Chemistry - Atomic Structure Question with Solution | TestHub

${\text{KE}}_{\text{final}}\text{=}{\text{E}}_{\text{photon}}\text{+}{\text{E}}_{\text{potential}}-\varphi$
where$\varphi$= threshold energy for metal surface
${\text{E}}_{\text{photon}}\text{=}\frac{\text{hc}}{\text{λ}}=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{3000\times {10}^{-10}}$
$\simeq 6.6\times {10}^{-19}\text{J}=4.12\text{eV}$
${\text{E}}_{\text{potential}}\text{=}\text{2}\text{eV}$
${\text{KE}}_{\text{final}}=\frac{8\times {10}^{-19}\text{J}}{1.6\times {10}^{-19}\text{J}}=5\text{eV}$
Hence,$\varphi =(4.12+2-5)\text{eV}=1.125\text{eV}$
Numerical value =$8\times 1.125=9$