If `lambda_1, lambda_2, lambda_3` are wavelengths of first 3 lines of balmer series. Find `lambda_1/lambda_3`

Answer

To solve the problem of finding the ratio \( \frac{\lambda_1}{\lambda_3} \) for the first three lines of the Balmer series, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the transitions for the first three lines of the Balmer series**:
- The Balmer series corresponds to transitions to \( n_1 = 2 \).
- The first three lines correspond to:
- \( \lambda_1 \): Transition from \( n_2 = 3 \) to \( n_1 = 2 \)
- \( \lambda_2 \): Transition from \( n_2 = 4 \) to \( n_1 = 2 \)
- \( \lambda_3 \): Transition from \( n_2 = 5 \) to \( n_1 = 2 \)

2. **Use the Rydberg formula for wavelengths**:
The Rydberg formula for the wavelength of emitted light 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.

3. **Calculate \( \lambda_1 \)**:
For \( n_2 = 3 \) and \( n_1 = 2 \):
\[ \frac{1}{\lambda_1} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) \]
Finding a common denominator (36):
\[ \frac{1}{\lambda_1} = R \left( \frac{9 - 4}{36} \right) = \frac{5R}{36} \]
Thus,
\[ \lambda_1 = \frac{36}{5R} \]

4. **Calculate \( \lambda_2 \)**:
For \( n_2 = 4 \) and \( n_1 = 2 \):
\[ \frac{1}{\lambda_2} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{4} - \frac{1}{16} \right) \]
Finding a common denominator (16):
\[ \frac{1}{\lambda_2} = R \left( \frac{4 - 1}{16} \right) = \frac{3R}{16} \]
Thus,
\[ \lambda_2 = \frac{16}{3R} \]

5. **Calculate \( \lambda_3 \)**:
For \( n_2 = 5 \) and \( n_1 = 2 \):
\[ \frac{1}{\lambda_3} = R \left( \frac{1}{2^2} - \frac{1}{5^2} \right) = R \left( \frac{1}{4} - \frac{1}{25} \right) \]
Finding a common denominator (100):
\[ \frac{1}{\lambda_3} = R \left( \frac{25 - 4}{100} \right) = \frac{21R}{100} \]
Thus,
\[ \lambda_3 = \frac{100}{21R} \]

6. **Calculate the ratio \( \frac{\lambda_1}{\lambda_3} \)**:
\[ \frac{\lambda_1}{\lambda_3} = \frac{\frac{36}{5R}}{\frac{100}{21R}} = \frac{36 \cdot 21}{5 \cdot 100} \]
Simplifying this:
\[ \frac{\lambda_1}{\lambda_3} = \frac{756}{500} \]

### Final Answer:
\[ \frac{\lambda_1}{\lambda_3} = \frac{756}{500} \]