The element which has a `K_(alpha)` X-rays line of wavelength `1.8 Å` is
`(R = 1.1 xx 10^(7) m^(-1), b = 1 and sqrt(5//33) = 0.39)`

Options

A. `C o, Z = 27`
B. `Iron, Z = 26`
C. `M n, Z = 25`
D. `N i, Z = 28`

Explanation

To solve the problem, we need to find the atomic number \( Z \) of the element that has a \( K_{\alpha} \) X-ray line with a wavelength of \( 1.8 \, \text{Å} \). We will use the formula relating the wavelength \( \lambda \) to the atomic number \( Z \):

\[ \frac{1}{\lambda} = R(Z - b)^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Where:
- \( R \) is the Rydberg constant, given as \( 1.1 \times 10^7 \, \text{m}^{-1} \)
- \( b \) is given as \( 1 \)
- \( n_1 = 1 \) and \( n_2 = 2 \) for the \( K_{\alpha} \) transition.

### Step-by-Step Solution

1. **Convert Wavelength to Meters:**
\[ \lambda = 1.8 \, \text{Å} = 1.8 \times 10^{-10} \, \text{m} \]

2. **Calculate \( \frac{1}{n_1^2} - \frac{1}{n_2^2} \):**
\[ n_1 = 1, \quad n_2 = 2 \]
\[ \frac{1}{n_1^2} - \frac{1}{n_2^2} = \frac{1}{1^2} - \frac{1}{2^2} = 1 - \frac{1}{4} = \frac{3}{4} \]

3. **Substituting into the Formula:**
\[ \frac{1}{\lambda} = R(Z - b)^2 \left( \frac{3}{4} \right) \]
Substituting \( R \) and \( b \):
\[ \frac{1}{1.8 \times 10^{-10}} = 1.1 \times 10^7 (Z - 1)^2 \left( \frac{3}{4} \right) \]

4. **Calculate \( \frac{1}{\lambda} \):**
\[ \frac{1}{1.8 \times 10^{-10}} \approx 5.56 \times 10^9 \, \text{m}^{-1} \]

5. **Rearranging the Equation:**
\[ 5.56 \times 10^9 = 1.1 \times 10^7 (Z - 1)^2 \left( \frac{3}{4} \right) \]
\[ (Z - 1)^2 = \frac{5.56 \times 10^9}{1.1 \times 10^7 \times \frac{3}{4}} \]

6. **Calculating the Right Side:**
\[ (Z - 1)^2 = \frac{5.56 \times 10^9}{8.25 \times 10^6} \approx 674.55 \]
\[ Z - 1 = \sqrt{674.55} \approx 25.94 \]

7. **Finding \( Z \):**
\[ Z \approx 25.94 + 1 \approx 26.94 \]
Rounding to the nearest whole number, we find:
\[ Z \approx 27 \]

### Conclusion
The element with a \( K_{\alpha} \) X-ray line of wavelength \( 1.8 \, \text{Å} \) corresponds to the atomic number \( Z = 27 \), which is **Cobalt (Co)**.