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The maximum kinetic energy of the photoe...

The maximum kinetic energy of the photoelectrons is found to be `6.63xx10^(-19)J`, when the metal is irradiated with a radiation of frequency `2xx10^(15)`Hz. The threshold frequency of the metal is about:

A

`1xx10^(15)s`

B

`1xx10^(15)s^(-1)`

C

`2.5xx10^(15)s^(-1)`

D

`4xx10^(15)s^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
B
`K.E_("max")=hv-hv_(0)`
`v-v_(0)=(K.E_("max"))/(h)=(6.63xx10^(-19)J)/(6.63xx10^(-34)"J. sec")=10^(15)"sec"^(-1)" "rArr" "v_(0)=(2-1)xx10^(15)Hz=10^(15)Hz`
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