The wavenumber of the first line $\left(n_2=3\right)$ in the Balmer series of hydrogen is $\overline{\mathrm{v}}_1 \mathrm{~cm}^{-1}$. What is the wavenumber (in $\mathrm{cm}^{-1}$ ) of the second line ( $\mathrm{n}_2=4$ ) in the Balmer series of $\mathrm{He}^{+}$?

• A

  $\frac{5 \bar{v}_1}{27}$
• Correct answer
  B

  $\frac{27 \bar{v}_1}{5}$
• C

  $\frac{27 \bar{v}_1}{20}$
• D

  $\frac{20 \bar{v}_1}{27}$

Hence, the correct answer is option: 'B'.