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When a 6000Å light falls on the cathode of a photo cell and produced photoemission. If a stopping potential of 0.8 V is required to stop emission of electron, then determine the net energy of the electron after it leaves the surface

Answer»

Solution :Given data `:`
WAVELENGTH of incident light `lambda =6000Å = 6 xx 10^(-7) m `
Stopping POTENTIAL `V_(0) =0.8 V`
To find `:`
(i) FREQUENCY of light v = ?
(ii) Energy of incident photon E =?
(iii) Work function of the cathode material W= ?
(iv) Threshold frequency `v_(0) = ?`
(v)Net energy of the electron after it LEAVES the surface E = ?
Net energy of the photo electrons `E =hv - hv_(0) = ( 2.07 -1.26) eV`
`E = 0.81 eV`


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