Atomic StructureHard
Question
Wave number of the first line in the Balmer series of Be3+ is 2.5 × 105 cm−1. Wave number of the second line of the Paschen series of Li2+ is
Options
A.7.2 × 104 cm–1
B.7.2 × 105 cm−1
C.7.2 × 10−4 cm−1
D.1.8 × 104 cm−1
Solution
$\frac{\overline{\upsilon_{2}}}{\overline{\upsilon_{1}}} = \frac{z_{2}^{2}\left( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right)_{1}}{z_{1}^{2}\left( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right)_{2}} \Rightarrow \frac{\overline{\upsilon_{2}}}{2.5 \times 10^{5}} = \frac{3^{2}\left( \frac{1}{3^{2}} - \frac{1}{5^{2}} \right)}{4^{2}\left( \frac{1}{2^{2}} - \frac{1}{3^{2}} \right)}$
$\therefore\overline{\upsilon_{2}} = 7.2 \times 10^{4}\text{ c}\text{m}^{- 1}$
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