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. $\frac{hc}{e}[\frac{1}{{\lambda }_3}+\frac{1}{{\lambda }_2}-\frac{1}{{\lambda }_1}]$
• B. $\frac{hc}{e}[\frac{1}{{\lambda }_3}+\frac{1}{{2\lambda }_2}-\frac{3}{{2\lambda }_1}]$
• C. $\frac{hc}{e}[\frac{1}{{\lambda }_3}-\frac{1}{{\lambda }_2}-\frac{1}{{\lambda }_1}]$
• D. $\frac{hc}{e}[\frac{1}{{\lambda }_3}+\frac{1}{2{\lambda }_2}-\frac{1}{{\lambda }_1}]$

Answer 1

Answer: (B)

Stopping potential is given by $e{V}_o=\frac{hc}{\lambda }-\varphi$

where

$h:$ Planck's constant, $c:$ speed of light, $\lambda :$ wavelength, $\varphi :$ work function

Hence,

$eV=\frac{hc}{{\lambda }_1}-\varphi ...........\left(i\right)$

$3eV=\frac{hc}{{\lambda }_2}-\varphi ...............\left(ii\right)$

From (i) and (ii),

$\varphi =\frac{3hc}{2e{\lambda }_1}-\frac{hc}{2e{\lambda }_2}.........\left(iii\right)$

When photons of wavelength ${\lambda }_3$ is incident, stopping potential is equal to

$e{V}^{\prime }=\frac{hc}{{\lambda }_3}-\varphi .........\left(iv\right)$

From (iii) and (iv),

$V^{\prime }=-\frac{3hc}{2e{\lambda }_1}+\frac{hc}{2e{\lambda }_2}+\frac{hc}{e{\lambda }_3}$