The minimum energy required to overcome the attractive force between an electron and the surface of Ag metal is 7.52×10−19J . What will be the maximum kinetic energy of electrons ejected out from Ag which is being exposed to UV light of λ=360∘A.
A
47.68×1019J
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B
47.68×10−19J
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C
47.68×10−18J
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D
47.68×1018J
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Solution
The correct option is B47.68×10−19J We know, from photoelectric effect, Kinetic energy (K.E.) of photoelectron = Energy of photon - Work function K.E=hν−ϕ = hcλ−ϕ Given: Work function = 7.52×10−19J where h = Planck's constant ν = incident frequency Put up the values, K.E = 6.626×10−34×3×108360×10−10−7.52×10−19 On solving, K.E = 47.68×10−19 J