# A gas of identical H like atom 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 transition to a higher energy level by absorbing monochromatic light of photon of energy 2.7eV.Subsequently,the atoms emit radiation of only six different photon energies.Some of the emmited photons have energy 2.7eV.Some have more and some have less than 2.7eV.(i)Find the principal quantum number of initially excited level B.(ii)Find the ionozation energies of the gas atoms(iii)Find the maximum and minimum energies of the emmited photons.

Arun Kumar IIT Delhi
askIITians Faculty 256 Points
8 years ago
Since its 6 different emission
$n_{2}=4\\ 4 \to 3\\ 4 \to 2\\ 4 \to 1\\ 3 \to 2\\ 3 \to 1\\ 2 \to 1$
$13.6(1/n^2-1/4^2)=2.7\\ =>1/n^2=1/5+1/16=21/80 \sim 1/4\\=>n=2$

Transition of 2.7eV is 4 to 2
for
$\Delta E<2.7eV$
n=3
$\Delta E>2.7eV$
n=1

Now its fairly simple to do all 3 parts.

Arun Kumar
IIT Delhi