Question

A hydrogen-like atom (described by the Bohr model) is observed to emit six wavelengths, originating from all possible transitions between a group of levels. These levels have energies between $-0.85 \, eV$ and $ \, -0.544 \, eV$ (including both these values). Find the atomic number of the atom. (Take $hc=1240 \, eV \, nm$ , ground state energy of hydrogen atom = $-13.6 \, eV$ )

Answer 1

Total $6$ lines are emitted. Therefore,
$\frac{n \left(\right. n - 1 \left.\right)}{2} = 6$
or $n=4$
So, transition is taking place between $m^{t h}$ energy state and $(m + 3)^{th}$ energy state.
$E_{m}=-0.85\,eV$
$\Rightarrow -13.6\left(\right. \frac{z^{2}}{m^{2}} \left.\right)=-0.85$
$\Rightarrow \frac{z}{m}=0.25...\left(\right.i\left.\right)$
Similarly, $E_{m + 3}=-0.544\,eV$
$\Rightarrow -13.6\frac{z^{2}}{\left( m + 3 \right)^{2}}=-0.544$
$\Rightarrow \frac{z}{\left( m + 3 \right)}=0.2\ldots \left( ii \right)$
Solving Eq.(i) and Eq. (ii) , we get
$m=12$ and $Z=3$