question_answer

The ratio of wavelengths of the last line of Balmer series and the last line of Lyman series is                  [NEET-2017]

A)  0.5                  

B) 2

C)  1                    

D)       4

[d] For last Balmer series Correct Answer: D Solution :

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

\[\frac{1}{{{\lambda }_{l}}}=R\left[ \frac{1}{{{1}^{2}}}-\frac{1}{{{\infty }^{2}}} \right]\]

\[{{\lambda }_{1}}=\frac{1}{R}\]

\[\frac{{{\lambda }_{b}}}{{{\lambda }_{l}}}=\frac{\frac{4}{R}}{\frac{1}{R}}\]

\[\frac{{{\lambda }_{b}}}{{{\lambda }_{l}}}=4\]