We have given that $n_{f}-n_{i}=2$ And $n_{f}+n_{i}=4$
Solving these equation $n_{f}=3 \quad n_{i}=1$
For helium, $Z=2-$
Using rydberg's formula, $\frac{1}{\lambda}=R \times 2^{2}\left(\frac{1}{1^{2}}-\frac{1}{3^{2}}\right)$
$\bar{v}=\frac{32 R}{9}$
Therefore, the answer is B.
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| $n$ | $l$ | $m_l$ | |
| $A$ | $3$ | $3$ | $-3$ |
| $B$ | $3$ | $2$ | $-2$ |
| $C$ | $2$ | $1$ | $+1$ |
| $D$ | $2$ | $2$ | $+2$ |
The number of correct sets of quantum numbers is ..... .