Question

Determine the shortest wavelengths of Balmer and Paschen series. Given the limit for Lyman series is 912 Å.

Hint:

Shortest wavelength corresponds to maximum energy transition (\( n_2 = \infty \)); use Rydberg formula.

Answer 1

The shortest wavelength (series limit) occurs when the electron falls from \( n_2 = \infty \) to the lowest level of the series. Given Lyman series limit (\( n_2 = \infty \) to \( n_1 = 1 \)) is 912 Å.
Rydberg formula: \( \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \), where \( R \) is Rydberg constant.
For Lyman limit:
\[ \frac{1}{912} = R \left( 1 - 0 \right) \quad \Rightarrow \quad R = \frac{1}{912} \, \text{Å}^{-1}. \] Balmer Series (\( n_1 = 2 \), \( n_2 = \infty \)):
\[ \frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - 0 \right) = \frac{R}{4} = \frac{1}{912 \times 4} \quad \Rightarrow \quad \lambda_B = 4 \times 912 = 3648 \, \text{Å}. \] Paschen Series (\( n_1 = 3 \), \( n_2 = \infty \)):
\[ \frac{1}{\lambda_P} = R \left( \frac{1}{3^2} - 0 \right) = \frac{R}{9} = \frac{1}{912 \times 9} \quad \Rightarrow \quad \lambda_P = 9 \times 912 = 8208 \, \text{Å}. \] Answer: Balmer: 3648 Å, Paschen: 8208 Å.