Question

When a metallic surface is illuminated with monochromatic light of wavelength l, the stopping potential is 5V₀. When the same surface is illuminated with light of wavelength 3l, the stopping potential is V₀. Then the work function of the metallic surface is-

• A. $\frac{hc}{6\lambda}$
• B. $\frac{hc}{5\lambda}$
• C. $\frac{hc}{4\lambda}$
• D. $\frac{2hc}{4\lambda}$

Answer 1

Answer: (A)

$\frac{hc}{6\lambda}$

Answer 2

5eV₀ = $\frac { \mathrm { hc } } { \lambda }$ – W ...(1)

eV₀ = $\frac { \mathrm { hc } } { 3 \lambda }$ – W ...(2)

Solving equation (1) & (2)

– W = $\frac { 5 \mathrm { hc } } { 3 \lambda } - 5 \mathrm {~W}$

4W = ̃ W =