1. Use the hydrogen-like ion formula

For any hydrogen-like species,

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

where:

• $R$ = Rydberg constant
• $Z$ = atomic number
• $n_2 \to n_1$ is the transition

For $\mathrm{He}^+$, we have $Z=2$.

2. Find the wavenumber of the given transition in $\mathrm{He}^+$

The Balmer-type transition from $n=4$ to $n=2$ means:

$n_2=4, \quad n_1=2$

So,

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

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

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

Thus, the required wavelength satisfies:

$\frac{1}{\lambda} = \frac{3R}{4}$

3. Check which hydrogen transition gives the same value

For hydrogen, $Z=1$, so

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

We need:

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

So,

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

Now evaluate options:

• A: $n=3$ to $n=4$
  This is absorption, and even if checked: $\left|\frac{1}{3^2}-\frac{1}{4^2}\right| = \frac{1}{9}-\frac{1}{16} = \frac{7}{144} \neq \frac{3}{4}$ Not correct.

• B: $n=2$ to $n=1$ $\frac{1}{1^2}-\frac{1}{2^2} = 1-\frac{1}{4} = \frac{3}{4}$ This matches exactly.

• C: $n=1$ to $n=2$
  Same levels as option B but absorption direction. Spectral line wavelength depends on the energy gap magnitude, so the wavelength is the same.

• D: $n=1$ to $n=3$ $1-\frac{1}{9}=\frac{8}{9} \neq \frac{3}{4}$ Not correct.

4. Important note about the options

Physically, the wavelength depends on the energy difference between two levels, not on whether the process is emission or absorption. Therefore, both transitions $2 \leftrightarrow 1$ in hydrogen correspond to the same wavelength.

So mathematically and physically, both B and C give the same wavelength.

However, since this is a single-correct MCQ, such questions usually intend the emission transition, and hence option B is taken as the expected answer.

5. Final answer

$\boxed{\text{B}}$

with the caveat that option C corresponds to the same wavelength as well, if absorption is also allowed.