Class 12th Physics 2017 Set3 Delhi Board Paper Solution
Radiation of frequency 1015 Hz is incident on two photosensitive surfaces P and Q. There is no photoemission from surface P. Photoemission occurs from surface Q but photoelectrons have zero kinetic energy. Explain these observations and find the value of work function for surface Q.
h = c / f = 3 x 106 / 1015 = 300nm
Energy of rad. = 1240/300 = 4.13 eV
The work function of P is higher so there is no ejection from P.
Q emits electron so its work function is 1015j as there is 380 KE of emitted electron from Q. Work function of Q = 4.13eV
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