In hydrogen spectrum, the shortest wavelength in Balmer series is the `lambda`. The shortest wavelength in the Brackett series will be

Options

A. `2 lambda`
B. `4 lambda`
C. `9 lambda`
D. `16 lambda`

Explanation

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

### Step 1: Understand the Series
The Balmer series corresponds to transitions where the final energy level (n1) is 2. The Brackett series corresponds to transitions where the final energy level (n1) is 4.

### Step 2: Identify the Wavelength Formula
The wavelength of the emitted light during a transition in the hydrogen atom can be calculated using the Rydberg formula:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \( R \) is the Rydberg constant,
- \( n_1 \) is the lower energy level,
- \( n_2 \) is the higher energy level.

### Step 3: Determine the Values for the Brackett Series
For the Brackett series, the lowest energy level is \( n_1 = 4 \) and the transitions can go to higher levels starting from \( n_2 = 5, 6, \ldots \) up to infinity. The shortest wavelength corresponds to the transition from \( n_2 = \infty \) to \( n_1 = 4 \).

### Step 4: Apply the Rydberg Formula
Substituting the values into the Rydberg formula for the Brackett series:
\[ \frac{1}{\lambda_{\text{Brackett}}} = R \left( \frac{1}{4^2} - \frac{1}{\infty^2} \right) \]
Since \( \frac{1}{\infty^2} = 0 \), the equation simplifies to:
\[ \frac{1}{\lambda_{\text{Brackett}}} = R \left( \frac{1}{16} \right) \]

### Step 5: Solve for the Wavelength
Rearranging the equation gives:
\[ \lambda_{\text{Brackett}} = \frac{16}{R} \]

### Step 6: Relate to the Wavelength in the Balmer Series
From the Balmer series, we know that the shortest wavelength \( \lambda_{\text{Balmer}} \) corresponds to the transition from \( n_2 = \infty \) to \( n_1 = 2 \):
\[ \frac{1}{\lambda_{\text{Balmer}}} = R \left( \frac{1}{2^2} - 0 \right) = \frac{R}{4} \]
Thus,
\[ \lambda_{\text{Balmer}} = \frac{4}{R} \]

### Step 7: Relate the Two Wavelengths
Since we have:
\[ \lambda_{\text{Brackett}} = \frac{16}{R} = 4 \cdot \frac{4}{R} = 4 \cdot \lambda_{\text{Balmer}} \]

### Conclusion
The shortest wavelength in the Brackett series is:
\[ \lambda_{\text{Brackett}} = 4 \lambda_{\text{Balmer}} \]

### Final Answer
The shortest wavelength in the Brackett series will be \( 4\lambda \).

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