Correct Answer - Option 2 : 9 : 4
Concept:
The Balmer series is the name given to a series of spectral emission lines of the hydrogen atom that result from electron transitions from higher levels down to the energy level with principal quantum number 2.
There are four transitions that are visible in the optical waveband that are empirically given by the Balmer formula.
The spectral line for
- Lyman series, the transitions from n ≥ 2 to the n = 1 state,
- Balmer series, the transitions from n ≥ 3 to the n = 2 state
- Paschen series, the transitions from n ≥ 4 to n = 3 state.
\(\bar v \propto \Delta {\rm{E}}\)
The wavelength (or wave number) of any line of the series can be given by using the relation.
For H-atom Z = 1
\(\bar v = {\rm{R}}\left[ {\frac{1}{{{\rm{n}}_1^2}} - \frac{1}{{{\rm{n}}_2^2}}} \right]\)
Calculation:
Rydberg constant R = 13.6 eV/hc
For Lyman series,
\(\bar v\left( {{\rm{max}}} \right) = 13.6\left( {\frac{1}{{{1^2}}} - \frac{1}{\infty }} \right)\)
\(\bar v\left( {{\rm{min}}} \right) = 13.6\left( {1 - \frac{1}{4}} \right)\)
\(\therefore {\bar v_{{\rm{max}}}} - {\bar v_{{\rm{min}}}} = 13.6\left( {\frac{1}{4}} \right)\)
\(\Delta {\bar v_{Lyman}} = 13.6\left( {\frac{1}{4}} \right)\)
For Balmer series,
\({\bar v_{{\rm{max}}}} = 13.6\left( {\frac{1}{{{2^2}}} - \frac{1}{\infty }} \right)\)
\({\bar v_{{\rm{min}}}} = 13.6\left( {\frac{1}{4} - \frac{1}{9}} \right)\)
\(\therefore {\bar v_{{\rm{max}}}} - {\bar v_{{\rm{min}}}} = 13.6\left( {\frac{1}{9}} \right)\)
\(\Delta {\bar v_{Balmer}} = 13.6\left( {\frac{1}{9}} \right)\)
\(\frac{{\Delta {{\bar v}_{Lyman}}}}{{\Delta {{\bar v}_{Balmer}}}} = \frac{{13.6\left( {\frac{1}{4}} \right)}}{{13.6\left( {\frac{1}{9}} \right)}} = \frac{{\frac{1}{4}}}{{\frac{1}{9}}} = \frac{9}{4}\)
\(\frac{{\Delta {{\bar v}_{Lyman}}}}{{\Delta {{\bar v}_{Balmer}}}} = \frac{9}{4}\)
Thus, the
ratio \(\frac{{\Delta {{\bar v}_{Lyman}}}}{{\Delta {{\bar v}_{Balmer}}}}\) is 9 : 4
