Question

A parallel beam of light of intensity $I$ and cross section area $S$ is incident on a plate at normal incidence. The photoelectric emission efficiency is $100\%$, the frequency of beam is $v$ and the work function of the plate is $\varphi (hv>\varphi )$. Assuming all the electrons are ejected normal to the plane and with same maximum possible speed. The net force exerted on the plate only due to striking of photons and subsequent emission of electron is

• A. $\frac{IS}{hv}\left(\frac{2h}{\lambda }\sqrt{hv-\varphi }\right)$
• B. $\frac{2IS}{hv}(\frac{h}{\lambda }+\sqrt{2m(hv-\varphi )})$
• C. $\frac{IS}{hv}(\frac{h}{\lambda }+\sqrt{2m(hv-\varphi )})$
• D. $\frac{2IS}{hv}(\frac{h}{\lambda }+\sqrt{m(hv-\varphi )})$

Answer 1

The correct option is C $\frac{IS}{hv}(\frac{h}{\lambda }+\sqrt{2m(hv-\varphi )})$

$F_{net}=F_{photons}+F_{photelectrons}$
$=\frac{\Delta p_{photons}}{\Delta t}+\frac{\Delta p_{photoelectrons}}{\Delta t}$
$=p_{photon}\frac{\Delta n_{photon}}{\Delta t}+p_{photoelectron}\frac{\Delta n_{photoelectron}}{\Delta t}$ ----- (A)
As it is given that it has 100% efficiency we can say that
$p_{photon}\frac{\Delta n_{photon}}{\Delta t}=p_{photoelectron}\frac{\Delta n_{photoelectron}}{\Delta t}$
So
$\frac{\Delta n_{photon}}{\Delta t}=\frac{IS}{h\nu }=\frac{\Delta n_{photoelectron}}{\Delta t}$ ...(i)
$p_{photon}=\frac{h}{{\lambda }_{light}}$ ...(ii)
and
$p_{photoelectron}=\sqrt{2m_{e}K.E.}\Rightarrow p_{photoelectron}=\sqrt{2m_e(h\nu -\varphi )}$ ...(iii)

Putting (i),(ii) and (iii) in equation (A)
$F_{net}=\frac{h}{\lambda }\frac{IS}{h\nu }+\sqrt{2m_e(h\nu -\varphi )}\frac{IS}{h\nu }$
$F_{net}=\frac{IS}{h\nu }(\frac{h}{\lambda }+\sqrt{2m_e(h\nu -\varphi )})$