# What transition in the hydrogen spectrum would have

Question.

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

Solution:

For $\mathrm{He}^{+}$ion, the wave number $(\bar{v})$ associated with the Balmer transition, $n=4$ to $n$

$=2$ is given by:

$\bar{v}=\frac{1}{\lambda}=R Z^{2}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$

Where, $n_{1}=2$

$n_{2}=4$

$Z=$ atomic number of helium

$\bar{v}=\frac{1}{\lambda}=R(2)^{2}\left(\frac{1}{4}-\frac{1}{16}\right)$

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

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

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

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

$\Rightarrow R(1)^{2}\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]=\frac{3 R}{4}$

$\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]=\frac{3}{4}$ .....(i)

By hit and trail method, the equality given by equation (1) is true only when

$n_{1}=1$ and $n_{2}=2$

$\therefore$ 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 $\mathrm{He}^{+}$spectrum.