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

Level C must correspond to n=4 since the total number of possible transitions from level 4 to lower levels is 1+2+3+4 =6.The level corresponds either to n=2or n=3.The energies of all the photons involved in the transitions are (in eV):`E_(4),_(2)=13.6 Z^(2)((1)/(2^(2))-(1)/(4^(2)))=2.55Z^(2)`,`E_(4),_(3)=0.66Z^(2)`,`E_(4),_(1)=12.75Z^(2)`,`E_(3),_(2)=1.89Z^(2)`,`E_(3),_(1)=12.1Z^(2)`,`E_(2),_(1)=10.2Z^(2)`.The only possible choices are (a) Z=1 and `E_(4),_(2)=2.55ZeV approx 2.7eV`(b) Z=2 ,`E_(4),_(3)=2.64ZeV approx 2.7eV`.The choice Z=2 however is consistent with the fact that some products have energy less than 2.7 eV.(i) Thus, Z=1 and the quantum number of level B is n=2.The ionisation energy is ,`13.6eVxx1^(2)=13.6eV`.(iii) The maximum energy of the photons is `12.75 eV (E_(4),_(1)) while the minimum is `0.66eV(E_(4),_(3))`.