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Q.

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

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a

2.0 V

b

4.04 V

c

5.04 V

d

6.04 V

answer is B.

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Detailed Solution

${E=W+eV}_{S} \Rightarrow V_{S} = \frac{h ν - h ν_{0}}{e} = \frac{h (ν - ν_{0})}{e} = 4 .04 V$

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