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 $-0.85eV$ and $-0.544eV$ (including both these values). Find the atomic number of the atom. (Take $hc=1240eVnm$ , ground state energy of hydrogen atom = $-13.6eV$ )

Answer 1

Answer: 3

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