Wave number of the first line in the Balmer series of is . Wave number of the second line of the Pas — Atomic Structure Chemistry Question
Question
Wave number of the first line in the Balmer series of $Be^{3+}$ is $2.5 \times 10^5 \text{ cm}^{-1}$. Wave number of the second line of the Paschen series of $Li^{2+}$ is
💡 Solution & Explanation
First Balmer of $Be^{3+}$ ($Z=4, 3 \to 2$): $\bar{\nu}_1 = R(16)(\frac{1}{4} - \frac{1}{9}) = \frac{80R}{36} = \frac{20R}{9} = 2.5 \times 10^5 \text{ cm}^{-1}$, so $R = \frac{9 \times 2.5 \times 10^5}{20} = 1.125 \times 10^5 \text{ cm}^{-1}$. Second Paschen of $Li^{2+}$ ($Z=3, 5 \to 3$): $\bar{\nu}_2 = R(9)(\frac{1}{9} - \frac{1}{25}) = R(9)(\frac{16}{225}) = \frac{16R}{25}$. Substituting $R$, we get $\bar{\nu}_2 = \frac{16}{25} \times 1.125 \times 10^5 = 0.64 \times 1.125 \times 10^5 = 0.72 \times 10^5 = 7.2 \times 10^4 \text{ cm}^{-1}$. Therefore, correct answer is A.