The shortest wavelength of H atom in the Lyman series is $$\lambda_1$$. The longest wavelength in the Balmer series of He$$^+$$ is:

For any hydrogen-like species, the line-spectrum obeys the Rydberg formula

$$\frac{1}{\lambda}=R\,Z^{2}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right),\qquad n_{2}>n_{1},$$

where $$R$$ is the Rydberg constant, $$Z$$ is the atomic number, $$n_{1}$$ is the lower level and $$n_{2}$$ is the upper level.

We first connect this formula with the given quantity $$\lambda_{1}$$, the shortest wavelength of the Lyman series of the hydrogen atom (H, so $$Z=1$$). In the Lyman series $$n_{1}=1$$, and the shortest wavelength arises when $$n_{2}\rightarrow\infty$$, because the term $$\frac{1}{n_{2}^{2}}$$ then vanishes:

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

Thus we have the handy relation

$$R=\frac{1}{\lambda_{1}}.$$

Next we deal with the Balmer series of the singly ionised helium atom He$$^{+}$$, whose nuclear charge is $$Z=2$$. For the Balmer series $$n_{1}=2$$. The longest wavelength in any series corresponds to the smallest energy (and hence wavenumber) difference, which occurs for the transition having the smallest possible $$n_{2}$$, i.e. $$n_{2}=3$$.

Substituting $$Z=2,\;n_{1}=2,\;n_{2}=3$$ in the Rydberg formula we obtain

$$\frac{1}{\lambda_{\,\text{Balmer}}}=R\,(2)^{2}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right).$$

Carrying out every arithmetic step,

$$\frac{1}{\lambda_{\,\text{Balmer}}}=R\cdot4\left(\frac{1}{4}-\frac{1}{9}\right).$$

Inside the brackets, bring to a common denominator 36:

$$\frac{1}{4}-\frac{1}{9}=\frac{9}{36}-\frac{4}{36}=\frac{5}{36}.$$

So

$$\frac{1}{\lambda_{\,\text{Balmer}}}=4R\left(\frac{5}{36}\right)=\frac{20R}{36}=\frac{5R}{9}.$$

Now replace $$R$$ by $$\dfrac{1}{\lambda_{1}}$$ using the earlier relation:

$$\frac{1}{\lambda_{\,\text{Balmer}}}=\frac{5}{9}\,\frac{1}{\lambda_{1}}.$$

Taking reciprocals on both sides gives the required wavelength explicitly in terms of $$\lambda_{1}$$:

$$\lambda_{\,\text{Balmer}}=\frac{9\,\lambda_{1}}{5}.$$

This matches Option C.

Hence, the correct answer is Option C.