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The threshold frequency of a certain met...

The threshold frequency of a certain metal is `3.3xx10^(14)Hz`. If light of frequency `8.2xx10^(14)Hz` is incident on the metal, predict the cut off voltage for photoelectric emission. Given Planck's constant, `h=6.62xx10^(-34)Js`.

A

1V

B

2V

C

3V

D

5V

Text Solution

Verified by Experts

The correct Answer is:
B
`V_0=(E_(ph)-W)/(e)=(h(v-v_0))/(e)`
= `(6.62xx10^-34(8.2xx10^14-3.3xx10^14))/(1.6xx10^-19)`
`=(6.62xx10^-34)/(1.6)xx4.9xx10^(14+19)`
`=(6.62xx4.9)/(1.6)xx10^-1`=2volt
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