Which transition in the hydrogen spectrum would have the same wavelength as the Balmer type transition from n = 4 to n = 2 of He$$^+$$ spectrum

Using the Rydberg formula,

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

For the (He^+) transition with (Z=2), (n_2=4), and (n_1=2),

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

Simplifying,

$$\frac{1}{\lambda}=4R\left(\frac{1}{4}-\frac{1}{16}\right)=4R\left(\frac{3}{16}\right)=\frac{3R}{4}.$$

For hydrogen ((Z=1)),

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

Equating the two expressions,

$$\frac{3R}{4}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right),$$

$$\frac{3}{4}=\frac{1}{n_1^2}-\frac{1}{n_2^2}.$$

Taking (n_1=1) and (n_2=2),

$$\frac{1}{1^2}-\frac{1}{2^2}=1-\frac{1}{4}=\frac{3}{4},$$

which satisfies the required condition.

Hence, the corresponding transition in the hydrogen spectrum is

$$\boxed{n=2\rightarrow n=1}.$$