For a hydrogen atom, the ratio of the largest wavelength of Lyman series to that of the Balmer series is

The wavelengths of hydrogen spectral lines are obtained from the Rydberg‐formula,

$$\frac{1}{\lambda}=R\left(\frac{1}{n_f^{2}}-\frac{1}{n_i^{2}}\right)$$

where $$R$$ is the Rydberg constant, $$n_f$$ is the lower (final) level of the transition and $$n_i$$ is the upper (initial) level with $$n_i \gt n_f$$.

For any given series, the largest wavelength (smallest energy difference) comes from the transition having the smallest possible gap, that is $$n_i = n_f+1$$.

For the Lyman series, $$n_f = 1$$. Therefore take $$n_i = 2$$.

$$\frac{1}{\lambda_L}=R\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)=R\left(1-\frac{1}{4}\right)=R\left(\frac{3}{4}\right)$$

Hence $$\lambda_L=\frac{4}{3R}$$ $$-(1)$$

For the Balmer series, $$n_f = 2$$. Therefore take $$n_i = 3$$.

$$\frac{1}{\lambda_B}=R\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=R\left(\frac{1}{4}-\frac{1}{9}\right)=R\left(\frac{5}{36}\right)$$

Hence $$\lambda_B=\frac{36}{5R}$$ $$-(2)$$

Taking the ratio of $$(1)$$ to $$(2)$$:

$$\frac{\lambda_L}{\lambda_B}=\frac{\dfrac{4}{3R}}{\dfrac{36}{5R}}=\frac{4}{3R}\times\frac{5R}{36}=\frac{20}{108}=\frac{5}{27}$$

Therefore, the required ratio is $$5 : 27$$.

Correct option: Option B (5 : 27)