Electron in a hydrogen-like atom `(Z = 3)` make transition from the forth excited state to the third excited state and from the third excited state to the second excited state. The resulting radiations are incident potential for photoelectrons ejested by shorter wavelength is `3.95 e V`.
Calculate the work function of the metal and stopping potiential for the photoelectrons ejected by the longer wavelength.

Explanation

Energy of the photon corresponding to transition from `n = 5` (forth excited state) to `n = 4` (third excited state) is
`h v = 13.6 (3) ^(2) [(1)/(4^(2)) - (1)/(5^(2))] = 2.75 e V` …(i)
Similarly, energy of the photon corresponding to transition from `n = 4` to `n = 3` is `h v = 13.6 (3) ^(2) [(1)/(3^(2)) - (1)/(4^(2))] = 5.95 e V` …(ii)
From Einstein's photoelectron effect equation, `h v = phi + KE_(max)`
`KE_(max) = e V_(s) = h v - phi`
The shorter wevelength corresponds to greater energy different between energy level involved in the transition . So, shorter wevelength photon are emitted for transition `n = 4 to n = 3`. Thus we have
`3.95 = 5.95 - phi implies phi = 2 e V`
and for longer wevelength photon , `e V_(s) = 2.75 - 2 = 0.75 e V`
So, stopping potential `= ((0.75 e V)/(e)) = 0.75 V`