Question

When light of frequency $2.2\times {10}^{15}Hz$ is incident on a metal surface, photoelectrons emitted can be stopped by a retarding potential of $6.6V$. For light of frequency $4.6\times {10}^{15}Hz$, the reverse potential is $16.5V$. Find $h$

Answer 1

We know that-

$eV=h\nu -\varphi$

$⟹e(V_2-V_1)=h({\nu }_1-{\nu }_2)$

$⟹h=\frac{e(V_2-V_1)}{{\nu }_2-{\nu }_1}$

$⟹h=\frac{1.6\times {10}^{-19}\times 9.9}{2.4\times {10}^{15}}$

$⟹h=6.6\times {10}^{-34}Js$