Question

What transition in $H{{e}^{+}}$ ion shall have the same wavenumber as the first line in the Balmer series of H atoms?
A) $7\to 5$
B) $6\to 4$
C) $5\to 3$
D) $4\to 2$

Answer 1

Answer to this question is based on the formula used for the Balmer line series for H- like particle which is given by $\overline{\gamma } = R{{Z}^{2}}\left[ \dfrac{1}{{{n}_{1}}^{2}}-\dfrac{1}{{{n}_{2}}^{2}} \right]$ . Applying values for $H{{e}^{+}}$ atom gives the required answer.

Complete Solution :
Now, one among those series is the Balmer series which is the series of visible lines in the spectra of hydrogen atoms.
- The emission values for the Balmer series ranges from 383 nm to 656 nm that is from violet to red coloured spectra respectively.
- The Balmer series is more useful in astronomy as these lines appear in various stellar objects due to abundance of hydrogen present in the universe and these lines are relatively strong compared to that of other lines from other elements.
- Balmer series of spectral lines occur when the electron undergoes transitions from higher energy to lower energy levels.
- Other series of spectral lines were obtained for hydrogen atoms and is likely to be said to be a series for hydrogen like atoms.
The series for hydrogen like atom is given by,
$\overline{\gamma }=R{{Z}^{2}}\left[ \dfrac{1}{{{n}_{1}}^{2}}-\dfrac{1}{{{n}_{2}}^{2}} \right]$
where, Z = atomic number
${{n}_{1}}$= initial orbit of electron
${{n}_{2}}$= final orbit of electron
Thus, first line of Balmer series that is for H atom, ${{n}_{1}}$=2, ${{n}_{2}}$=3 and z=1
Therefore,

$\overline{\gamma }=R\times {{1}^{2}}\left[ \dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{3}^{2}}} \right]$= $R\left[ \dfrac{1}{4}-\dfrac{1}{9} \right]$
$\Rightarrow \overline{\gamma }=0.138R$

For $H{{e}^{+}}$ ion, ${{n}_{1}}$ = 4, ${{n}_{2}}$ = 6 and Z = 2
Thus, $\overline{\gamma } = R\times {{\left( 2 \right)}^{2}}\left[ \dfrac{1}{{{4}^{2}}}-\dfrac{1}{{{6}^{2}}} \right]$=$4\times R\left[ \dfrac{1}{16}-\dfrac{1}{36} \right]=0.138R$
$\Rightarrow \overline{\gamma }=0.138R$

Thus, the first line of the Balmer series for H atoms is the same as that of transition from $6\to 4$ for $H{{e}^{+}}$ ion.
So, the correct answer is “Option B”.

Note: Note that the Balmer series of equations is valid for all the hydrogen like species that are atoms which have only one single electron. The Balmer series equation is called the Rydberg formula and do not be confused if a question is asked by this name.