Question

When a surface is irradiated with light of wavelength $4950$ , a photocurrent appears which vanishes if a retarding potential greater than $0.6V$ is applied across the phototube. When different source of light is used, it is found that the critical retarding potential is changed to $1.1V$. Find the work function of the emitted surface and the wavelength of the second source.

Answer 1

Given, wavelength of light used, $\lambda =4950\dot{A}$

Stopping potential, $V_{o} =0.6V$

According to Einstein potential equation we have;

$\frac{1}{2}m{v}^2=e{V}_o=hv-V_o$........ (1)

ie, $e{V}_o=\frac{hc}{\lambda }-{\varphi }_o$

Where, ${\varphi }_o$ = work function

$V_o$ = stopping potential

For the first source

${\lambda }_1=4950\dot{A}=4950\times {10}^{-10}$

$V_o=0.6v$

Putting these values in equation (1), we get

$1.6\times {10}^{-19}\times 0.6=\frac{6.6\times {10}^{-34\times 3\times {10}^8}}{495\times {10}^{-9}}-{\varphi }_o$

${\varphi }_o=3.04\times {10}^{-19}J$

$=\frac{3.04\times {10}^{-19}}{1.6\times {10}^{-19}}eV$

$=1.9eV$

Let ${\lambda }_2$ be the wavelength of the second source.

Stopping potential $V_o$ =1.1 v (given)

Therefore,

$1.6\times {10}^{-19}\times 1.1=\frac{6.6\times {10}^{-34\times 3\times {10}^8}}{{\lambda }_2}-3.04\times {10}^{-19}J$

$1.76\times {10}^{-19}=\frac{19.8\times {10}^{-27}}{{\lambda }_2}-3.04\times {10}^{-19}$

$\frac{19.8\times {10}^{-26}}{{\lambda }_2}=4.8\times {10}^{-19}$

$\therefore {\lambda }_2=\frac{19.8\times {10}^{-26}}{4.8\times {10}^{-19}}=4.125\times {10}^{-7}m$

$=4125\dot{A}$