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 use the Rydberg formula, which 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 (\( R = 1.09678 \times 10^7 \, \text{m}^{-1} \)),
- \( n_1 \) is the lower energy level,
- \( n_2 \) is the higher energy level.

### Step 1: Identify the Energy Levels
For the Brackett series, the lower energy level \( n_1 \) is 4 (since it transitions to the fourth level). The higher energy levels \( n_2 \) can be 5, 6, 7, etc.

### Step 2: Determine the Maximum Wavelength
To find the maximum wavelength, we need to minimize the term \( \frac{1}{n_2^2} \). This occurs when \( n_2 \) is at its minimum value, which is 5.

### Step 3: Substitute Values into the Rydberg Formula
Substituting \( n_1 = 4 \) and \( n_2 = 5 \) into the Rydberg formula:

\[ \frac{1}{\lambda} = R \left( \frac{1}{4^2} - \frac{1}{5^2} \right) \]

Calculating the squares:

\[ 4^2 = 16 \quad \text{and} \quad 5^2 = 25 \]

So:

\[ \frac{1}{\lambda} = R \left( \frac{1}{16} - \frac{1}{25} \right) \]

### Step 4: Find a Common Denominator
The common denominator for 16 and 25 is 400. Thus:

\[ \frac{1}{16} = \frac{25}{400} \quad \text{and} \quad \frac{1}{25} = \frac{16}{400} \]

Now we can rewrite the equation:

\[ \frac{1}{\lambda} = R \left( \frac{25 - 16}{400} \right) = R \left( \frac{9}{400} \right) \]

### Step 5: Substitute the Rydberg Constant
Now substituting \( R \):

\[ \frac{1}{\lambda} = 1.09678 \times 10^7 \left( \frac{9}{400} \right) \]

Calculating this gives:

\[ \frac{1}{\lambda} = \frac{9.87002 \times 10^6}{400} \]

### Step 6: Calculate \( \lambda \)
Taking the reciprocal to find \( \lambda \):

\[ \lambda = \frac{400}{9.87002 \times 10^6} \]

Calculating this gives:

\[ \lambda \approx 4.0519 \times 10^{-7} \, \text{m} \]

### Step 7: Convert to Angstroms
To convert meters to angstroms (1 m = \( 10^{10} \) angstroms):

\[ \lambda \approx 4.0519 \times 10^{-7} \times 10^{10} \, \text{angstroms} \approx 4051.9 \, \text{angstroms} \]

### Conclusion
Thus, the maximum wavelength for the Brackett series of the hydrogen spectrum is approximately **4051.9 angstroms**.