Question

A hydrogen-like atom (atomic number $Z$ ) is in a higher excited state of quantum number $n$ . By successively emitting two photons of energies $10.20eV$ and $17.00eV$ respectively this excited atom can transition to the first excited state. By successively emitting two photons of energies $4.25eV$ and $5.95eV$ respectively the atom can also transition to the second excited state. If the value of $\frac{Z}{n}$ is $\frac{3}{x}$ then find $x$ . (ionization energy of hydrogen atom $=13.6eV$ )

Answer 1

For the transition from a higher state n to the first excited state $\left(\right.n_{1}=2\left.\right)$
$\text{ Total energy }=10.2+17=27.2eV$
$\text{ For }\Delta E=27.2eV,n_{1}=2\text{ and }n_{2}=n$
$\therefore 27.2=13.6Z^{2}\left[\frac{1}{4} - \frac{1}{n^{2}}\right]\ldots .\left(\right.i\left.\right)$
For the eventual transition to the second excited
state $\left(\right.n_{1}=3\left.\right),$
Thus,
$10.2=13.6Z^{2}\left[\frac{1}{9} - \frac{1}{n^{2}}\right]\ldots .\left(\right.ii\left.\right)$
Dividing equations (i) and (ii), we get
$\frac{27 . 2}{10 . 2}=\frac{9 n^{2} - 36}{4 n^{2} - 36}$
$Solving,wegetn^{2}=36orn=6$
Substituting n=6 in any of the above equations,
we obtain
$Z^{2}=9$
$\therefore Z=3Thus,n=6andZ=3\therefore \frac{Z}{n}=\frac{3}{6}=0.5$