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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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Knowledge Check

  • The minimum energy required for the emission of photoelectron from the surface of a metal is 4.95 xx10^(-19)J . Calculate the critical frequency of the photon required to eject the electron . h = 6.6 xx10^(-34) J sec

    A
    `7.5xx10^(14)s^(-1)`
    B
    `7.5xx10^(13)s^(-1)`
    C
    `7.5xx10^(16)s^(-1)`
    D
    `7.5xx10^(19)s^(-1)`

  • If an electron in a hydrogen atom is moving with a kinetic energy of 5.45 xx10^(-19) j then what will be the energy level for this electron ?

    A
    1
    B
    2
    C
    3
    D
    6

  • The energy required to remove an electron from metal X is E = 3.31 xx 10^(-20)J . Calculate the maximum wavelength of light that can be photoeject an electron from metal X :

    A
    `4 mum`
    B
    `6 mum`
    C
    `7 mum`
    D
    `5 mum`