1.

The energy required to stop ejection of electrons from a Cu plate is `0.24` eV.Calculate the work function Cu when a radiation of wavelength `lambda = 250` nm strikes the plate.A. 24.3 eVB. 24 eVC. 4.65 eVD. 4.95 eV

Answer» Correct Answer - B
Energy of Photon =work function + K.E. =work function + `eV_0`…(1)
where e=electronic charge
`V_0`=Stopping potential and `eV_0` is equal to energy required to stop ejection of electron.
E/photon =hc/`lambda`
`=(6.626xx10^(-34)xx3xx10^8)/(253.7xx10^(-9))`
`=7.835xx10^(-19)J`
`=(7.835xx10^(-19))/(1.602xx10^(-9))`eV
=4.89 eV
So, from (1),
4.89=Work function + 0.24
work function = 4.65 eV


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