First slide

Photo Electric Effect

Question

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

Moderate

A

30 volt
B

45 volt
C

59 volt
D

Information is insufficient

Solution

$\begin{array}{l} v_{\max} = 4 \times 10^{8} cm / \sec = 4 \times 10^{6} m / \sec . \\ ∴ K_{\max} = \frac{1}{2} {mv}_{\max}^{2} = \frac{1}{2} \times 9 \times 10^{- 31} \times {(4 \times 10^{6})}^{2} \\ = 7 .2 \times 10^{- 18} J = 45 eV . \\ Hence, stopping potential |V_{o}| = \frac{K_{\max}}{e} = \frac{45 eV}{e} = 45 volt . \end{array}$

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