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

the energy ( E) of a 30 nm photon is given by
`hc = hc//lambda`
`(6.626xx10^(-34)Jsxx3xx10^(8)ms^(-1))/(300xx10^(-9)m)`
`6.626 xx 10^(-19)j`
the energy of one mole of photons
`= 6.626zz10^(-19)Jxx6.022xx10^(23)mol^(-1)`
` 3.99xx10^(5)J mol^(-1)`
the minimum energy needed to remove one mole of electrons from sodium
`(3.99-1.68)10^(5)J mol^(-1)`
`2.31xx10^(5) J mol^(-1)`
the minimum energy for one electron `(2.31xx10^(5)J mol^(-1))/(6.022xx10^(23)"electrons mol"^(-1))`
`3.84xx10^(-19)J `
this corresponds to the wavelength
`lambda= (hc)/E`
`(6.626xx10^(-34)Jsxx3.0 xx 10^(8) ms^(-1))/(3.84xx10^(-19)J) `
517nm
( this corresponds to green light)