Question

The wavelength of light incident on a photo – cathode is changed from 500 nm to 400 nm. The change in the stopping potential is, given $\frac{hc}{e}=1.25\times {10}^{-6}Jm{C}^{-1}$

• A. 0.1v
• B. 10 v
• C. 62.5 v
• D. 0.625 v

Answer 1

The correct option is D 0.625 v

We have $\frac{hc}{{\lambda }_1}=w+e{V}_{01}$
And $\frac{hc}{{\lambda }_2}=w+e{V}_{02}$
$\Rightarrow hc[\frac{1}{{\lambda }_2}-\frac{1}{{\lambda }_1}]=e(V_{02}-V_{01})$
$\to V_{02}-V_{01}=\frac{hc}{e}[\frac{1}{{\lambda }_2}-\frac{1}{{\lambda }_1}]=1.25\times {10}^{-6}\times \frac{{10}^9}{2000}=0.625$