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

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