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Question

The minimum energy required to overcome the attractive forces between an electron and the surface of Ag metal is 552×1019J. What will be the maximum kinetic energy of electrons ejected out from Ag which is being exposed to UV light of λ=360˚A?

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Solution

Energy of the photon absorbed =hcλ=6.625×1027×3×1010360×108
=5.52×1011erg=5.52×1018J
E(photon)= Work function +KE
KE=5.52×10185.52×1019
=49.68×1019J.

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