Question

The voltage at anode is $+3V$. Incident radiation has frequency $1.4\times {10}^{15}Hz$ and work function of the photocathode is $2.8eV$. The minimum and maximum kinetic energy of photo electrons in $eV$ respectively are

• A. $3,6$
• B. $0,3$
• C. $0,6$
• D. $2.8,5.8$

Answer 1

The correct option is A $3,6$

$PD=3V$ , V = $1.4\times {10}^{15}Hz$

work function $\varphi$ = 2.8 eV

In case, when an electron is just able to escape out of the metal, its KE = 0.

$\therefore$ it will be then accelerated by anode voltage

$\therefore$ its total $K{E}_{min}=3eV$

energy of radiation = $4.14\times {10}^{-15}\times 1.4\times {10}^{15}$

$=5.8eV$

$\varphi =2.8eV$

max KE of electron coming out of cathode = $KE=5.8-2.8=3eV$

It is then accelerated by anode voltage.

$K{E}_{max}=3+3$

$=6eV$