What would be maximum wavelength for Brackett series of hydrogen-spectrum?

Options

A. ` 74583 Å`
B. `22790Å`
C. `40519Å`
D. `18753Å`

Explanation

To find the maximum wavelength for the Brackett series of the hydrogen spectrum, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Brackett Series**:
The Brackett series corresponds to the transitions of electrons from higher energy levels (n = 5, 6, 7, ...) to the third energy level (n = 4) in a hydrogen atom.

2. **Identify the Transition for Maximum Wavelength**:
The maximum wavelength occurs when the electron transitions from the lowest possible energy level in the series to n = 4. This means the transition from n = 5 to n = 4 will give us the maximum wavelength.

3. **Use the Rydberg Formula**:
The Rydberg formula for the wavelength of light emitted during electron transitions 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, \( n_1 \) is the lower energy level (4 for Brackett series), and \( n_2 \) is the higher energy level (5 for maximum wavelength).

4. **Substitute Values into the Formula**:
- Rydberg constant \( R = 1.09678 \times 10^7 \, \text{m}^{-1} \)
- For the Brackett series, \( n_1 = 4 \) and \( n_2 = 5 \):
\[ \frac{1}{\lambda} = R \left( \frac{1}{4^2} - \frac{1}{5^2} \right) \]

5. **Calculate the Values**:
- Calculate \( \frac{1}{4^2} = \frac{1}{16} \)
- Calculate \( \frac{1}{5^2} = \frac{1}{25} \)
- Now, find the difference:
\[ \frac{1}{16} - \frac{1}{25} = \frac{25 - 16}{400} = \frac{9}{400} \]

6. **Substitute Back into the Rydberg Formula**:
\[ \frac{1}{\lambda} = 1.09678 \times 10^7 \left( \frac{9}{400} \right) \]

7. **Calculate \( \lambda \)**:
- First, calculate \( 1.09678 \times 10^7 \times \frac{9}{400} \):
\[ \frac{1.09678 \times 9}{400} \times 10^7 = \frac{9.87102}{400} \times 10^7 = 2.467755 \times 10^5 \, \text{m}^{-1} \]
- Now, take the reciprocal to find \( \lambda \):
\[ \lambda = \frac{1}{2.467755 \times 10^5} \approx 4.05 \times 10^{-6} \, \text{m} = 40519 \, \text{Å} \]

### Final Answer:
The maximum wavelength for the Brackett series of the hydrogen spectrum is approximately **40519 Å**.