The wavelength of the first line of the He ion spectral series whose interval between the extreme li — Atomic Structure Chemistry Question
Question
The wavelength of the first line of the He$^+$ ion spectral series whose interval between the extreme lines is $2.725 \times 10^6 \text{ m}^{-1}$ is ($R = 1.09 \times 10^7 \text{ m}^{-1}$)
💡 Solution & Explanation
The interval is $\Delta\bar{\nu} = R Z^2 \left(\frac{1}{n_1^2} - 0\right) - R Z^2 \left(\frac{1}{n_1^2} - \frac{1}{(n_1+1)^2}\right) = R Z^2 \left(\frac{1}{(n_1+1)^2}\right)$. Given $2.725 \times 10^6 = (1.09 \times 10^7)(2^2) \frac{1}{(n_1+1)^2} \implies (n_1+1)^2 = \frac{43.6 \times 10^6}{2.725 \times 10^6} = 16 \implies n_1+1 = 4 \implies n_1 = 3$. This is the Paschen series. First line is $4 \to 3$: $\frac{1}{\lambda} = 4R\left(\frac{1}{9} - \frac{1}{16}\right) = \frac{7R}{36}$. $\lambda = \frac{36}{7R} = \frac{36}{7 \times 1.09 \times 10^7} \approx 471.82$ nm. Therefore, correct answer is A.