Step 1: Understanding the Concept:
The wavelengths of spectral lines in the hydrogen emission spectrum are calculated using the Rydberg formula.
The 'last line' of any spectral series, also known as the series limit, corresponds to the shortest possible wavelength in that series.
This shortest wavelength occurs when an electron transitions from the highest possible energy level (\(n_2 = \infty\)) down to the base principal quantum number (\(n_1\)) for that series.
Step 2: Key Formula or Approach:
The Rydberg formula is given by:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \(R\) is the Rydberg constant, \(n_1\) is the lower energy level, and \(n_2\) is the higher energy level from which the electron transitions.
Step 3: Detailed Explanation:
For the Lyman series, the base energy level is \(n_1 = 1\).
For the last line of the Lyman series, the transition is from \(n_2 = \infty\).
Let its wavelength be \(\lambda_L\). Substituting these values into the formula:
\[ \frac{1}{\lambda_L} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R (1 - 0) = R \]
\[ \implies \lambda_L = \frac{1}{R} \]
For the Balmer series, the base energy level is \(n_1 = 2\).
For the last line of the Balmer series, the transition is from \(n_2 = \infty\).
Let its wavelength be \(\lambda_B\). Substituting these values into the formula:
\[ \frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = R \left( \frac{1}{4} - 0 \right) = \frac{R}{4} \]
\[ \implies \lambda_B = \frac{4}{R} \]
Now, we must find the ratio of these two wavelengths:
\[ \text{Ratio} = \frac{\lambda_L}{\lambda_B} = \frac{\frac{1}{R}}{\frac{4}{R}} = \frac{1}{4} = 0.25 \]
Step 4: Final Answer:
The ratio of the wavelengths is \(0.25\).