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Q.

Radiation of wavelength λ , is incident on a photocell. The fastest emitted electron has speed v  . If the wavelength is changed to 3λ4,  the speed of the fastest emitted electron will be: 

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a

=v4312

b

=v3412

c

>v4312

d

<v4312

answer is C.

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Detailed Solution

hcλ−hv0=12mv2 ∴            43hcλ−hv0=12mv'2 ∴          v'2v2=43v−v0v−v0∴v'=v43v−v0v−v0  ∴    v'>v43

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