The shortest wavelength of the Brackett series of a hydrogen-like atom (atomic number of `Z` ) is the same as the shortest wavelength of the Balmer series of hydrogen atom. The value of `z` is

Options

A. 2
B. 3
C. 4
D. 6

Answer

To solve the problem, we need to find the atomic number \( Z \) of a hydrogen-like atom such that the shortest wavelength of the Brackett series of this atom is the same as the shortest wavelength of the Balmer series of a hydrogen atom.

### Step-by-Step Solution:

1. **Understanding the Series**:
- The **Balmer series** of hydrogen corresponds to transitions from higher energy levels (n = 3, 4, ...) to n = 2.
- The **Brackett series** corresponds to transitions from higher energy levels (n = 5, 6, ...) to n = 4.

2. **Finding the Shortest Wavelength of the Balmer Series**:
- The shortest wavelength in the Balmer series occurs when the electron transitions from \( n = \infty \) to \( n = 2 \).
- The formula for the wavelength is given by:
\[ \frac{1}{\lambda_1} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = R \left( \frac{1}{4} - 0 \right) = \frac{R}{4} \]
- Here, \( R \) is the Rydberg constant.

3. **Finding the Shortest Wavelength of the Brackett Series**:
- The shortest wavelength in the Brackett series occurs when the electron transitions from \( n = \infty \) to \( n = 4 \).
- The formula for the wavelength is:
\[ \frac{1}{\lambda_2} = R Z^2 \left( \frac{1}{4^2} - \frac{1}{\infty^2} \right) = R Z^2 \left( \frac{1}{16} - 0 \right) = \frac{R Z^2}{16} \]

4. **Setting the Wavelengths Equal**:
- Since the shortest wavelengths are equal, we set \( \lambda_1 = \lambda_2 \):
\[ \frac{R}{4} = \frac{R Z^2}{16} \]

5. **Canceling Rydberg Constant**:
- We can cancel \( R \) from both sides (assuming \( R \neq 0 \)):
\[ \frac{1}{4} = \frac{Z^2}{16} \]

6. **Cross-Multiplying**:
- Cross-multiplying gives:
\[ 16 = 4 Z^2 \]

7. **Solving for Z**:
- Dividing both sides by 4:
\[ Z^2 = 4 \]
- Taking the square root:
\[ Z = 2 \]

### Conclusion:
The value of \( Z \) is \( 2 \).