Question

A hydrogen-like atom (atomic number $Z$) is in a higher excited state of the quantum number $n$. This excited atom can make a transition to the first excited state by successfully emitted to photons of energies $10.20eV$ and $17.00eV$ respectively.

Alternatively, the atom from the same excited state can make a transition to the second excites state by the successively emitting two photons of energy $4.25ev$ and $5.95ev$ respectively. Determine the values of $n$ and $Z$ (ionization energy of hydrogen atom $=13.6eV$).

• A. $Z=3,n=4$
• B. $Z=2,n=3$
• C. $Z=3,n=6$
• D. $Z=6,n=3$

Answer 1

The correct option is C $Z=3,n=6$

Total energy liberated during transition of electron from $n^{th}$ shell to first excited state (i.e., $2$nd shell)

$=10.20+17.0=27.20eV$

$=27.20\times 1.602\times {10}^{-12}erg$

$\because \frac{hc}{\lambda }=R_H\times Z^2\times hc[\frac{1}{2^2}-\frac{1}{n^2}]$

$\because 27.20\times 1.602\times {10}^{-12}=R_H\times Z^2\times h\times c[\frac{1}{2^2}-\frac{1}{n^2}]$ ......(i)

Similarly, total energy liberated during transition of electron from $n$th shell to second excited state (i.e., $3rd$ shell)

$=4.25\times 5.95=10.20eV$

$=10.20\times 1.602\times {10}^{-12}erg$

$\therefore 10.20\times 1.602\times {10}^{-12}=R_H\times Z^2\times h\times c[\frac{1}{3^2}-\frac{1}{n^2}]$...............(ii)

Dividing Eq.(i) by Eq. (ii),

$n=6$

On substituting the value of $n$ in Eqs. (i) or (ii),
$Z=3$