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 electrons being present in (I) shell and another shell ` n_1` . These are excited to higher level ` n_2` by absorbing ` 2, 7 eV` and on de-exctation emits is ( `lambda`) and thus excited state ` n_2` comes to be 940.
` [ 6 = sum Delta n = sum ( n_2 -1) :. n_2 = 4 ]`
Now` E_1 =- (R_H .c.h)/1^2 , E_(n1) =- (R_H .c.h)/(n_1^2),`
` E-4 =- (R_H .c.h)/4^@`
Since de-exctation leads to different `( lambda)` having photon energy ` 2. 76 eV` and thus absorpton of ` 2, 7 eV` energy casuing excitiaon to ` 4th` shell and them re-emitting photons of `._(lt)^(gt)2.76eV`. is possible only when ` n_ 1 = 2` the de-exctation from ` 4th` shell occurs in I, II and III shell) .
` E_4 - E-2 = 2 . 7 eV`
` E_4 - E-3 lt 2.7 eV`
` E_4 - E-1 gt 2.7 eV`
` :. E_(n_1) = E_2 = ( R_H . ch)/2^2 = E_1/2^2`
since ` ._1 =2` ( as obtained by discussion )
Also ` E_4 - E-2 = 2.7 eV`
` :. - E_1/4^2 + E_1 /2^2 = 2. 7 eV`
` :. E_1 = - 124. 4 eV`
` :. I. P. = 14 .4 eV`
` W_("max"). = E_4 - E_1`
`=-0 E_1/4_2 + E^1/1^2`
` =- ( 14 . 4 )/( 16) + 14 . 4 = 13 . 5 eV`
` E_("min") = E_ 4 - E_3`
` =- E-1 /4^2 + E-1 /3^2 =0. 7 eV`
Note : It is` ._1 H^2` atom .