First slide

Photo Electric Effect

Question

In the following arrangement y = 1.0 mm, d = 0.24 mm and D = 1.2 m. The work function of the material of the emitter is 2.2 eV. The stopping potential V needed to stop the photo current will be

Difficult

A

0.9 V
B

0.5 V
C

0.4 V
D

0.1 V

Solution

As we know in Young's double slit experiment fringe width = separation between two consecutive fringe or dark fringes = $b = \frac{λD}{d}$ .
$\begin{array}{l} Here β = 2 y \Rightarrow \frac{λD}{d} \Rightarrow λ = \frac{2 yd}{D} \\ \Rightarrow λ = \frac{2 \times 1 \times 10^{- 3} \times 0 .24 \times 10^{- 3}}{1 .2} = 4 \times 10^{- 7} m = 4000 \overset{o}{A} \\ Energy of light incident on photo plate \\ E (eV) = \frac{12375}{4000} = 3 .1 eV \\ According to Einstein photoelectric equation, \\ E = W_{o} + {eV}_{o} \Rightarrow V_{o} = \frac{(E - W_{o})}{e} = \frac{(3 + 2 .2)}{e} eV \approx 0 .9 V \end{array}$

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