Q.

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

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a

>v4312

b

=v4312

c

=v3512

d

<v4312

answer is B.

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Detailed Solution

Initially,  12mv2=hcλ-ϕ0 When wavelength changes to 3λ/4,           12mv'2=4hc3λ-ϕ0   12mv'2=4hc3λ-4ϕ03+ϕ03   12mv'2=43hcλ-ϕ0+ϕ03

   12mv'2=4312mv2+ϕ03

Since  ϕ03>0,

     12mv'2>4312mv2 

   v'>v4312

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