Electrons in hydrogen like atoms (Z = 3) make transitions form the fifth to the fourth orbit and from the fourth to the third orbit. The resulting radiations are incident normally on a metal plate and eject photoelectrons. The stopping potential for the metal and the stopping potential 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`