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=hv−ϕ where v = frequency of incidence light New kinetic energy KE′=2hv−ϕ So, KE′=2hv−2ϕ+ϕ=2KE+ϕ So, the new kinetic energy is more than twice its initial value.