When a certain metal was irradiated with light of frequency `1.6 xx 10^(16) Hz` , kinetic energy of the photoelectron emitted was twice the kinetic energy of the photoelectron emitted when the same metal was irradiated with light of frequency `1.0 xx 10^(16) Hz` . Calculate the threshold frequency `(nu_(0))` for the metal.

Answer

4

Explanation

By photoelectric effect `hv=hv_(0)+KE`
`:. KE_(1)=h(v_(1)-v_(0)) …(1)`
`KE_(2)=h(v_(2)-v_(0))=KE_(1)//2 …(2)`
Dividing equation (2) by (1) we have `(v^(2)-v_(0))/(v_(1)-v_(0))=1/2`
`(1.0xx10^(16))/(1.6xx10^(16)-v_(0))=1/2 2.0 xx10^(16)-2v_(0)=1.6xx10^(16)-v_(0)`
`v_(0)=4xx10^(15) Hz`