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Q.

In a photo electric experiment, the work function of the illuminated metal surface is equal to half of the energy of individual photon of the light used and stopping potential is found to be 1.5 volt. Then wavelength of the light used is

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a

4140 Ao

b

5230 Ao

c

6250 Ao

d

5500 Ao

answer is A.

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Detailed Solution

hv=ϕ+kmax⁡⇒hv=12hv+kmax⁡∴ kmax⁡=hv2=hc2λ, But kmax⁡=ev0∴ hc2λ=ev0⇒λ=hc2(ev0)=124202×1.5  Ao⇒λ=4140 Ao
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