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

Energy of a photon of radiation of wavelength 300 nm will be
`E = hv = h (c)/(lamda) = ((6.626 xx 10^(-34) Js) (3.0 xx 10^(8) ms^(-1)))/((300 xx 10^(-9) m)) = 6.626 xx 10^(-19) J`
`:.` Energy of 1 mole of photons `= (6.626 xx 10^(-19)J) xx (6.022 xx 10^(23) mol^(-1)) = 3.99 xx 10^(5) J mol^(-1)`
As `E = E_(0) +` K.E. of photoelectrons emitted.
`:.` Minimum energy `(E_(0))` required to remove 1 mole of electrons from sodium `= E -` K.E.
`= (3.99 - 1.68) 10^(5) J mol^(-1) = 2.31 xx 10^(5) J mol^(-1)`
`:.` Minimum energy required 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`
The wavelength corresponding to this energy can be calculated using the formula, `E = hv = h (c)/(lamda)`
`:. lamda = (hc)/(E) = ((6.626 xx 10^(-34) Js) (3.0 xx 0^(8) ms^(-1)))/(3.84 xx 10^(-19)J) = 5.17 xx 10^(-7) m = 517 xx 10^(-9) m = 517 nm`
which corresponds to the green light. As `lamda prop (1)/(E)`, hence when E is minimum, `lamda` is maximum