Q36 / 120 · VITEEE 2013 · Chemistry
What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition, $n=4$ to $n=2$ of $\mathrm{He}^{+}$spectrum?
- A$n=4$ to $n=2$
- B$n=3$ to $n=2$
- Correct answerC$n=2$ to $n=1$
- D$n=4$ to $n=3$
Solution
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.
$\begin{array}{l}
\because \lambda_{\mathrm{H}}=\lambda_{\mathrm{He}^{+}} \\
\therefore \mathrm{R}_{\mathrm{H}} Z_{\mathrm{H}}^{2}\left[\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right]=\mathrm{R}_{\mathrm{H}} Z_{\mathrm{He}^{+}}^{2}\left[\frac{1}{2^{2}}-\frac{1}{4^{2}}\right] \\
1 \times\left[\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right]=4 \times\left(\frac{1}{4}-\frac{1}{16}\right) \\
{\left[\begin{array}{l}
\left.\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right]=4 \times \frac{4-1}{16} \\
{\left[\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right]=\frac{3}{4}}
\end{array}\right.}
\end{array}$
If $\mathrm{n}_{1}=1$, then $\mathrm{n}_{2}=2,3, \ldots$
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 $\mathrm{He}^{+}$
Hence, the correct answer is option: 'C'.