Question

A gas of identical hydrogenic atoms has some atoms in ground state and same atoms in certain excited states and there are no atom in any other energy level. The atoms of the gas makes transition to higher energy level by absorbing a light of photon energy of $2.7eV$.
Subsequently, the atoms emit radiations of only $6$ different photon energy some of them have emitted photons of energy $2.7eV$, some have energy more and some have less than $2.7eV$. Find :

(i) Principle Quantum no. of atoms in excited state
(ii) Find ionisation energy of gaseous atom.
(iii) Find maximum and minimum energies of emitted photons.

Answer 1

(i) Let ground state $=n_1$ and excited state $=n_2$. Energy absorbed $=2.7eV$. When electrons fall from higher orbit $n2$ to lower orbit $n_1$, the six photon energies are emitted.

$\sum n_2-1=\frac{n_2(n_2-1)}{2}$ or $\frac{n_2(n_2-1)}{2}=6$

$n_2^2-n_2=12$ or $n_2^2-n_2=0$ or

$n_2^2-4n_2+3n_2-12=0$ (using factors)

$n_2(n_2-4)+3(n_2-4)=0$ or $(n_2-4)(n_2+3)=0$

$n_2-4=0$ or $n_2=4$ [When $n_2+3=0$ $,n_2=-3$ which is not possible.]

Principal quantum number $=4$

But,

$E_n=\frac{-R_{H}ch}{n^2}$ for hydrogen like atoms

$E_4=\frac{-R_{H}ch}{4^2}=\frac{-R_{H}ch}{16}$

$E_3=\frac{-R_{H}ch}{3^2}=\frac{-R_{H}ch}{9}$

$E_2=\frac{-R_{H}ch}{2^2}=\frac{-R_{H}ch}{4}$

$E_1=\frac{-R_{H}ch}{1^2}=-R_{H}ch$

But, $E_4-E_2=2.7eV$ (given)

$E_4-E_3<2.7eV$ and $E_4-E_1>2.7eV$

(ii) We know,

$E_4-E_2=2.7eV$

$\frac{-R_{H}ch}{16}-\left(\frac{-R_{H}ch}{4}\right)=2.7eV$

$\frac{-R_{H}ch}{16}+\left(\frac{R_{H}ch}{4}\right)=2.7eV$

$\frac{R_{H}ch}{4}[\frac{1}{4}-1]=2.7eV$

$\frac{E_1}{4}\left[\frac{-3}{4}\right]=2.7eV$

$\frac{-3E_1}{16}=2.7eV$

Hence, $E_1=\frac{2.7eV\ast 16}{13}=-14.4eV$

Ionization energy of gas atoms $=14.4eV$

(iii) Minimum energy of emitted photons

$=E_4-E_3=\frac{-E_1}{16}-\frac{-E_1}{9}$

$=E_4-E_3=\frac{E_1}{9}-\frac{E_1}{16}=E_1(\frac{1}{9}-\frac{1}{16})$

$=E_1\left(\frac{16-9}{16\ast 9}\right)=E_1$ X $\frac{7}{144}$

$E_4-E_3=14.4$ X $\frac{7}{144}=0.7eV$

Maximum energy of emitted photons

$E_4-E_1=\frac{-E_1}{16}-\frac{-E_1}{1}$

$\frac{E_1}{1}-\frac{E_1}{16}=E_1(1-\frac{1}{16})=E_1$ X $\frac{15}{16}$

$E_4-E_1=14.4$ X $\frac{15}{16}=13.5eV$