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Question

If the frequency of the incident radiation is increased from 4×1015Hz to 8×1015Hz, by how much will the stopping potential for a given photosensitive surface go up?

Solution
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eV0=hvhv0
=h(vv0)
=6.6×1034(8×10154×1015)
eV0=2.64×1018
V0=2.64×10181.6×1019
V0=16.5volt

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