The threshold frequency for a photosensitive metal is 3.3 10^14 Hz. If light of frequency 8.2 10^14 Hz is incident on this metal, the cut-off voltage for the photoelectric emission is nearly

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Correct answer: A. (a) According to Einstein's photoelectric equation eV_0= hv-hv_0 where, v= Incident frequency v_0= Threshold frequency V_0= Cut-off or stopping potential or V_0= he(v-v_0) Substituting the given values, we get V_0= 6.63 10^-34(8.2 10^14-3.3 10^14)1.6 10^-19 ≈2 V

$The threshold frequency for a photosensitive metal is 3.3 \times 10^{14} Hz . If light of frequency 8.2 \times 10^{14} Hz is incident on this metal, the cut-off voltage for the photoelectric emission is nearly$

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