When electromagnetic radiation of wavelength 300 nm falls on the surface of sodium, electrons are emitted with kinetic energy of `1.68 xx 10^(5) J "ml"^(-1)`. What is the minimum energy needed to remove an electron from sodium ? What is the maximum wavelength that will cause a photoelectron to be emitted.

Explanation

Wavelength of photon `(lambda)=300nm=300xx 10^(-9)m`
Energy of photon (hv) `=(hc)/(lambda)=((6.626 xx 10^(-34)Js) xx (3xx10^(8) ms^(-1)))/((300 xx 10^(-9)m))=6.626 xx 10^(-19)J`
Energy of one mole of photons (E)`=(6.626 xx 10 ^(-19)J) xx (6.022 xx 10^(23) "mol" ^(-1))`
`=3.99 xx 10^(5)J "mol"^(-1)`
Kinetic energy (K.E.)`=1.68 xx 10^(5) J"mol"^(-1)`
The minimum energy needed or threshold energy `hv^(@)` may be calculated as follows :
`hv^(@)=hv-"K.E."`
`=(3.99xx10^(5)-1.68 xx 10^(5)) J "mol"^(-1)`
`=2.31 xx 10^(5)J "mol"^(-1)`
Minimum energy needed to remove one electron `=((2.31 xx 10^(5) J "mol"^(-1)))/((6.022 xx 10^(23) "mol"^(-1)) )=3.84 xx 10^(-19)J`
Wavelength corresponding to this energy `(lambda)=(hc)/(E)=((6.626 xx 10^(-34)Js)xx(3 xx 10^(8) ms^(-1)))/((3.84 xx 10^(-19)J))`
`=5.17 xx 10^(-7)m=517m`