A hydrogen atom makes a transition from `n=2` to `n=1` and emits a photon. This photon strikes a doubly ionized lithium atom `(Z=3)` in excited state and completely removes the orbiting electron. The least quantum number for the excited stated of the ion for the process is:

Options

A. 2
B. 3
C. 4
D. 5

Answer

C

Explanation

Incident energy of photon by H-atom
Transition take place from n=2 to n=1
`DeltaE=E_2-E_1`
=-3.4+13.6
`DeltaE`=10.2 eV
For emitting the electron from Lithium atom.
`DeltaE ge` The energy required to remove electron from Li-atom.
`DeltaE ge 13.6/n^2 Z^2`
`n^2 ge (13.6 Z^2)/(DeltaE)`eV
`n^2 ge (13.6xx(3)^2)/(10.2 eV)` eV
`n^2 ge 12`
`n ge 2 sqrt3`
`n ge 3.4`
So, n must be 4 which is greater than 3.4 .