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Q.

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
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a

0.9 V

b

0.5 V

c

0.4 V

d

0.1 V

answer is A.

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Detailed Solution

As we know in Young's double slit experiment fringe width = separation between two consecutive fringe or dark fringes = \beta = \frac{{\lambda D}}{d}
Here\,\beta = 2y\, \Rightarrow 2y = \frac{{\lambda D}}{d} \Rightarrow \lambda = \frac{{2yd}}{D}
\Rightarrow \lambda = \frac{{2 \times 1 \times {{10}^{ - 3}} \times 0.24 \times {{10}^{ - 3}}}}{{1.2}} = 4 \times {10^{ - 7}}m = 4000\,{\AA}
Energy of light incident on photo plate
E\,(eV) = \frac{{12375}}{{4000}} = 3.1\,eV
According to Einstein photoelectric equation
E = W0 + eV0   \Rightarrow {V_0} = \frac{{(E - {W_0})}}{e} = \frac{{(3 + 2.2)}}{e}eV\, \approx 0.9\,V

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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