Which transition in the hydrogen spectrum would have the same...

For ${He}^{+}$ ion, the wave number associated with the Balmer transition, $n = 4$ to $n = 2$ is given by:

$\dfrac{1}{\lambda} = {RZ}^{2}\left(\dfrac{1}{{{n}_{1}}^{2}} - \dfrac{1}{{{n}_{2}}^{2}}\right)$

Where, ${n}_{1} = 2$

${n}_{2} = 4$

$Z$ = atomic number of helium

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

$⇒ \dfrac{1}{\lambda} = 4R\left[\dfrac{4 - 1}{16}\right]$

$⇒ \dfrac{1}{\lambda} = \dfrac{3R}{4}$

$⇒ \lambda = \dfrac{4}{3R}$

According to the question, the desired transition for hydrogen will have the same wavelength as that of ${He}^{+}$.

$\lambda ( H ) = \lambda\left({He}^{+}\right)$

$R ( 1{)}^{2}\left[1{n}_{1}^{2} - 1{n}_{2}^{2}\right] = \dfrac{3R}{4}$

By hit and trial method,

$1{n}_{1}^{2} - 1{n}_{2}^{2} = \dfrac{1}{{1}^{2}} - \dfrac{1}{{2}^{2}}$

On comparing ${n}_{1} = 1 \text{ and } {n}_{2} = 2$

So, the correct answer is n = 2 to n = 1.