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Question
In the arrangement shown in the figure, 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. Find the stopping potential V needed to stop the photocurrent.
(Use h = 6.63 × 10-34J-s = 4.14 × 10-15 eV-s, c = 3 × 108 m/s and me = 9.1 × 10-31kg)
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Solution
Given :-
Fringe width, y = 1 mm `xx` 2 = 2 mm
Work function, W0 = 2.2 eV
D = 1.2 m
d = 0.24 mm
Fringe width,
`y = (lambdaD)/d`,
where `lambda` = wavelength of light
`therefore lambda = (2 xx 10^-3 xx 0.24 xx 10^-3)/1.2`
`= 4 xx 10^-7 "m"`
Energy , `E = (hc)/lambda`
`= (4.14 xx 10^-15 xx 3 xx 10^8)/(4 xx 10^-7)`
= 3.105 `"eV"`
From Einstein's photoelectric equation,
`eV_0 = E - W_0`
where V0 is the stopping potential and e is charge of electron .
`therefore eV_0 = 3.105 - 2.2 = 0.905 "eV"`
`V_0 = 0.905/(1.6 xx 10^-19) xx 1.6 xx 10^-19 V`
= 0.905 V
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