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Question
Find the maximum magnitude of the linear momentum of a photoelectron emitted when a wavelength of 400 nm falls on a metal with work function 2.5 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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Solution
Given :-
wavelength of light , `λ = 400 "nm" = 400 xx 10^-9 "m"`
Work function of metal, `phi = 2.5 "eV"`
From Einstein's photoelectric equation,
Kinetic energy = `(hc)/λ - phi`
Here, c = speed of light
h = Planck's constant
`therefore K.E. = (6.63 xx 10^-34 xx 3 xx 10^8)/(4 xx 10^-7 xx 1.6 xx 10^-19) - 2.5 "eV"`
`= 0.605 "eV"`
Also , `K.E. = p^2/(2m)`
where p is momentum and m is the mass of an electron.
`therefore p^2 = 2"m" xx K.E.`
`⇒ p^2 = 2 xx 9.1 xx 10^-31 xx 0.605 xx 1.6 xx 10^-19`
`⇒ p = 4.197 xx 10^-25 "kg - m/s"`
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