Question

When a beam of $10.6\text{eV}$ photons of intensity falls on a platinum surface of area $1.0\times {10}^{-4}{\text{m}}^2$ and work function $5.6\text{eV}.$ If $53\%$ of the incident photon eject photoelectrons. Total photoelectrons emitting per second is $n\times {10}^{11}$. Find $n$(integer only)

Answer 1

Given, $E=10.6\text{eV};I=2.0{\text{W m}}^{-2}\varphi =5.6\text{eV};A=1.0\times {10}^{-4}{\text{m}}^2\eta =53\%$

The energy incident on the platinum surface per second is,

$\text{Power}=IA=2.0\times 1\times {10}^{-4}=2\times {10}^{-4}{\text{J s}}^{-1}$

The energy of each incident photon is,

$E=10.6\text{eV}=10.6\times 1.6\times {10}^{-19}\text{J}$

The number of photon incident on the surface per second is,

$n=\frac{\text{Power}}{\text{Energy of a photon}}$

$=\frac{2\times {10}^{-4}}{10.6\times 1.6\times {10}^{-19}}$

As only $53\%$ of incident photons can eject electrons, thus photo electrons ejected per second is,

$\text{Photon flux}\left(\eta \right)=\frac{\text{No.of photoelectrons}\left(n_e\right)}{\text{No.of photons}\left(n\right)}$

$n_e=0.53\left[\frac{2\times {10}^{-4}}{10.6\times 1.6\times {10}^{-19}}\right]=0.0625\times {10}^{15}$

$n_e=625\times {10}^{11}$

$\therefore n=625$