Question

A light of wavelength $3000 \, \overset{o}{\text{A}}$ 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 $\text{'}\text{ϕ}^{'}eV$ . Then find the numerical value of $8\phi$

Answer 1

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

where $\phi$ = threshold energy for metal surface

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

$\sim eq 6.6 \times 10^{- 19} \text{J} = 4.12 \, \text{eV}$

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

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

Hence, $\phi = \left(\right. 4.12 + 2 - 5 \left.\right) \, \text{eV} = 1.125 \, \text{eV}$

Numerical value = $8 \times 1.125 = 9$