A) 0.5
B) 2
C) 1
D) 4
Correct Answer: D
Solution :
For last Balmer series For last Lyman series \[\frac{1}{{{\lambda }_{b}}}=R\left[ \frac{1}{{{2}^{2}}}-\frac{1}{{{\infty }^{2}}} \right]\] \[{{\lambda }_{b}}=\frac{4}{R}\] For last Lyman series \[\frac{1}{{{\lambda }_{l}}}=R\left[ \frac{1}{{{1}^{2}}}-\frac{1}{{{\infty }^{2}}} \right]\] \[{{\lambda }_{1}}=\frac{1}{R}\] \[\frac{{{\lambda }_{b}}}{{{\lambda }_{l}}}=\frac{\frac{4}{R}}{\frac{1}{R}}\] \[\frac{{{\lambda }_{b}}}{{{\lambda }_{l}}}=4\]
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