A) 4 : 1
B) 4 : 3
C) 4 : 9
D) 5 : 9
Correct Answer: A
Solution :
[a] For Lyman series \[\frac{1}{\lambda }=R\left[ \frac{1}{{{1}^{2}}}-\frac{1}{{{n}^{2}}} \right]\] For shortest wavelength, \[n=\infty ,\] \[\therefore \] \[\theta =\frac{64}{2}=32{}^\circ C.\]\[\Rightarrow \]\[{{\lambda }_{1}}=\frac{1}{R}\] For Banner series \[\frac{1}{{{\lambda }_{2}}}=R\left[ \frac{1}{{{2}^{2}}}-\frac{1}{{{n}^{2}}} \right]\] For shortest wavelength, \[n=\infty ,\] \[\therefore \] \[\frac{1}{{{\lambda }_{2}}}=\frac{R}{4}\]\[\Rightarrow \]\[{{\lambda }_{2}}=\frac{4}{R}\] The ratio \[\frac{{{\lambda }_{2}}}{{{\lambda }_{1}}}=\frac{4}{1}.\]
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