For any given series of spectral lines of atomic hydrogen, let `Deltavecv = Deltavecv_("max") - vecv_("min")` be the difference in maximum and minimum frequencies in `cm^(-1)`. The ratio `Deltavecv_("Lyman")//Deltavecv_("Balmer")` is:

Options

A. `4:1`
B. `9:4`
C. `27:5`
D. `5:4`

Answer

To solve the problem of finding the ratio of the frequency differences for the Lyman and Balmer series of atomic hydrogen, we can follow these steps:

### Step 1: Understand the Series
The Lyman series corresponds to transitions from higher energy levels (n ≥ 2) to n = 1, while the Balmer series corresponds to transitions from higher energy levels (n ≥ 3) to n = 2.

### Step 2: Write the Formula for Frequency
The frequency of the spectral lines can be calculated using the Rydberg formula:
\[ \Delta \nu = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R \) is the Rydberg constant, \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level.

### Step 3: Calculate for the Lyman Series
For the Lyman series:
- Maximum frequency occurs when \( n_2 \to \infty \) (transition from n = 1 to n = ∞):
\[ \nu_{\text{max}} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R \left( 1 - 0 \right) = R \]
- Minimum frequency occurs for the transition from \( n = 2 \) to \( n = 1 \):
\[ \nu_{\text{min}} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right) \]
- Therefore, the difference in frequencies for the Lyman series is:
\[ \Delta \nu_{\text{Lyman}} = \nu_{\text{max}} - \nu_{\text{min}} = R - \frac{3R}{4} = \frac{R}{4} \]

### Step 4: Calculate for the Balmer Series
For the Balmer series:
- Maximum frequency occurs when \( n_2 \to \infty \) (transition from n = 2 to n = ∞):
\[ \nu_{\text{max}} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = R \left( \frac{1}{4} - 0 \right) = \frac{R}{4} \]
- Minimum frequency occurs for the transition from \( n = 3 \) to \( n = 2 \):
\[ \nu_{\text{min}} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{9 - 4}{36} \right) = R \left( \frac{5}{36} \right) \]
- Therefore, the difference in frequencies for the Balmer series is:
\[ \Delta \nu_{\text{Balmer}} = \nu_{\text{max}} - \nu_{\text{min}} = \frac{R}{4} - \frac{5R}{36} \]
To simplify this, we need a common denominator:
\[ \Delta \nu_{\text{Balmer}} = \frac{9R}{36} - \frac{5R}{36} = \frac{4R}{36} = \frac{R}{9} \]

### Step 5: Calculate the Ratio
Now we can find the ratio of the frequency differences:
\[ \frac{\Delta \nu_{\text{Lyman}}}{\Delta \nu_{\text{Balmer}}} = \frac{\frac{R}{4}}{\frac{R}{9}} = \frac{R}{4} \cdot \frac{9}{R} = \frac{9}{4} \]

### Final Answer
The ratio \( \frac{\Delta \nu_{\text{Lyman}}}{\Delta \nu_{\text{Balmer}}} \) is \( \frac{9}{4} \).

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