Question

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 transition to a higher energy level by absorbing monochromatic light of photon energy 2.7 e V. Subsequently , the atom emits radiation of only six different photon energies. Some of the emitted photons have energy of 2.7 eV, 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 1

a) Since six energy different energies can be emitted, that is six transitions can take place, the atom's final level is $n=4$ and the six transitions are, $4\to 3$, $4\to 2$, $4\to 1$, $3\to 2$, $3\to 1$, $2\to 1$.

$E_f=-\frac{13.6}{4^2}eV$

Let intial state be $n_i$

$⟹2.7=-\frac{13.6}{4^2}-(-\frac{13.6}{n^2})$

$⟹n_i=2$

b) Let ionization energy be $E_i$

$E_4=\frac{E_i}{16},E_2=\frac{E_i}{4}$

$E_4-E_2=2.7$

$⟹E_i(\frac{1}{16}-\frac{1}{4})=2.7$

$⟹-\frac{3}{16}{E}_i=2.7$

$⟹E_i=-14.4eV$

So ionization energy is 14.4 eV

c) Maximum kinetic energy will be for transition $4\to 1$

$E_{4\to 1}=E_4-E_1=\frac{E_i}{16}-E_i=-\frac{15}{16}{E}_i=13.5eV$

Minimum kinetic energy will be for transition $4\to 3$

$E_{4\to 3}=E_4-E_3=\frac{E_i}{16}-\frac{E_i}{9}=-\frac{7}{144}{E}_i=0.7eV$