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 for which the shortest wavelength of the Brackett series is the same as that of the Balmer series of hydrogen.

### Step-by-Step Solution:

1. **Understand the Wavelength Formula**:
The wavelength \( \lambda \) for a transition in a hydrogen-like atom can be expressed using the formula:
\[ \frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R \) is the Rydberg constant, \( Z \) is the atomic number, \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level.

2. **Calculate Wavelength for the Balmer Series of Hydrogen**:
For the Balmer series in hydrogen (\( Z = 1 \)):
- The lowest transition occurs from \( n_2 = \infty \) to \( n_1 = 2 \).
- Plugging in the values:
\[ \frac{1}{\lambda_{\text{Balmer}}} = R \cdot 1^2 \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = R \left( \frac{1}{4} - 0 \right) = \frac{R}{4} \]
Thus, we have:
\[ \lambda_{\text{Balmer}} = \frac{4}{R} \]

3. **Calculate Wavelength for the Brackett Series of Hydrogen-like Atom**:
For the Brackett series:
- The lowest transition occurs from \( n_2 = \infty \) to \( n_1 = 4 \).
- Using the formula:
\[ \frac{1}{\lambda_{\text{Brackett}}} = R Z^2 \left( \frac{1}{4^2} - \frac{1}{\infty^2} \right) = R Z^2 \left( \frac{1}{16} - 0 \right) = \frac{R Z^2}{16} \]
Thus, we have:
\[ \lambda_{\text{Brackett}} = \frac{16}{R Z^2} \]

4. **Set the Wavelengths Equal**:
According to the problem, the wavelengths are equal:
\[ \lambda_{\text{Balmer}} = \lambda_{\text{Brackett}} \]
Therefore:
\[ \frac{4}{R} = \frac{16}{R Z^2} \]

5. **Solve for \( Z \)**:
Cancel \( R \) from both sides:
\[ 4 = \frac{16}{Z^2} \]
Rearranging gives:
\[ Z^2 = \frac{16}{4} = 4 \]
Taking the square root:
\[ Z = 2 \]

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

---