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Question

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

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Solution

Given: The threshold frequency of the metal 3.3× 10 14 Hz, the frequency of light incident on the metal is 8.2× 10 14 Hz.

The cutoff voltage is given as,

V 0 = h( ν ν 0 ) e

Where, the threshold frequency of metal is ν 0 , the frequency of the light is ν, Planck’s constant is h, the cut off voltage is V 0 and the charge on an electron is e.

By substituting the given values in the above equation, we get V 0 = 6.626× 10 34 ( 8.2× 10 14 3.3× 10 14 ) 1.6× 10 19 = 6.626× 10 34 ( 4.9× 10 14 ) 1.6× 10 19 =2.03V

Thus, the cutoff voltage for the photoelectric emission is 2.03V.


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