Wavelength of the first line of Paschen series is `- (R = 109700 cm^-1`)

Options

A. `[18750 Å]`
B. `[2854 Å]`
C. `[3452 Å]`
D. `[6243 Å]`

Explanation

To find the wavelength of the first line of the Paschen series, we will use the Rydberg formula for hydrogen spectral lines. The Paschen series corresponds to transitions where the electron falls to the n=3 energy level. The first line of the Paschen series occurs when the electron transitions from n=4 to n=3.

### Step-by-Step Solution:

1. **Identify the Rydberg Formula**:
The Rydberg formula is given by:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \( \lambda \) is the wavelength,
- \( R \) is the Rydberg constant,
- \( n_1 \) is the lower energy level,
- \( n_2 \) is the higher energy level.

2. **Assign Values for the Paschen Series**:
For the first line of the Paschen series:
- \( n_1 = 3 \)
- \( n_2 = 4 \)

3. **Substitute Values into the Formula**:
The Rydberg constant \( R \) is given as \( 109700 \, \text{cm}^{-1} \). Substituting the values into the formula:
\[ \frac{1}{\lambda} = 109700 \left( \frac{1}{3^2} - \frac{1}{4^2} \right) \]

4. **Calculate \( \frac{1}{3^2} \) and \( \frac{1}{4^2} \)**:
- \( \frac{1}{3^2} = \frac{1}{9} \)
- \( \frac{1}{4^2} = \frac{1}{16} \)

5. **Find a Common Denominator**:
The common denominator for 9 and 16 is 144. Thus:
\[ \frac{1}{9} = \frac{16}{144}, \quad \frac{1}{16} = \frac{9}{144} \]
Therefore:
\[ \frac{1}{3^2} - \frac{1}{4^2} = \frac{16}{144} - \frac{9}{144} = \frac{7}{144} \]

6. **Substitute Back into the Formula**:
Now substituting back:
\[ \frac{1}{\lambda} = 109700 \left( \frac{7}{144} \right) \]

7. **Calculate \( \frac{1}{\lambda} \)**:
\[ \frac{1}{\lambda} = \frac{109700 \times 7}{144} \]
\[ \frac{1}{\lambda} = \frac{768900}{144} \approx 5332.64 \, \text{cm}^{-1} \]

8. **Find \( \lambda \)**:
To find \( \lambda \), take the reciprocal:
\[ \lambda = \frac{1}{5332.64} \approx 0.00018752 \, \text{cm} \]

9. **Convert to Meters**:
Since \( 1 \, \text{cm} = 10^{-2} \, \text{m} \):
\[ \lambda \approx 0.00018752 \, \text{cm} = 1.8752 \times 10^{-5} \, \text{m} = 18752 \times 10^{-10} \, \text{m} \]

### Final Answer:
The wavelength of the first line of the Paschen series is approximately \( 1.8752 \times 10^{-5} \, \text{m} \) or \( 18752 \, \text{Å} \).