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.

Options

A. `2.0 V`
B. `0.75 V`
C. `0.6 V`
D. none of these

Answer

B

Explanation

`Delta E_(1) , (for 4^(th) to 3^(rd)` excited state)
`= 13.6 xx 3^(2) [(1)/(4^(2)) - (1)/(5^(2))] = 2.75 eV`
`Delta E_(2) , (for 3^(rd) to 2^(nd)` excited state)
`= 13.6 xx 3^(2) [(1)/(3^(2)) - (1)/(4^(2))] = 5.95 eV`
for shorter wavelength i.e., for `Delta E_(2), V_(0 1) = 3.95 volt From eV_(0 1) = h c- phi `.
`3.95 = 5.95 - phi`
`:. phi = 2 eV`
For longer wavelength ,
`eV_(0 2) = 2.75 - 2 = 0.75 eV`
So, `V_(0 1) = 0.75 V`