The ratio of the shortest wavelength of two spectral series of hydrogen spectrum is found to be about 9. The spectral series are:

We have to compare the shortest (limit) wavelengths of two spectral series in hydrogen. For any transition in hydrogen, the Rydberg formula gives

$$\frac{1}{\lambda}=R\left(\frac{1}{n_1^{\,2}}-\frac{1}{n_2^{\,2}}\right),$$

where $$R$$ is the Rydberg constant, $$n_1$$ is the lower (fixed) orbit for the series, and $$n_2$$ is the upper orbit with $$n_2\gt n_1$$.

The “shortest wavelength” (i.e. the series limit) is obtained by letting the electron fall from $$n_2\to\infty$$. In that limit the term $$1/n_2^{\,2}$$ vanishes, so

$$\frac{1}{\lambda_{\min}} = R\left(\frac{1}{n_1^{\,2}}-0\right)=\frac{R}{n_1^{\,2}}.$$

Taking the reciprocal we obtain an explicit expression for the shortest wavelength in a series:

$$\lambda_{\min}= \frac{n_1^{\,2}}{R}.$$

Now we are told that the ratio of the shortest wavelengths of two different series is about $$9$$. Let the two lower quantum numbers be $$n_{1a}$$ and $$n_{1b}$$. Then

$$\frac{\lambda_{\min,a}}{\lambda_{\min,b}} =\frac{\dfrac{n_{1a}^{\,2}}{R}}{\dfrac{n_{1b}^{\,2}}{R}} =\frac{n_{1a}^{\,2}}{n_{1b}^{\,2}} =9.$$

Cancelling the common factor $$R$$ and taking the square root on both sides, we find

$$\frac{n_{1a}}{n_{1b}} = \sqrt{9}=3.$$

Hence the two series must have lower quantum numbers whose ratio is exactly $$3:1$$.

The known values of $$n_1$$ for the named hydrogen series are:

$$\text{Lyman: }n_1=1,\quad \text{Balmer: }n_1=2,\quad \text{Paschen: }n_1=3,\quad \text{Brackett: }n_1=4,\quad \text{Pfund: }n_1=5.$$

To satisfy $$n_{1a}/n_{1b}=3$$, the only possible pair from this list is $$n_1=3$$ (Paschen) and $$n_1=1$$ (Lyman). Indeed,

$$\frac{\lambda_{\min,\text{Paschen}}}{\lambda_{\min,\text{Lyman}}} =\frac{3^{2}/R}{1^{2}/R}=9,$$

exactly as required.

Therefore, the two spectral series are Lyman and Paschen.

Hence, the correct answer is Option C.