Question

Suppose the wavelength of the incident light in photoelectric effect experiment is increased from $3000A^o$ to $3040A^o$. Find the corresponding change in the stopping potential. [Take the product $hc=12.4\times {10}^{-7}Vm]$

Answer 1

$\Delta V=\frac{\Delta E_{\text{incident}}}{e}$

$(\frac{hc}{{\lambda }_1}-\frac{hc}{{\lambda }_2})$

$=4.133-4.078$

$=5.5\times {10}^{-2}V$