Question

The atomic structure of $H{e}^{+}$ 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 $H{e}^{+}$ in nm.

Answer 1

Answer: 30

$n_1+n_2=3$ …(i)

$n_2-n_1=1$ …(ii)

$2n_2=4$

So $n_2=2$

Putting the value of $n_2$ in (i) $n_1=3-n_2=3-2=1$

$\frac{1}{\lambda }=R{z}^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)=R.2^2\left(\frac{1}{1^2}-\frac{1}{2^2}\right)$

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

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