A series of lines in the spectrum of atomic hydrogen lines at wavelength 656.46, 486.27, 434.17, 410 — Atomic Structure Chemistry Question

💡 Solution & Explanation

The given wavelengths belong to the Balmer series ($3\to 2, 4\to 2, 5\to 2, 6\to 2$). The next line is $7\to 2$. $\frac{1}{\lambda} = 1.097 \times 10^7 \left(\frac{1}{4} - \frac{1}{49}\right) = 1.097 \times 10^7 \times \frac{45}{196} \approx 2.518 \times 10^6 \text{ m}^{-1}$. $\lambda = \frac{1}{2.518 \times 10^6} \approx 397.1$ nm. Therefore, correct answer is A.