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On shining light of wavelength 6.2xx10^(-6) m on a metal surface photo-electrons are emitted. The work function of the metal is 0.1 eV. Find the kinetic energy of a photo-electron (in eV)

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Solution :Here, `lambda=6.2xx10^(-6) m , phi_0`= 0.1 EV
Energy of the incident photon , `E=hupsilon=(hc)/lambda`
`E=((6.6xx10^(-34))(3xx10^8))/(6.2xx10^(-6))`J
`=(6.6xx3xx10^(-26))/((6.2xx10^(-6))(1.6xx10^(-19)))` eV=0.2 eV
As `E=K+phi_0 , K=E- phi_0`=0.2 eV - 0.1 eV =0.1 eV


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