For any given series of spectral lines of atomic hydrogen, let $$\Delta\bar{v} = \bar{v}_{max} - \bar{v}_{min}$$ be the difference in maximum and minimum wave number in cm$$^{-1}$$. The ratio $$\Delta\bar{v}_{Lyman}/\Delta\bar{v}_{Balmer}$$ is:

We recall the general Rydberg formula for hydrogen:

$$\bar{v}=R\left(\frac{1}{n_1^{\,2}}-\frac{1}{n_2^{\,2}}\right),\qquad n_2>n_1,$$

where $$\bar{v}$$ is the wave-number, $$R$$ is the Rydberg constant and $$n_1$$ is fixed for a given spectral series while $$n_2$$ varies.

For any series, the maximum wave-number $$\bar{v}_{\text{max}}$$ occurs when the electron is removed to infinity, i.e. $$n_2\to\infty$$. In that limit $$1/n_2^{\,2}\to0$$, so

$$\bar{v}_{\text{max}}=R\left(\frac{1}{n_1^{\,2}}-0\right)=\frac{R}{n_1^{\,2}}.$$

The minimum wave-number $$\bar{v}_{\text{min}}$$ occurs for the first possible transition in that series, i.e. when $$n_2=n_1+1$$. Therefore

$$\bar{v}_{\text{min}}=R\left(\frac{1}{n_1^{\,2}}-\frac{1}{(n_1+1)^{2}}\right).$$

We define $$\Delta\bar{v}=\bar{v}_{\text{max}}-\bar{v}_{\text{min}}$$. Substituting the two expressions just written, we get

$$\Delta\bar{v}= \frac{R}{n_1^{\,2}}-R\left(\frac{1}{n_1^{\,2}}-\frac{1}{(n_1+1)^{2}}\right) = R\left[\frac{1}{n_1^{\,2}}-\frac{1}{n_1^{\,2}}+\frac{1}{(n_1+1)^{2}}\right] = \frac{R}{(n_1+1)^{2}}.$$

Thus in every series the difference $$\Delta\bar{v}$$ depends only on $$n_1+1$$:

$$\Delta\bar{v}=\frac{R}{(n_1+1)^{2}}.$$

Lyman series. Here $$n_1=1$$, so $$n_1+1=2$$. Therefore

$$\Delta\bar{v}_{\text{Lyman}}=\frac{R}{2^{2}}=\frac{R}{4}.$$

Balmer series. Here $$n_1=2$$, so $$n_1+1=3$$. Hence

$$\Delta\bar{v}_{\text{Balmer}}=\frac{R}{3^{2}}=\frac{R}{9}.$$

We now form the required ratio:

$$\frac{\Delta\bar{v}_{\text{Lyman}}}{\Delta\bar{v}_{\text{Balmer}}} =\frac{R/4}{R/9} =\frac{9}{4}.$$

Therefore the ratio is $$9:4$$.

Hence, the correct answer is Option D.