Consider a hydrogen-like atom whose energy in nth excited state is given by
`E_(n) = (13.6 Z^(2))/(n^(2))`
When this excited makes a transition from excited state to ground state , most energetic photons have energy
`E_(max) = 52.224 eV`. and least energetic photons have energy
`E_(max) = 1.224 eV`
Find the atomic number of atom and the intial state or excitation.

Options

A. `2`
B. `5`
C. `4`
D. None of these

Answer

A

Explanation

Maximum enegry is liberated for transition `E_(n) rarr 1` and minimum energy for `E_(n) rarr E_(n-1)`
Hence `(E_(1))/(n^(2)) - E_(1) = 52.224 eV ………(i)`
and `(E_(1))/(n^(2)) - (E_(1))/((n-1)^(2)) = 1.224eV …….(ii)`
Solving equations (i) and (ii), we get
`E_(1) =- 54.4eV` and `n = 5`
Now `E_(1)=- (13.6Z^(2))/(1^(2)) =- 54.4 eV` Hence `Z = 2`