A certain species of ionized atoms produces emission line spectral according to the Bohr's model. A group of lines in the spectrum is forming a series in which in the shortest wavelength is `22.79 nm` and the longest wavelength is `41.02 nm`. The atomic number of atom is `Z`.
Based on above information, answer the following question:
The series belongs to

Options

A. Lyman
B. Balmer
C. Paschen
D. Brackett

Explanation

To solve the problem, we need to determine which series the given wavelengths belong to based on the information provided. We have the shortest wavelength (λ_min) of 22.79 nm and the longest wavelength (λ_max) of 41.02 nm. We can use the Bohr model to find the atomic number (Z) and identify the series.

### Step-by-Step Solution:

1. **Identify the Wavelengths**:
- Shortest wavelength (λ_min) = 22.79 nm
- Longest wavelength (λ_max) = 41.02 nm

2. **Use the Bohr Model Formulas**:
The energy levels in the Bohr model for hydrogen-like atoms are given by:
\[ E_n = -\frac{13.6 Z^2}{n^2} \text{ eV} \]
The wavelength of emitted radiation can be calculated using:
\[ \frac{hc}{\lambda} = E_i - E_f \]
where \(E_i\) and \(E_f\) are the energies of the initial and final states, respectively.

3. **Calculate the Maximum Wavelength**:
The longest wavelength corresponds to the smallest energy transition. This typically occurs when the transition is from \(n = 2\) to \(n = 1\):
\[ \frac{hc}{\lambda_{max}} = E_1 - E_2 \]
Substituting the values:
\[ \frac{hc}{41.02 \times 10^{-9}} = \left(-\frac{13.6 Z^2}{1^2}\right) - \left(-\frac{13.6 Z^2}{2^2}\right) \]
Simplifying gives:
\[ \frac{hc}{41.02 \times 10^{-9}} = 13.6 Z^2 \left(\frac{1}{1} - \frac{1}{4}\right) = 13.6 Z^2 \cdot \frac{3}{4} \]

4. **Calculate the Minimum Wavelength**:
The shortest wavelength corresponds to the largest energy transition, typically from \(n = \infty\) to \(n = 2\):
\[ \frac{hc}{\lambda_{min}} = E_2 - E_\infty \]
Substituting the values:
\[ \frac{hc}{22.79 \times 10^{-9}} = \left(-\frac{13.6 Z^2}{2^2}\right) - 0 \]
Simplifying gives:
\[ \frac{hc}{22.79 \times 10^{-9}} = -\left(-\frac{13.6 Z^2}{4}\right) = \frac{13.6 Z^2}{4} \]

5. **Equate the Two Equations**:
Now we have two equations:
\[ \frac{hc}{41.02 \times 10^{-9}} = 10.2 Z^2 \]
\[ \frac{hc}{22.79 \times 10^{-9}} = 3.4 Z^2 \]

6. **Solve for Z**:
Dividing the two equations:
\[ \frac{22.79}{41.02} = \frac{3.4 Z^2}{10.2 Z^2} \]
Simplifying gives:
\[ Z^2 = \frac{22.79 \times 10.2}{41.02 \times 3.4} \]
Calculate \(Z\):
\[ Z^2 = \frac{232.638}{139.468} \approx 1.67 \implies Z \approx 4 \]

7. **Identify the Series**:
Since the final state corresponds to \(n = 2\), this series is the **Balmer series**.

### Final Answer:
The series belongs to the **Balmer series**.