Question

Anode voltage is at $+3V$. Incident radiation has frequency 1.4$\times$10${}^{15}Hz$ and work function of the photo cathode is 2.8 $eV$. Find the minimum and maximum $KE$ of photo electrons in $eV$

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

Answer 1

The correct option is A 3, 6

Energy of the incident radiation is:
$hf=\frac{6.625\times {10}^{-34}\times 1.4\times {10}^{15}}{1.6\times {10}^{-19}}=5.8eV$

So, maximum KE of the emitted electron is given by:
$(KE{)}_{max}=hf-\varphi =5.8-2.8=3$

Since anode voltage is 3$V$, the electrons emitted with zero $KE$ will acquire an energy=3 $eV$ and the electrons emitted with 3 $eV$ will acquire $3+3=6$ $eV$
$\therefore$ Minimum $KE=$3 $eV$ and maximum $KE$$=6$ $eV$