Question

In a photo-electric cell, the wavelength of incident light is changed from 4000$A^0$ to 3600$A^0$ . The change in stopping potential will be

• A. 0.14v
• B. 0.24v
• C. 0.35v
• D. 0.44v

Answer 1

The correct option is C 0.35v

$eV=\frac{hc}{\lambda }-\phi$
$e{V}_1=\frac{20\times {10}^{-26}}{4000\times {10}^{-10}}-\phi$
$e{V}_1=5\times {10}^{-19}-\phi$
$e{V}_2=\frac{20\times {10}^{-26}}{3600\times {10}^{-10}}-\phi$
$e{V}_2=5.5\times {10}^{-19}-\phi$
$V_2-V_1=0.35V$