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.

Answer

(i) `4` , (ii) `23.04 xx 10^(-19)J`
(iii) `E_(max) = 13.5 eV, E_(min) = 0.7 eV`

Explanation

Let `n =` no. of level of excited state
`rArr nC_(2) = 6` (spectral lines)
`rArr n= 3` (3rd excited states)
`:.` No. of excited state level `= n+ 1 = 4`

(i) Principle quantum no. of initially excited level `B = 4`
(ii) `2.7 = k((1)/(4) - (1)/(16))`
`rArr k = (16)/(3) xx 2.7 = 14.4 eV`
`:.` Ionisation energy
`= k (1 - (1)/(oo)) = k`
`= 14.4eV`
`= 23.04 xx 10^(-19)J`
(iii) `DeltaE_("max") = 4 rarr 1 = 13.5 eV`
`DeltaE_("min") = 4 rarr 3 = 0.7 eV`