When electromagnetic radiaiton of wavelength `300 nm` falls on the surface of sodium electrons are emitted with a kinetic enegry of `1.68xx10^(5)Jmol^(-)`. What is the minimum enegry needed to remove an electorn from sodium?
Strategy: The minimum enegry required to remove an electron from target metal is called work function `W_(0)` of the metal. It can be calculated from Eq., provided we know the energy of the incident photon and kinetic enegry of a single photoelectorn.

Explanation

Step `1:` Calculating the enegry `(E)` of incident photon:
`E = hv = (hc)/(lambda)`
`=((6.6xx10^(-34)Js)(3.0xx10^(8)ms^(-1)))/((300xx10^(-9)m))`
`= 6.6 xx 10^(-19)J`
Step `2:` Calculating the kinetic enegry of a single photoelecton:
`KE = ((1.68xx10^(5)Jmol^(-1)))/(6.022xx10^(23)e mol^(-1))`
`0.279 xx 10^(18)Je^(-1)`
`= 2.79 xx 10^(-19) Je^(-1)`
Step `3:` Calculating the minimun energy needed to remove an electorn:
`hv = hv_(0) +kE`
`:. hv_(0) = hv - kE`
`=(6.6 xx 10^(19)J)-(2.79 xx 10^(-19)J)`
`= 3.81 xx 10^(-19) J`