Question

When photons of wavelength $\lambda_1$ are incident on an isolated sphere, the corresponding stopping potential is found to be $V$. When photons of wavelength $\lambda_2$ are used, the corresponding stopping potential was thrice that of the above value. If light of wavelength $\lambda_3$ is used then find the stopping potential for this case :

• A. $\frac{hc}{e} \left[ \frac{1}{\lambda_{3}} - \frac{1}{\lambda_{2}} - \frac{1}{\lambda_{1}}\right] $
• B. $\frac{hc}{e} \left[ \frac{1}{\lambda_{3}} + \frac{1}{\lambda_{2}} - \frac{1}{\lambda_{1}}\right] $
• C. $\frac{hc}{e} \left[ \frac{1}{\lambda_{3}} + \frac{1}{ 2 \lambda_{2}} - \frac{3}{2 \lambda_{1}}\right] $
• D. $\frac{hc}{e} \left[ \frac{1}{\lambda_{3}} + \frac{1}{2 \lambda_{2}} - \frac{1}{\lambda_{1}}\right] $

Answer 1

Answer: (C)

$\frac{hc}{\lambda_{1}}=\frac{hc}{\lambda_{0}}+eV ... \left(1\right)$
$\frac{hc}{\lambda_{2}}=\frac{hc}{\lambda_{0}}+eV ... \left(2\right)$
$\frac{hc}{\lambda_{3}}=\frac{hc}{\lambda_{0}}+3eV ' ... \left(3\right) $
Equation (1) & (2)
$\frac{3}{2\lambda_{1}}-\frac{2}{2\lambda_{2}}=\frac{1}{\lambda_{0}}$
$\frac{\lambda c}{\lambda_{3}}hc\left[\frac{3}{2\lambda_{1}}-\frac{1}{2\lambda_{2}}\right]eV $'
$\frac{hc}{e}\left[\frac{1}{\lambda_{3}}-\frac{3}{2\lambda_{1}}+\frac{1}{2\lambda_{2}}\right]=V '$