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प्रश्न
The electric field associated with a monochromatic beam is 1.2 × 1015 times per second. Find the maximum kinetic energy of the photoelectrons when this light falls on a metal surface whose work function is 2.0 eV.
(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 :-
Electric field of the monochromatic beam, `E = 1.2 xx 10^15` times per second
Frequency, v = `(1.2 xx 10^15)/2 = 0.6 xx 10^15 "Hz"`
Work function of the metal surface, `phi` = 2.0 eV
From Einstein's photoelectric equation, kinetic energy,
`K = hv - phi_0`
`⇒ K = (6.63 xx 10^-34 xx 0.6 xx 10^15)/(1.6 xx 10^-19) - 2`
`= 0.486 "eV"`
Thus, the maximum kinetic energy of a photon is 0.486 eV.
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