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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3 V

2 V

1 V

5 V

answer is A.

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Detailed Solution

Using Einstein's photoelectricity equation,

${(KE)}_{max} = hν - {hν}_{threshold} = {eV}_{0} (V_{0} = cut-off voltage)$

$\Rightarrow V_{0} = \frac{h}{e} (8.2 \times 10^{14} - 3.3 \times 10^{14})$

$= \frac{(6.6 \times 10^{- 34}) \times (4.9 \times 10^{14})}{(1.6 \times 10^{- 19})} \approx 2 V$

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