Question

Wavelength of incident light falling on a photosensitive surface is changed from $2000\stackrel{o}{A}$to $2050\stackrel{o}{A}$. Corresponding change in stopping potential is [Take $hc=1240eV-nm$] (Write upto two digits after the decimal point.)

Answer 1

$e{V}_o=\frac{hc}{\lambda }-\varphi \Rightarrow V_1=\frac{hc}{e{\lambda }_1}-\frac{\varphi }{e}and{V}_2=\frac{hc}{e{\lambda }_2}-\frac{\varphi }{e}$
$V_1-V_2=\frac{hc}{e}(\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2})=1240Vnm(\frac{1}{200}-\frac{1}{205})\frac{1}{nm}$
$V_1-V_2=1240\times \frac{5}{200\times 205}=0.15V$