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The fringe width in a YDSE is 2mm, distance between slits and screen is 1.2 m and separation between the slits is 0.24 mm. The radiation of same source is also incident on a photo cathode of work- function 2.2 eV. The stopping potential is

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0.9 V

3.1 V

2.2 V

5.3 V

answer is C.

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Detailed Solution

$β = 2 m m, D = 1.2 m$

$d = 0.24 m m$

$\frac{λ D}{d} = β$

$λ = \frac{2 \times 10^{- 3} \times 24 \times 10^{- 5}}{12 \times 10^{- 1}}$

$λ = 4 \times 10^{- 7} m$

$e v_{0} = \frac{12400}{4000} - 2.2 = 0.9 e v$

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