Question

The atomic structure of $He^{+}$ arises due to transition from $n_{2}$ to $n_{1}$ level. If $n_{1}+n_{2}$ is 3 and $n_{2}-n_{1}$ is 1. Find the $\lambda $ in nm of transition for this series in $He^{+}$ in nm.

Answer 1

$n_{1}+n_{2}=3$ …(i)

$n_{2}-n_{1}=1$ …(ii)

$2n_{2}=4$

So $n_{2}=2$

Putting the value of $n_{2}$ in (i) $n_{1}=3-n_{2}=3-2=1$

$\frac{1}{\lambda }=Rz^{2}\left(\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}\right)=R.2^{2}\left(\frac{1}{1^{2}} - \frac{1}{2^{2}}\right)$

$\lambda =\frac{1}{3 R}=\frac{1}{3 \times 109678 \text{c} \text{m}^{- 1}}$

$=3\times 10^{- 6}cm=3\times 10^{- 6}\times 10^{7}=30$ nm.