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प्रश्न
Find the maximum kinetic energy of the photoelectrons ejected when light of wavelength 350 nm is incident on a cesium surface. Work function of cesium = 1.9 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:-
Wavelength of light, λ = 350 nm = 350 × 10−9 m
Work-function of cesium, ϕ = 1.9 eV
From Einstein's photoelectric equation,
`E = phi` + Kinetic energy of electron
`⇒ K E = E - phi`
`⇒ K E = (hc)/λ - phi`,
Where λ = wavelength of light
h = Planck's constant
Maximum kinetic energy of electrons,
`E_max = (hc)/λ - phi`
`E_max = (6.63 xx 10^-34 xx 3 xx 10^8)/(350 xx 10^-9 xx 1.6 xx 10^-19) - 1.9`
`E_max = (6.63 xx 3 xx 10^2)/(350 xx 1.6) - 1.9`
`E_max = 1.65 "eV" = 1.6 "eV"`
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