the given parameters:
Incident photon energy, \( E = 3 \text{ eV} \)
Work function of metal 1, \( \Phi_1 = 2.75 \text{ eV} \)
Work function of metal 2, \( \Phi_2 = 2 \text{ eV} \)
Calculate the maximum kinetic energy for the first metal (\( K_1 \)): \[ K_1 = E - \Phi_1 \] \[ K_1 = 3 \text{ eV} - 2.75 \text{ eV} = 0.25 \text{ eV} \]
Calculate the maximum kinetic energy for the second metal (\( K_2 \)): \[ K_2 = E - \Phi_2 \] \[ K_2 = 3 \text{ eV} - 2 \text{ eV} = 1 \text{ eV} \]
Calculate the ratio of their maximum momenta: \[ \frac{p_1}{p_2} = \sqrt{\frac{K_1}{K_2}} \] \[ \frac{p_1}{p_2} = \sqrt{\frac{0.25 \text{ eV}}{1 \text{ eV}}} \] \[ \frac{p_1}{p_2} = \sqrt{0.25} \] \[ \frac{p_1}{p_2} = 0.5 = \frac{1}{2} \]
Step 4: Final Answer:
The ratio of the maximum momenta of the photoelectrons is \( 1:2 \).