Question

A hydrogen-like atom (atomic number $Z$ ) is in a higher excited state of quantum number $n$ . This excited atom can make a transition to the first excited state by successively emitting two photons of energies $\text{10.20}eV$ and $\text{17.00}eV$ respectively. Alternatively, the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energy $\text{4.25}eV$ and $\text{5.95}eV$ respectively. Determine the value of $Z$ . [Ionization energy of hydrogen atom $=\text{13.6}eV]$

Answer 1

$\text{10.2}+17=\text{13.6}\times Z^{2}\left(\frac{1}{2^{2}} - \frac{1}{n^{2}}\right);\text{4.25}+\text{5.95}=\text{13.6}\times Z^{2}\left(\frac{1}{3^{2}} - \frac{1}{n^{2}}\right)$
Solving the above two equation we get, $Z=3$ and $n=6$ .