Question

The wavelength of photons in two cases are $4000\stackrel{\circ }{A}$ and $3600\stackrel{\circ }{A}$ respectively what is difference in stopping potential for these two?

• A. $0.02V$
• B. $0.03V$
• C. $0.04V$
• D. $0.01V$

Answer 1

The correct option is A $0.02V$

$V_{s_1}=\frac{hc}{e}[\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_0}]$

and $V_{s_2}=\frac{hc}{e}[\frac{1}{{\lambda }_2}-\frac{1}{{\lambda }_0}]$

$V_{s_2}-V_{s_1}=\frac{hc}{e}[\frac{1}{{\lambda }_2}-\frac{1}{{\lambda }_0}]-\frac{hc}{e}[\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_0}]$

$=\frac{hc}{e}[\frac{1}{{\lambda }_2}-\frac{1}{{\lambda }_1}]$

$=\frac{6.624\times {10}^{-34}\times 3\times {10}^8}{1.6\times {10}^{-19}}[\frac{1}{3600}-\frac{1}{4000}]\times {10}^{10}$

$=0.02$V on simplification