Question:medium
The ratio of the longest wavelength of the Lyman series to that of Ballmer series of the hydrogen spectrum is
The ratio of the longest wavelength of the Lyman series to that of Ballmer series of the hydrogen spectrum is
Show Hint
Longest wavelength transitions are always between adjacent levels ($n+1 \to n$). For Lyman it's $2 \to 1$ and for Balmer it's $3 \to 2$.
Updated On: Jun 26, 2026
- 3:29
- 7:27
- 5:27
- 4:29
- 5:29
Show Solution
The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
The longest wavelength in a spectral series corresponds to the smallest energy transition (i.e., the electron falling from the state immediately above the series' base state).
Step 2: Key Formula or Approach:
Rydberg Formula: \(\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)\).
Lyman series base state: \(n_1 = 1\). Longest wavelength transition: \(n_2 = 2\).
Balmer series base state: \(n_1 = 2\). Longest wavelength transition: \(n_2 = 3\).
Step 3: Detailed Explanation:
Find longest wavelength of Lyman series (\(\lambda_L\)):
\[ \frac{1}{\lambda_L} = R\left(\frac{1}{1^2} - \frac{1}{2^2}\right) = R\left(1 - \frac{1}{4}\right) = \frac{3R}{4} \] \[ \lambda_L = \frac{4}{3R} \] Find longest wavelength of Balmer series (\(\lambda_B\)):
\[ \frac{1}{\lambda_B} = R\left(\frac{1}{2^2} - \frac{1}{3^2}\right) = R\left(\frac{1}{4} - \frac{1}{9}\right) = R\left(\frac{9 - 4}{36}\right) = \frac{5R}{36} \] \[ \lambda_B = \frac{36}{5R} \] Now find the ratio \(\frac{\lambda_L}{\lambda_B}\):
\[ \text{Ratio} = \frac{\frac{4}{3R}}{\frac{36}{5R}} = \frac{4}{3R} \times \frac{5R}{36} \] The Rydberg constant \(R\) cancels out:
\[ \text{Ratio} = \frac{4 \times 5}{3 \times 36} = \frac{20}{108} \] Simplify by dividing numerator and denominator by 4:
\[ \text{Ratio} = \frac{5}{27} \] Step 4: Final Answer:
The ratio is 5:27.
The longest wavelength in a spectral series corresponds to the smallest energy transition (i.e., the electron falling from the state immediately above the series' base state).
Step 2: Key Formula or Approach:
Rydberg Formula: \(\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)\).
Lyman series base state: \(n_1 = 1\). Longest wavelength transition: \(n_2 = 2\).
Balmer series base state: \(n_1 = 2\). Longest wavelength transition: \(n_2 = 3\).
Step 3: Detailed Explanation:
Find longest wavelength of Lyman series (\(\lambda_L\)):
\[ \frac{1}{\lambda_L} = R\left(\frac{1}{1^2} - \frac{1}{2^2}\right) = R\left(1 - \frac{1}{4}\right) = \frac{3R}{4} \] \[ \lambda_L = \frac{4}{3R} \] Find longest wavelength of Balmer series (\(\lambda_B\)):
\[ \frac{1}{\lambda_B} = R\left(\frac{1}{2^2} - \frac{1}{3^2}\right) = R\left(\frac{1}{4} - \frac{1}{9}\right) = R\left(\frac{9 - 4}{36}\right) = \frac{5R}{36} \] \[ \lambda_B = \frac{36}{5R} \] Now find the ratio \(\frac{\lambda_L}{\lambda_B}\):
\[ \text{Ratio} = \frac{\frac{4}{3R}}{\frac{36}{5R}} = \frac{4}{3R} \times \frac{5R}{36} \] The Rydberg constant \(R\) cancels out:
\[ \text{Ratio} = \frac{4 \times 5}{3 \times 36} = \frac{20}{108} \] Simplify by dividing numerator and denominator by 4:
\[ \text{Ratio} = \frac{5}{27} \] Step 4: Final Answer:
The ratio is 5:27.
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