A gas of identical hydrogen-like atoms has some atoms in the lowest in lower (ground) energy level `A` and some atoms in a partical upper (excited) energy level `B` and there are no atoms in any other energy level.The atoms of the gas make transition to higher energy level by absorbing monochromatic light of photon energy `2.7 e V`.
Subsequenty , the atom emit radiation of only six different photon energies. Some of the emitted photons have energy `2.7 e V` some have energy more , and some have less than `2.7 e V`.
a Find the principal quantum number of the intially excited level `B`
b Find the ionization energy for the gas atoms.
c Find the maximum and the minimum energies of the emitted photons.

Explanation

The energy level of identical hydrogen-like atom are given by
`E_(n) = - (B)/(n^(2)), B` being a constant.
a. When a hydrogen-like atom absorbs energy `2.7 e V` This is only possible if transion of electron is from state `B` to any higher energy state. Six radiations are emitted only ?If the final state is `n = 4`.
The energy of emitted radition is also more than `2.7 e V`, therefore , initial state connot be `n = 3`. Hence , the only possibility is that initially state `B` has `n = 2`.
That is the principal quantum number of initially excited state `B` is `n = 2`
b. Given `E_(4) - E_(2) = 2.7 e V`
`:. - (B)/(4^(2)) -(- (B)/(2^(2))) = 2.7 e V`
Therefore , this gives `B = 14.4 e V`
the transition of electron should be from ground state `(n = 1)` to `n = oo`.
Lonization energy of identical hydrogen-like gas atoms
`Delta E = E_(1) = - (B)/(oo^(2)) - (- (B)/(1^(2)))`
c. Maximum energy of emitted photon is obtained when transition of electron is from `n = 4` to `n = 1`.
`:. E_(max) = - (B)/(4^(2)) - (- (B)/(1^(2)))`
Maximum energy of emitted photon is obtained when transition of electron is from `n = 4 to n = 3`.
i.e. ,` E_(max) = B ((1)/(3^(2)) - (1)/(4^(2))) = 14.4 (1/9 - (1)/(16)) = 0.4 e V`