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When a certain metal was irradiated wit...

When a certain metal was irradiated with light of frequency `3.2 xx 10^(16)s^(-1)` the photoelectrons emitted had twice the KE as did photoelectrons emitted when the same metal was irradiated with light of frequency`2.0 xx 10^(16)s^(-1)` . Calculate the thereshold frequency of the metal.

Text Solution

Verified by Experts

Applying photoelectric equation,
`KE=hv-hv_(0)`
or `(v-v_(0))=(KE)/(h)`
Given `KE_(2)=2KE_(1)`
`v_(2)-v_(0)=(KE^(2))/(h)` . . . (i)
and `v_(1)-v_(0)=(KE_(1))/(h)` . . .(ii)
Dividing equation (i) by equation (ii),
`(v_(2)-v_(0))/(v_(1)-v_(0))=(KE_(2))/(KE_(1))=(2KE_(1))/(KE_(1))=2`
or `v_(2)-v_(0)=2v_(1)-2v_(0)`
or `v_(0)=2v_(1)v_(2)=2(2.0xx10^(16))-(3.2xx10^(16))`
`=8.0xx10^(15)Hz`.
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