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_(min) = 1.224 eV`
Find the atomic number of atom and the initial state or excitation.

Options

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

Answer

A

Explanation

Maximum energy is liberated for transition `E_(n)rarr1` and minimum energy for `E_(n)rarrE_(n-1)`
Hence `(E_(1))/(n^(2))-E_(1)=52.224eV " " ....(i)`
and `(E_(1))/(n^(2))-(E_(1))/((n-1))=1.224eV" " ....(ii)`
Solving Eqs. (i) and (ii), we get
`E_(1)=-54.4eV and n=5`
But `E_(1)=-(13.6Z^(2))/(1^(2))`
`:. -54.4=(13.)/(1^(2))Z^(2)rArrZ=2`