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 $\nu$ and the work function of the plate is $\Phi (h\nu >\Phi ).$ Assume 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 electrons is

• A. $\frac{IS}{h\nu }(\frac{2h}{\lambda }+\sqrt{2m(h\nu -\Phi )})$
• B. $\frac{2IS}{h\nu }(\frac{h}{\lambda }+\sqrt{2m(h\nu -\Phi )})$
• C. $\frac{IS}{h\nu }(\frac{h}{\lambda }+\sqrt{2m(h\nu -\Phi )})$
• D. $\frac{2IS}{h\nu }(\frac{h}{\lambda }+\sqrt{m(h\nu -\Phi )})$

Answer 1

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

The net force on the plate is due to incidence of photons $+$ due to emission of electrons. The number of photons incidents per second on the plate
$=$ number of electrons emitted per second (Given efficiency $=$ 100 %) $=\frac{IS}{h\nu }$

The momentum of each photon and electron respectively are $\frac{h}{\lambda }$ and $\sqrt{2m(h\nu -\Phi )}$ where m is the mass of the electron.

Hence the net force exerted on the metal plate is $\frac{IS}{h\nu }(\frac{h}{\lambda }+\sqrt{2m(h\nu -\Phi )}).$

So, the answer is option (C).