The wavelength`  K_(alpha)` X-rays produced by an X-ray tube is 0.76 A The atomic number of anticathode materials is

Options

A. 82
B. 41
C. 20
D. 10

Explanation

To find the atomic number of the anticathode material based on the given wavelength of Kα X-rays, we can use the formula derived from the Rydberg formula for X-rays:

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

Where:
- \(\lambda\) is the wavelength of the emitted X-ray,
- \(R\) is the Rydberg constant (\(1.097 \times 10^7 \, \text{m}^{-1}\)),
- \(Z\) is the atomic number of the anticathode material,
- \(n_1\) and \(n_2\) are the principal quantum numbers of the energy levels involved in the transition.

### Step-by-step Solution:

1. **Convert Wavelength to Meters**:
The given wavelength is \(0.76 \, \text{Å}\) (angstroms). We need to convert this to meters:
\[ \lambda = 0.76 \, \text{Å} = 0.76 \times 10^{-10} \, \text{m} \]

2. **Identify the Quantum Numbers**:
For Kα X-rays, the transitions occur from \(n_2 = 2\) to \(n_1 = 1\).

3. **Substitute Values into the Formula**:
Substitute \(\lambda\), \(R\), \(n_1\), and \(n_2\) into the formula:
\[ \frac{1}{\lambda} = R \cdot (Z - 1)^2 \cdot \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]
Simplifying the right side:
\[ \frac{1}{1^2} - \frac{1}{2^2} = 1 - \frac{1}{4} = \frac{3}{4} \]
Thus, the equation becomes:
\[ \frac{1}{\lambda} = R \cdot (Z - 1)^2 \cdot \frac{3}{4} \]

4. **Calculate \(\frac{1}{\lambda}\)**:
\[ \frac{1}{\lambda} = \frac{1}{0.76 \times 10^{-10}} \approx 1.3158 \times 10^{10} \, \text{m}^{-1} \]

5. **Set Up the Equation**:
\[ 1.3158 \times 10^{10} = 1.097 \times 10^7 \cdot (Z - 1)^2 \cdot \frac{3}{4} \]

6. **Rearranging for \((Z - 1)^2\)**:
\[ (Z - 1)^2 = \frac{1.3158 \times 10^{10} \cdot 4}{1.097 \times 10^7 \cdot 3} \]

7. **Calculate the Right Side**:
\[ (Z - 1)^2 = \frac{5.2632 \times 10^{10}}{3.291 \times 10^7} \approx 1600 \]

8. **Take the Square Root**:
\[ Z - 1 = \sqrt{1600} = 40 \]

9. **Solve for \(Z\)**:
\[ Z = 40 + 1 = 41 \]

### Final Answer:
The atomic number of the anticathode material is \(Z = 41\).