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प्रश्न
A monochromatic light source of intensity 5 mW emits 8 × 1015 photons per second. This light ejects photoelectrons from a metal surface. The stopping potential for this setup is 2.0 V. Calculate the work function of the metal.
(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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उत्तर
Given:-
Intensity of light, I = 5 mW
Number of photons emitted per second, n = 8 × 1015
Stopping potential, V0 = 2 V
Energy, `E = hv = I/n = (5 xx 10^-3)/(8 xx 10^15)`
From Einstein's photoelectric equation, work function,
`W_0 = hv - eV_0`
Here, h = Planck's constant
`e = 1.6 xx 10^-19 C`
On substituting the respective values, we get :-
`W_0 = (5 xx 10^-3)/(8 xx 10^15) - 1.6 xx 10^-19 xx 2`
`= 6.25 xx 10^-19 - 3.2 xx 10^-19`
`= 3.05 xx 10^-19`
`= (3.05 xx 10^-19)/(1.6 xx 10^-15) = 1.906 "eV"`
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