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In the spectrum of hydrogen, the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is:
A
$\frac{5}{27}$
B
$\frac{4}{9}$
C
$\frac{9}{4}$
D
$\frac{27}{5}$
Detailed Solution
Lyman (2 → 1): $\frac{1}{\lambda_L} = R\left(\frac{1}{1^2} - \frac{1}{2^2}\right) \Rightarrow \lambda_L = \frac{4}{3R}$
Balmer (3 → 2): $\frac{1}{\lambda_B} = R\left(\frac{1}{2^2} - \frac{1}{3^2}\right) \Rightarrow \lambda_B = \frac{36}{5R}$
$\frac{\lambda_L}{\lambda_B} = \frac{4}{3R}\times\frac{5R}{36} = \frac{5}{27}$
Balmer (3 → 2): $\frac{1}{\lambda_B} = R\left(\frac{1}{2^2} - \frac{1}{3^2}\right) \Rightarrow \lambda_B = \frac{36}{5R}$
$\frac{\lambda_L}{\lambda_B} = \frac{4}{3R}\times\frac{5R}{36} = \frac{5}{27}$