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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4140 Ao

5230 Ao

6250 Ao

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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