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Q.

The threshold frequency for a photosensitive metal is 3.3×1014 Hz. If light of frequency 8.2×1014 Hz is incident on this metal, the cut-off voltage for the photoelectric emission is nearly 

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a

3 V

b

2 V

c

1 V

d

5 V

answer is A.

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Detailed Solution

Using Einstein's photoelectricity equation,

KEmax=hν-hνthreshold =eV0    V0=cut-off voltage 

⇒   V0=he8.2×1014-3.3×1014

            =6.6×10-34×4.9×10141.6×10-19≈2 V

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