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प्रश्न
UV light of wavelength 1800 Å is incident on a lithium surface whose threshold wavelength is 4965 Å. Determine the maximum energy of the electron emitted.
उत्तर
λ = 1800 × 10−10 m
λ0 = 4965 × 10−10m
h = 6.6 × 10−34 Js
c = 3 × 108 ms−1
Maximum kinetic energy of electron,
Kmax = `"hc" (1/λ - 1/λ_0)`
= `(6.6 xx 10^-34 xx 3 xx 10^8)/(10^-10) xx (1/1800 - 1/4965)`
= `19.8 xx 10^-16 xx 3165/(8937 xx 10^3)`
= 7.01208 × 10−19 J
= `(7.01208 xx 10^-19)/(1.6 xx 10^-19)`
= 4.38 eV
Kmax = 4.40 eV
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