The shortest wavelength of hydrogen atom in Lyman series is $$\lambda$$. The longest wavelength in Balmer series of He$$^+$$ is

We are given that $$\lambda$$ is the shortest wavelength in the Lyman series of hydrogen and we need the longest wavelength in the Balmer series of He$$^+$$.

The shortest wavelength in the Lyman series corresponds to the series limit (transition from $$n = \infty$$ to $$n = 1$$):

$$\frac{1}{\lambda} = R_H\left(\frac{1}{1^2} - \frac{1}{\infty^2}\right) = R_H$$

So $$\lambda = \frac{1}{R_H}$$.

Now, the longest wavelength in the Balmer series of He$$^+$$ corresponds to the transition from $$n = 3$$ to $$n = 2$$. For He$$^+$$ with $$Z = 2$$:

$$\frac{1}{\lambda'} = R_H \times 4 \times \left(\frac{1}{4} - \frac{1}{9}\right) = R_H \times 4 \times \frac{5}{36} = \frac{20R_H}{36} = \frac{5R_H}{9}$$

Hence $$\lambda' = \frac{9}{5R_H} = \frac{9}{5} \times \frac{1}{R_H} = \frac{9\lambda}{5}$$.

Hence, the correct answer is Option 2: $$\frac{9\lambda}{5}$$.