1.

`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.

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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