Question

What is the ratio of the shortest wavelength present in the Lyman series of hydrogen spectral emissions to the shortest wavelength present in the Balmer series?

• A. \( 1 : 4 \)
• B. \( 4 : 1 \)
• C. \( 1 : 2 \)
• D. \( 2 : 1 \)

Hint:

The shortest wavelength limit for any hydrogen series can be quickly found using the shortcut formula \(\lambda_{\text{min}} = \frac{n_1^2}{R_\infty}\). This shows that the minimum wavelength scales directly with the square of the destination orbit number (\(n_1^2\)).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The hydrogen spectrum is produced when an electron transitions between discrete energy levels within an atom.
According to the Bohr model, the wavelength of the emitted light depends on the difference in energy between the initial orbit (\(n_2\)) and the final orbit (\(n_1\)).
Spectral series are defined by their destination orbit (\(n_1\)):
- Lyman series: Ends at \(n_1 = 1\) (Ultraviolet region).
- Balmer series: Ends at \(n_1 = 2\) (Visible region).
Since energy and wavelength are inversely proportional (\(E = hc/\lambda\)), the "shortest wavelength" in any series corresponds to the transition with the "maximum energy change."
The largest possible energy jump occurs when an electron drops from the ionization limit at infinity (\(n_2 = \infty\)) down to the base orbit of the series (\(n_1\)). This is called the series limit wavelength.
Step 2: Key Formula or Approach:
The wavelength \(\lambda\) for any hydrogen transition is given by the Rydberg formula:
\[ \frac{1}{\lambda} = R_{\infty} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
For the shortest wavelength (\(\lambda_{min}\)), we set \(n_2 = \infty\). This simplifies the formula to:
\[ \frac{1}{\lambda_{min}} = R_{\infty} \left( \frac{1}{n_1^2} - 0 \right) \implies \lambda_{min} = \frac{n_1^2}{R_{\infty}} \]
Step 3: Detailed Explanation:
1. Calculate the shortest wavelength for the Lyman series (\(\lambda_L\)):
For the Lyman series, the final orbit is \(n_1 = 1\).
Substituting into our simplified formula:
\[ \lambda_L = \frac{1^2}{R_{\infty}} = \frac{1}{R_{\infty}} \]
2. Calculate the shortest wavelength for the Balmer series (\(\lambda_B\)):
For the Balmer series, the final orbit is \(n_1 = 2\).
Substituting into our simplified formula:
\[ \lambda_B = \frac{2^2}{R_{\infty}} = \frac{4}{R_{\infty}} \]
3. Determine the ratio of the two wavelengths:
We are asked for the ratio \(\lambda_L : \lambda_B\):
\[ \text{Ratio} = \frac{\lambda_L}{\lambda_B} = \frac{1/R_{\infty}}{4/R_{\infty}} \]
The Rydberg constant cancels out from both the numerator and denominator:
\[ \text{Ratio} = \frac{1}{4} = 1 : 4 \]
This result shows that the shortest wavelength in the Lyman series is exactly one-fourth the length of the shortest wavelength in the Balmer series.
This makes physical sense because the Lyman series transitions involve jumps to the ground state, which are much higher in energy and thus much shorter in wavelength compared to Balmer transitions.
Step 4: Final Answer:
The ratio of the shortest Lyman wavelength to the shortest Balmer wavelength is \(1 : 4\).