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Question

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 :

A
<v(43)1/2
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B
=v(43)1/2
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C
=v(34)1/2
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D
>v(43)1/2
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Solution

The correct option is D >v(43)1/2According to the Einstein's photoelectric equation, E=ϕ+K.Emax For case (1), hcλ=ϕ+12mv2−−−(1) For case (2), hcλ′=ϕ+12m(v′)2 ⇒hc(3λ4)=ϕ+12m(v′)2−−−(2) [(1)×43]−(2) 4hc3λ−43hcλ=43ϕ+43(12mv2)−ϕ−12m(v′)2 ⇒43ϕ+43(12mv2)=ϕ+12m(v′)2 ⇒12m(v′)2=ϕ3+4312mv2 ⇒12m(v′)2>43(12mv2) ⇒v′>√43 v Hence, option (D) is correct.

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