Question

If maximum velocity with which an electron can be emitted from a photo cell is $4 \times 10^{8} cm / \sec$, the stopping potential is (mass of electron $=9 \times 10^{-31} kg$ )

• A. 30 volt
• B. 45 volt
• C. 59 volt
• D. Information is insufficient

Answer 1

Answer: (B)

$v_{\max }=4 \times 10^{8} cm / \sec =4 \times 10^{6} m / \sec$
$\therefore K_{\max }=\frac{1}{2} m v_{\max }^{2}$
$=\frac{1}{2} \times 9 \times 10^{-31} \times\left(4 \times 10^{6}\right)^{2}$
$=7.2 \times 10^{-18} J =45\, eV$
Hence, stopping potential
$\left |V_{0}\right|=\frac{K_{\max }}{e}$
$=\frac{45\, e V}{e}=45$ volt .