For silver metal `mu_(0)` is `1.13 xx 10^(17)s^(-1)` .What is the maximum energy of the photoelectron produced by shiniong atraviolet light wavelength `1.5` nm on the metal by shining liught wavelength `1.5 nm` on the metal .

Explanation

`hv = hv_(0) + KE`
`KE = hv - hv_(0)`
`hc/(lambda) -hv_(0)`
`:. KE = ((6.626 xx 10^(-34) Js) xx 3 xx 10^(8) ms^(-1))/(1.5 xx 10^(-9) m)`
`-(6.626 xx 10^(-34) Js)(1.13 xx 10^(17) s^(-1))`

`= 1.325 xx 10^(-16) J - 7.487 xx 10^(-17) J`
`= 0.576 xx 10^(-16) J`