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Calculate the velocity of a photo-electr...

Calculate the velocity of a photo-electron if the work function of the target material is `1.24eV` and the wavelength of incident light is `4.36xx10^(-7)`m. What retarding potential is necessary to stop the emission of the electrons?

Text Solution

Verified by Experts

As `KE_(max) = hv + phi rArr 1/2 mv_(max)^(2) = hv - phi = (hc)/(lambda) - phi`
`v_(max) = sqrt((2(sqrt(hc)/(lambda)-phi))/(m)) = sqrt(2(((6.63 xx 10^(-34) xx 3 xx 10^(8))/(4.36 xx 10^(-7)) - 1.24 xx 1.6 xx 10^(19)))/(9.11 xx 10^(-31))) = 7.523 xx 10^(5) m//s`
`:.` The speed of a photoelectron acan be any value between `0` and `7.43 xx 10^(5) m//s`
If `V_(0)` is the stopping potential, then `eV_(0) = 1/2 mv_(max)^(2)`
`rArr V_(0) = 1/2 (mv_(max)^(2))/(e) = (hc)/(elambda) - phi/e = (2400)/(4360) - 1.24`
`= 1.60 V [:' (hc)/(e) = 12400 xx 10^(-10)V-m]`
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Knowledge Check

  • Work function of metal is phi wavelength of incident light is lambda .Which condition should be followed for not emitting the photo-electron?

    A
    `lambdagt(hc)/(phi)`
    B
    `lambda=(hc)/(phi)`
    C
    `lambdalt(hc)/(phi)`
    D
    `lambdale(hc)/(phi)`
  • The work function of a metal surface is 4.2 eV. The maximum wavelength which can eject photo-electron from this surface is =…….Å

    A
    2956
    B
    3076
    C
    4116
    D
    5088
  • The work function of caesium is 2.14 eV.Find (a)the threshold frequency for caesium,and (b)the wavelength of the incident light if the photocurrent is brought to zero by a stopping potential of 0.60 V.

    A
    `v_(0)=4.54xx10^(14)Hz,lambda=454nm`
    B
    `v_(0)=5.16xx10^(14)Hz,lambda=516nm`
    C
    `v_(0)=5.16xx10^(14)Hz,lambda=454nm`
    D
    `V_(0)5.16xx10^(14)Hz,lambda=414 nm`
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