Q19 / 160 · WBJEE 2011 · Chemistry
The electronic transitions from \(\mathrm{n}=2\) to \(\mathrm{n}=1\) will produce shortest wavelength in (where \(\mathrm{n}=\) principal quantum state)
- Correct answerA\(\mathrm{Li}^{+2}\)
- B\(\mathrm{He}^{+}\)
- C\(\mathrm{H}\)
- D\(\mathrm{H}^{+}\)
Solution
Hints: \(\frac{1}{\lambda}=\mathrm{z}^2 \cdot \mathrm{R}_{\mathrm{H}}\left[1 / \mathrm{n}_1^2-1 / \mathrm{n}_2^2\right]\)
\(\begin{aligned}
& \Rightarrow \frac{1}{\lambda}=(\mathrm{z})^2 \cdot \mathrm{R}_{\mathrm{H}}\left\{\frac{1}{1}-\frac{1}{4}\right\}=\frac{3}{4} \mathrm{R}_{\mathrm{H}} \mathrm{z}^2 \\
& \therefore \lambda \propto 1 / \mathrm{z}^2
\end{aligned}\)
Hence, for shortest \(\lambda, \mathrm{z}\) must be maximum, which is for \(\mathrm{Li}^{+2}\).
Hence, the correct answer is option: 'A'.