What is the ratio of `Deltav==v_("max")-v_("min")` for spectral lines corresponding to lyman `&` Balmer series for hydrogen

Options

A. `9:4`
B. `4:9`
C. `5:7`
D. `7:5`

Answer

To solve the problem of finding the ratio of \( \Delta \nu = \nu_{\text{max}} - \nu_{\text{min}} \) for spectral lines corresponding to the Lyman and Balmer series for hydrogen, we will follow these steps:

### Step 1: Understand the Formula
We will use the formula that relates the wavenumber (or frequency) to the energy levels of the hydrogen atom:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R \) is the Rydberg constant, \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level.

### Step 2: Convert to Frequency
Since we need to find the change in frequency \( \Delta \nu \), we can relate frequency \( \nu \) to wavelength \( \lambda \) using:
\[ \nu = \frac{c}{\lambda} \]
Thus, we can rewrite the formula in terms of frequency:
\[ \frac{\nu}{c} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

### Step 3: Calculate for Lyman Series
For the Lyman series, the minimum energy level \( n_1 = 1 \) and the maximum energy level \( n_2 \) approaches infinity.
- **Maximum frequency (\( \nu_{\text{max}} \)):**
\[ \nu_{\text{max}} = cR \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = cR \left( 1 - 0 \right) = cR \]
- **Minimum frequency (\( \nu_{\text{min}} \)):**
The minimum transition is from \( n_1 = 1 \) to \( n_2 = 2 \):
\[ \nu_{\text{min}} = cR \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = cR \left( 1 - \frac{1}{4} \right) = cR \left( \frac{3}{4} \right) \]
- **Calculate \( \Delta \nu \) for Lyman:**
\[ \Delta \nu_{\text{Lyman}} = \nu_{\text{max}} - \nu_{\text{min}} = cR - cR \left( \frac{3}{4} \right) = cR \left( \frac{1}{4} \right) \]

### Step 4: Calculate for Balmer Series
For the Balmer series, the minimum energy level \( n_1 = 2 \) and the maximum energy level \( n_2 \) approaches infinity.
- **Maximum frequency (\( \nu_{\text{max}} \)):**
\[ \nu_{\text{max}} = cR \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = cR \left( \frac{1}{4} - 0 \right) = \frac{cR}{4} \]
- **Minimum frequency (\( \nu_{\text{min}} \)):**
The minimum transition is from \( n_1 = 2 \) to \( n_2 = 3 \):
\[ \nu_{\text{min}} = cR \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = cR \left( \frac{1}{4} - \frac{1}{9} \right) = cR \left( \frac{9 - 4}{36} \right) = cR \left( \frac{5}{36} \right) \]
- **Calculate \( \Delta \nu \) for Balmer:**
\[ \Delta \nu_{\text{Balmer}} = \nu_{\text{max}} - \nu_{\text{min}} = \frac{cR}{4} - cR \left( \frac{5}{36} \right) = cR \left( \frac{9}{36} - \frac{5}{36} \right) = cR \left( \frac{4}{36} \right) = cR \left( \frac{1}{9} \right) \]

### Step 5: Calculate the Ratio
Now, we can find the ratio of \( \Delta \nu_{\text{Lyman}} \) to \( \Delta \nu_{\text{Balmer}} \):
\[ \text{Ratio} = \frac{\Delta \nu_{\text{Lyman}}}{\Delta \nu_{\text{Balmer}}} = \frac{cR \left( \frac{1}{4} \right)}{cR \left( \frac{1}{9} \right)} = \frac{\frac{1}{4}}{\frac{1}{9}} = \frac{9}{4} \]

### Final Answer
Thus, the ratio of \( \Delta \nu \) for the Lyman series to the Balmer series is:
\[ \text{Ratio} = 9:4 \]