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# Radiation of wavelength λ is incident on a photocell. The fastest emitted electron has a speed v. If the wavelength is changed to 3λ4, the speed of the fastest emitted electron will be :v>(43)1/2v<(43)1/2v=(43)1/2v=(34)1/2

A
v>(43)1/2
B
v<(43)1/2
C
v=(43)1/2
D
v=(34)1/2
Solution
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#### We know,hcλ−ϕ=12mv2⟶(i)4hc3λ−ϕ=12mv2⟶(ii)(ii) - (i) giveshc3λ=12m(v2−v1)⇒v1=√v2+2hc3λm⟶(iii)also from (i) hcλ=ϕ+mv22⇒2hcλm=2ϕmv2⇒2hc3λm=2ϕ3m+r23>v23 ⟶(iv)v>√4v23

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