Q111 / 120 · VITEEE 2013 · Physics
When a certain metallic surface is illuminated with monochromatic light of wavelength $\lambda$, the stopping potential for photoelectric current $3 V_{0}$. When the same surface is illuminated with a light of wave length $2 \lambda$, the stopping potential is $V_{0}$. The threshold wavelength for this surface to photoelectric effect is
- Correct answerA$4 \lambda$
- B$6 \lambda$
- C$8 \lambda$
- D$\frac{4}{3} \lambda$
Solution
Here, case (i) $\mathrm{e}\left(3 \mathrm{~V}_{0}\right)=\frac{\text { hc }}{\lambda}-\phi_{0}$
Case (ii) $\mathrm{eV}_{0}=\frac{\text { hc }}{2 \lambda}-$ From eqs. (i) and (ii),
Fromeqs. (i) and (ii), $\frac{3 h c}{2 \lambda}-3 \phi_{0}=\frac{h c}{\lambda}-\phi_{0}$ or $\frac{3 h c}{2 \lambda}-\frac{h c}{\lambda}=3 \phi_{0}-\phi_{0}=\frac{h c}{2 \lambda}=2 \phi_{0}$
or $\frac{3 h c}{2 \lambda}-\frac{h c}{\lambda}$ or, $\phi_{0}=\frac{h c}{4 \lambda}$
$\therefore$ Threshold wavelength
$\lambda_{0}=\frac{\mathrm{hc}}{\phi_{0}}=\frac{\mathrm{hc}}{\mathrm{hc}} \times 4 \lambda=4 \lambda$
Hence, the correct answer is option: 'A'.