A gas of identical hydrogen-like atoms has some atoms in the lowest (ground) energy level A and some atoms in a particular upper (excited) energy level B and there are no atoms in any other energy level. The atoms of the gas make the transition to a higher energy level by absorbing monochromatic light of photon energy 2.7eV. Subsequently, the atoms emit radiation of only six different photons more energies. Some of the emitted photons have energy of 2.7eV, some have more energy and some less than 2.7ev.

1. Find the principal quantum number of the initially excited level B.
2. Find the ionization energy for the gas atoms.
3. Find the maximum and the minimum energies of the emitted photons.

Let the other energy levl where atoms of the gas make the transition to a higher energy level by absorbing 

monochromatic light of photon energy 2.7eV = x

No. of transition of photon(energy level) = n(n-1) = 6

So the n=4 for “x”

So the quantum no. of initially excited level B will be

4>n_B>1

Energy req. to go e- from B to C=2.7eV

Ec-Eb=2.7eV

B can’t be 3^rd level of transition bcz. The excitation energy from B to C is ,Some of the emitted photons have energy of 2.7eV, some have more energy and some less than 2.7ev.

So ,

No. of energy leve should be=1

Since,

No. of transition of photon(energy level)=n(n-1)=n2-n=1

And 

N=2

Means the principal quantum no. is 2

2.

since,

Ec-Eb=2.7eV

E4-E2=2.7eV

We know

En=E1/n2

So,

(E1/16)-(E1/4)=2.7eV

E1= -14.4eV

And the I.E of gas atom is= -E1=14.4eV

3.

E1= -14.4eV,E3= -14.4/9=- -1.6eV

Here there is max. transition is up to n=4

So

E4= -14.4/16= -0.9eV

Emax.=E4-E1= -(-0.9)-(-14.4)=13.5eV

Emin.=E4-E3=-(-0.9)-(-1.6)=0.7eV