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Question

If the frequency of incident light falling on a photosensitive metal is doubled, the kinetic energy of the emitted photoelectron

A
unchanged
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B
halved
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C
doubled
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D
more than twice its initial value
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E
reduced to 14th
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Solution

The correct option is C more than twice its initial value
By Einstein's photoelectric equation, The kinetic energy of photoelectrons
KE=hvϕ
where v = frequency of incidence light
New kinetic energy KE=2hvϕ
So, KE=2hv2ϕ+ϕ=2KE+ϕ
So, the new kinetic energy is more than twice its initial value.

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