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 given that the shortest wavelength in the Balmer series is λ, we can follow these steps:

### Step 1: Understand the Formula for Wavelength
The formula for the wavelength of light emitted during an electron transition in a hydrogen atom 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, and \( n_2 \) is the higher energy level.

### Step 2: Determine the Shortest Wavelength in the Balmer Series
In the Balmer series, the transitions occur to \( n_1 = 2 \). The shortest wavelength corresponds to the transition from \( n_2 = \infty \) to \( n_1 = 2 \):
\[ \frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = R \left( \frac{1}{4} - 0 \right) = \frac{R}{4} \]
Thus, we can express the wavelength in the Balmer series as:
\[ \lambda_B = \frac{4}{R} \]
Given that \( \lambda_B = \lambda \), we have:
\[ \lambda = \frac{4}{R} \]

### Step 3: Determine the Shortest Wavelength in the Brackett Series
In the Brackett series, the transitions occur to \( n_1 = 4 \). The shortest wavelength corresponds to the transition from \( n_2 = \infty \) to \( n_1 = 4 \):
\[ \frac{1}{\lambda_Br} = R \left( \frac{1}{4^2} - \frac{1}{\infty^2} \right) = R \left( \frac{1}{16} - 0 \right) = \frac{R}{16} \]
Thus, we can express the wavelength in the Brackett series as:
\[ \lambda_Br = \frac{16}{R} \]

### Step 4: Relate the Wavelengths
Now, we can relate the shortest wavelength in the Brackett series to that in the Balmer series:
\[ \lambda_Br = 4 \cdot \lambda \]

### Final Answer
The shortest wavelength in the Brackett series is:
\[ \lambda_Br = 4\lambda \]

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