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`A` photon of wavelenth `3000` A strikes `a` metal surface, the work function of the metal being `2.20eV`. Calculate `(i)` the energy of the photon in `eV(ii)` the kinetic energy of the emitted photo electron and `(iii)` the velocity of the photo electron. |
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Answer» (i) Energy of the photon `E=hv=(hc)/(lambda)=((6.6xx10^(-34)Js)(3xx10^(8)ms^(-1)))/(3xx10^(-7)m)=6.6xx10^(-19)J` `1eV = 1.6xx10^(-19)J` Therefore `E =(6.6xx10^(-19)J)/(1.6xx10^(-19)J//eV)=4.125eV` (ii) Kinetic energy of the emitted photo electon `"Work function" " "=2.20 eV` Therefore, `KE " "=2.475-2.20` `" "=1.925eV=3.08xx10^(-19)J` `(iii)` Velocity of the photo electron `KE=(1)/(2) mv^(2)=3.08xx10^(-19)J` `"Therefore"`, velocity `(v) = sqrt((2xx3.08xx10^(-19))/(9.1xx10^(-31)))=8.22xx10^(5)ms^(-1)` |
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