Question

A hydrogen like atom with atomic number Z is in higher excited state of quantum number 'n' . This excited state atom can make a transition to the first excited state by successively emitting two photons of energies 10 eV and 68.2eV respectively. Alternatively the atom from the same the same excited state can make a transition to the 2nd excited state by emitting two photons of energies 4.25eV and 5.95eV respectively. Calculate the value of Z

Answer 1

We have, energy of the electron in the $n^{th}$ energy state in any atom is given by,

$E=\frac{-13.6z^2}{n^2}eV$

Let the intermediate energy state in the first case be $x$. So,

$\frac{z^2}{4}-\frac{z^2}{n^2}=\frac{10.2+17}{13.6}=2⟶\left(1\right)$

Let the intermediate energy state in the $2^{nd}$ case be $y$. So,

$\frac{z^2}{9}-\frac{z^2}{n^2}=\frac{4.25+5.95}{13.6}=\frac{3}{4}⟶\left(2\right)$

$\left(2\right)-\left(1\right),\frac{5}{36}{z}^2=\frac{5}{4}$

$\Rightarrow z^2=9,z=3$