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

The energy (E ) of a 300nm photon is given by
`= hv = (hc )/(lambda) = (6.626 xx 10^(-34) Js xx 3.0 xx 10^(8) ms^(-1))/(300 xx 10^(-9) m )`
The energy of one 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)`
`= 2.31 xx 10^(5) J mol^(-1)`
The minimum energy for one electron
`= (2.31 xx 10^(5) J "mol"^(-1))/(6.022 xx 10^(23) "electrons mol"^(-1))`
`=3.84 xx 10^(-19) J`
This corresponds to the wavelength
`therefore lambda = (hc )/( E)`
`= (6.626 xx 10^(-34) Js xx 3.0 xx 10^(8) ms^(-1))/(3.84 xx 10^(-19) J)`
= 517 nm this wavelength corresponds to green colour in visible spectrum.