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

For n=5 to n=4,
`DeltaE_(1)=13.6 Z^(2)((1)/(4^(2))-(1)/(5^(2)))`
`=13.6 (3)^(2)((25-16)/(16xx25))`
=2.75 eV
n=4 to n=3,
`DeltaE_(2)=13.6(3)^(2)((1)/(3^(2))-(1)/(4^(2)))`
`=13.6(3)^(2)((16-9)/(9xx16))`
=5.95 eV
For n=4 to n=3, wavelenth is smaller.
For n=4 to n=3,
`5.95=phi+3.95 implies phi=2 eV`
For n=5 to n=4
`2.75=2+eV_(s) implies V_(s)=0.75 V`