Question

A photon incident on a metal of work function 2 eV produced a photoelectron of maximum kinetic energy of 2 eV. The wavelength associated with the photon is:

• A. \( 6200 \) Å
• B. \( 3100 \) Å
• C. \( 9300 \) Å
• D. \( 2000 \) Å

Hint:

The energy of a photon is directly related to its frequency and inversely related to its wavelength.

Answer 1

The Correct Option is B

Step 1: Use Einstein’s Photoelectric Equation \[ E = W + K_{\text{max}} \] where: 
- \( W = 2 \) eV (work function), 
- \( K_{\text{max}} = 2 \) eV, 
- \( E = W + K_{\text{max}} = 4 \) eV. 
Step 2: Compute Wavelength Using the photon energy relation: \[ E = \frac{hc}{\lambda} \] \[ \lambda = \frac{hc}{E} \] Substituting \( h = 6.63 \times 10^{-34} \) Js, \( c = 3 \times 10^8 \) m/s, and \( 1 \) eV \( = 1.6 \times 10^{-19} \) J: \[ \lambda = \frac{12400}{4} = 3100 \text{ Å} \]