Chemistry - Atomic Structure Question with Solution | TestHub

We have to compare wavelength of transition in the $\mathrm{H}$-spectrum with the Balmer transition $\mathrm{n}=4$ to $\mathrm{n}=2$ of ${\mathrm{He}}^{+}$spectrum. $\begin{array}{l}\because {\lambda }_{\mathrm{H}}={\lambda }_{{\mathrm{He}}^{+}}\\ \therefore {\mathrm{R}}_{\mathrm{H}}{Z}_{\mathrm{H}}^2\left[\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2}\right]={\mathrm{R}}_{\mathrm{H}}{Z}_{{\mathrm{He}}^{+}}^2\left[\frac{1}{2^2}-\frac{1}{4^2}\right]\\ 1\times \left[\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2}\right]=4\times \left(\frac{1}{4}-\frac{1}{16}\right)\\ [\begin{array}{l}\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2}]=4\times \frac{4-1}{16}\\ \left[\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2}\right]=\frac{3}{4}\end{array}\end{array}$ If ${\mathrm{n}}_1=1$, then ${\mathrm{n}}_2=2,3,\dots$ For first line ${\mathrm{n}}_2=2,{\mathrm{n}}_1=1$ $\left[\frac{1}{1^2}-\frac{1}{2^2}\right]=\frac{1}{1}-\frac{1}{4}=\frac{3}{4}$ Hence, trnasition $n_2=2$ to $n_1=1$ will give spectrum of the same wavelength as that of Balmer transition, ${\mathrm{n}}_2=4$ to ${\mathrm{n}}_1=2$ in ${\mathrm{He}}^{+}$