The atomic number of hydrogen-like ion has the wavelength difference between the first line of Balme — Atomic Structure Chemistry Question

💡 Solution & Explanation

Using Rydberg's formula, $\frac{1}{\lambda} = R Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$. For the first line of Balmer ($3 \to 2$), $\lambda_B = \frac{36}{5 R Z^2}$. For the first line of Lyman ($2 \to 1$), $\lambda_L = \frac{4}{3 R Z^2}$. The difference is $\Delta \lambda = \frac{36}{5 R Z^2} - \frac{4}{3 R Z^2} = \frac{108 - 20}{15 R Z^2} = \frac{88}{15 R Z^2}$. Given $\frac{1}{R} \approx 91.15$ nm, $\Delta \lambda = \frac{88 \times 91.15}{15 Z^2} \approx \frac{534.7}{Z^2}$. Setting this equal to 59.3 nm gives $Z^2 = \frac{534.7}{59.3} \approx 9$, so $Z = 3$. Therefore, correct answer is 3.