Question

A hydrogen like atom (atomic number Z) is in a higher excited state of quantum number n. The excited atom can make a transition to the first excited state by successively emitting two photons of energy 10.2 eV and 17.0 eV respectively. Alternately, the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energies 4.25 eV and 5.95 eV respectively. Determine the values of n and Z.
(Ionization energy of H-atom = 13.6 eV)

Answer 1

From the given conditions
$E_n-E_2=(10.2+17)eV=27.2eV$....(i)
$E_n-E_2=(4.25+5.95)eV=10.2eV$....(ii)
Eq. (i) - Eq. (ii) gives
$E_3-E_2= 17.0 eV$
or $Z^2(13.6)\bigg(\frac{1}{4}-\frac{1}{9}\bigg)=17.0$
$\Rightarrow Z^2(13.6)(5/36)=17.0$
$\Rightarrow Z^2=9 or Z=3$
From Eq. (i)
$Z^2(13.6)\bigg(\frac{1}{4}-\frac{1}{n^2}\bigg)=27.2$
or $(3^2)(13.6)\bigg(\frac{1}{4}-\frac{1}{n^2}\bigg)=27.2$
or $\frac{1}{4}-\frac{1}{n^2}=0.222$
or $1/n^2=0.0278$
or $n^2=36$
$\therefore n=6$