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Ratio of longest wavelengths corresponding to Lyman and Balmer series in hydrogen spectrum is:
A
$\frac{9}{31}$
B
$\frac{5}{27}$
C
$\frac{3}{23}$
D
$\frac{7}{29}$
Detailed Solution
$\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$; the longest wavelength is for the smallest transition.
$\left(\frac{\lambda_{Lyman}}{\lambda_{Balmer}}\right)_{max} = \frac{\frac{1}{2^2} - \frac{1}{3^2}}{\frac{1}{1^2} - \frac{1}{2^2}} = \frac{5/36}{3/4} = \frac{5}{27}$
$\left(\frac{\lambda_{Lyman}}{\lambda_{Balmer}}\right)_{max} = \frac{\frac{1}{2^2} - \frac{1}{3^2}}{\frac{1}{1^2} - \frac{1}{2^2}} = \frac{5/36}{3/4} = \frac{5}{27}$