First slide

Photo Electric Effect

Question

The frequency of incident light falling on a photosensitive metal plate is doubled, the kinetic energy of the emitted photo-electron is

Easy

A

double the earlier value
B

unchanged
C

more than doubled
D

less than doubled

Solution

$hv = W + \frac{1}{2} {mv}^{2} or (K.E.) = h \lor - W Now (K.E.)^{'} = 2 hv - W > 2 (hv - W)$

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