MCQ
What is the shortest wavelength for Paschen series of $Li^{2+}$ ion
- A$\frac {R}{9}$
- B$\frac {9}{R}$
- ✓$\frac {1}{R}$
- D$\frac {9R}{4}$
✓
Answer
Correct option: C.
$\frac {1}{R}$
c
$\frac{1}{\lambda}=\mathrm{R} Z^{2}\left[\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right]=\mathrm{R} \times 3^{2}\left[\frac{1}{3^{2}}-\frac{1}{\infty^{2}}\right]$
$\frac{1}{\lambda}=\mathrm{R} Z^{2}\left[\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right]=\mathrm{R} \times 3^{2}\left[\frac{1}{3^{2}}-\frac{1}{\infty^{2}}\right]$
$\Rightarrow \mathrm{R}$ or $\lambda=\frac{1}{\mathrm{R}}$
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