Atoms and Nuclei Part 3

Concept:Rydberg formula gives the wavelength of spectral lines in hydrogen: $\frac{1}{\lambda} = R \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$, where $n_1 < n_2$, $R$ is Rydberg constant.Explanation:For Lyman series ($n_1 = 1$): Maximum wavelength (lowest energy) occurs for $n_2 = 2$: $\frac{1}{\lambda_{\max}} = R\left(\frac{1}{1^2} - \frac{1}{2^2}\right) = R\left(1 - \frac{1}{4}\right) = \frac{3R}{4}$ ⟹ $\lambda_{\max} = \frac{4}{3R}$.Statement A is true.Balmer series ($n_1 = 2$): transitions from $n = 3,4,\dots$ to $n=2$ produce wavelengths between about 400 nm and 700 nm — the visible range.Statement B is true.For Paschen series ($n_1 = 3$): Minimum wavelength (highest energy) occurs for $n_2 = \infty$: $\frac{1}{\lambda_{\min}} = R\left(\frac{1}{3^2} - \frac{1}{\infty^2}\right) = \frac{R}{9}$ ⟹ $\lambda_{\min} = \frac{9}{R}$.Statement C is true.For Lyman series, minimum wavelength (highest energy) uses $n_2 = \infty$: $\frac{1}{\lambda_{\min}} = R\left(\frac{1}{1^2} - \frac{1}{\infty^2}\right) = R$ ⟹ $\lambda_{\min} = \frac{1}{R} \neq \frac{5}{4R}$.Statement D is false.Thus, statements A, B, C are correct; D is incorrect.Answer:Option D (A, B and C only) is correct.