Question

What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of $H{e}^{+}$ spectrum?

Answer 1

For $H{e}^{+}$ ion, the wave number $\left(\overline{v}\right)$ associated with the Balmer transition, n = 4 to n = 2 is given by:
$\overline{v}=\frac{1}{\lambda }=R{Z}^2\left(\frac{1}{n_1^2-\frac{1}{n_2^2}}\right)where,n_1=2n_2=4$
Z = atomic number of helium
$\overline{v}=\frac{1}{\lambda }==R(2{)}^2(\frac{1}{4}-\frac{1}{16})=4R\left(\frac{4-1}{16}\right)\overline{v}=\frac{1}{\lambda }=\frac{3R}{4}\Rightarrow \lambda =\frac{4}{3R}$
According to the question, the desired transition for hydrogen will have the same wavelength as that of $H{e}^{+}$.
$\Rightarrow R(1{)}^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]=\frac{3}{4}.........(1)$
By hit and trail method, the equality given by equation (1) is true only when $n_1$ = 1and $n_2$ = 2.
$\therefore$The transition for $n_2$ = 2 to n = 1 in hydrogen spectrum would have the same wavelength as Balmer transition n = 4 to n = 2 of $H{e}^{+}$ spectrum.