Question

If $e/m$ of electron is $1.76 \times 10^{11}\, C \, kg^{-1}$ and the stopping potential is 0.71 V, then the maximum velocity of the photoelectron is

• A. $150 \, km \, s^{-1}$
• B. $200 \, km \, s^{-1}$
• C. $500 \, km \, s^{-1}$
• D. $250 \, km \, s^{-1}$

Answer 1

Answer: (C)

Here, stopping potential, $V_S=0.71 V$
For electron, $\frac{e}{m}=1 \cdot 76 \times 10^{11} C kg ^{-1}$
The maximum kinetic energy of the photoelectrons is
$K_{\max }=\frac{1}{2} m v_{\text {max }}^2=e V_s$
$\therefore v_{\text {max }}^2=2 \frac{e}{m} V_s$
$v_{\max }=\sqrt{2 \frac{e}{m} V_s}=\sqrt{2 \times 1.76 \times 10^{11} \times 0.71}$
$=5 \times 10^5 ms ^{-1}=500 \times 10^3 m s ^{-1} $
$=500 km s ^{-1}$