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)

$\frac{hc}{{\lambda }_1}=\frac{hc}{{\lambda }_0}+eV...\left(1\right)$
$\frac{hc}{{\lambda }_2}=\frac{hc}{{\lambda }_0}+eV...\left(2\right)$
$\frac{hc}{{\lambda }_3}=\frac{hc}{{\lambda }_0}+3e{V}^{\prime }...\left(3\right)$
Equation (1) & (2)
$\frac{3}{2{\lambda }_1}-\frac{2}{2{\lambda }_2}=\frac{1}{{\lambda }_0}$
$\frac{\lambda c}{{\lambda }_3}hc\left[\frac{3}{2{\lambda }_1}-\frac{1}{2{\lambda }_2}\right]eV$'
$\frac{hc}{e}\left[\frac{1}{{\lambda }_3}-\frac{3}{2{\lambda }_1}+\frac{1}{2{\lambda }_2}\right]=V^{\prime }$