Question

A light of wavelength $3000\stackrel{o}{A}$ falls on a metal surface. Ejected $e^{-}$ is further accelerated by a potential difference of $2V$ , then final K.E of the $e^{-}$ is found to be $8\times 10^{-19}J$ . If threshold energy for the metal surface is $\text{’}{\text{ϕ}}^{{}^{\prime }}eV$ . Then find the numerical value of $8\varphi$

Answer 1

Answer: 9.00

${\text{KE}}_{\text{final}}\text{=}{\text{E}}_{\text{photon}}\text{+}{\text{E}}_{\text{potential}}-\varphi$

where $\varphi$ = threshold energy for metal surface

${\text{E}}_{\text{photon}}\text{=}\frac{\text{hc}}{\lambda }=\frac{6.6\times 10^{-34}\times 3\times 10^8}{3000\times 10^{-10}}$

$\sim eq6.6\times 10^{-19}\text{J}=4.12\text{eV}$

${\text{E}}_{\text{potential}}\text{=}\text{2}\text{eV}$

${\text{KE}}_{\text{final}}=\frac{8\times 10^{-19}\text{J}}{1.6\times 10^{-19}\text{J}}=5\text{eV}$

Hence, $\varphi =(4.12+2-5)\text{eV}=1.125\text{eV}$

Numerical value = $8\times 1.125=9$