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The minimum energy requried to overcome the attractive forces between an electron and the surface of Ag metal is `5.52xx10^(-19)`J. what will be the maximum kinetic energy of electrons ejected out from Ag which is being exposed to UV light of `lamda=360`Ã…?

Text Solution

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Energy of the photon absorbed
`=(h*c)/(lamda)=(6.625xx10^(-27)xx3xx10^(10))/(360xx10^(-8))`
`=5.52xx10^(-11)erg=5.52xx10^(-18)J`
E(photon) Work function+KE
`KE=5.52xx10^(-18)-5.52xx10^(-19)`
`=49.68xx10^(-19)J`.
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The minimum energy required to overcome tha attractive forces between an electron and the surface of Ag metal is 5.52 xx 10^(19)J . What will be the maximum kinetic energy of electron ejected out from Ag which is being exposed to UV-light of lambda = 360 A^(@) ?

The minimum energy required overcoming the attractive forces between electron and the surface of a Ag metalis 7.52xx10^(-19)J . What will be the maximum kinetic enerlgy of electron ejected out from Ag which is beign exposed to UV light of lamda=360Å .

Knowledge Check

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