Atoms and Nuclei Part 3

Concept:The Rydberg formula gives the wavelength of spectral lines for hydrogen: $\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$.The smallest wavelength (largest energy) occurs when $n_2 = \infty$.The largest wavelength (smallest energy) occurs when $n_2 = n_1 + 1$.Explanation:Given: Smallest wavelength of Lyman series, $\lambda_{\text{L,min}} = 91\ \text{nm}$.For Lyman: $n_1 = 1$, $n_2 = \infty$.Using formula: $\frac{1}{91} = R\left(\frac{1}{1^2} - \frac{1}{\infty^2}\right) = R(1 - 0) = R$.So $R = \frac{1}{91}\ \text{nm}^{-1}$.For Balmer: $n_1 = 2$. Largest wavelength is from $n_2 = 3$ to $n_1 = 2$.$\frac{1}{\lambda_{\text{B,max}}} = R\left(\frac{1}{2^2} - \frac{1}{3^2}\right) = R\left(\frac{1}{4} - \frac{1}{9}\right) = R \cdot \frac{5}{36}$.Substitute $R = \frac{1}{91}$: $\frac{1}{\lambda_{\text{B,max}}} = \frac{1}{91} \cdot \frac{5}{36}$.Thus $\lambda_{\text{B,max}} = \frac{91 \times 36}{5} = \frac{3276}{5} = 655.2\ \text{nm}$.For Paschen: $n_1 = 3$. Largest wavelength is from $n_2 = 4$ to $n_1 = 3$.$\frac{1}{\lambda_{\text{P,max}}} = R\left(\frac{1}{3^2} - \frac{1}{4^2}\right) = R\left(\frac{1}{9} - \frac{1}{16}\right) = R \cdot \frac{7}{144}$.Substitute: $\frac{1}{\lambda_{\text{P,max}}} = \frac{1}{91} \cdot \frac{7}{144}$.Thus $\lambda_{\text{P,max}} = \frac{91 \times 144}{7} = \frac{13104}{7} = 1872\ \text{nm}$.Difference: $\lambda_{\text{P,max}} - \lambda_{\text{B,max}} = 1872 - 655.2 = 1216.8\ \text{nm} \approx 1217\ \text{nm}$.Answer:The difference is nearly $1217\ \text{nm}$, so the correct option is (C).