The wavelength of `K_(alpha)` X-rays produced by an X-ray tube is 1.785 Å. Find the atomic number of the anode material of the tube `(R=109737 cm^(-1))`

Options

A. 24
B. 32
C. 48
D. 27

Explanation

To find the atomic number of the anode material of the X-ray tube given the wavelength of K-alpha X-rays, we can use the modified Rydberg formula for X-rays. The steps are as follows:

### Step-by-Step Solution:

1. **Understand the Relationship**: The wavelength of K-alpha X-rays can be related to the atomic number (Z) of the anode material using the formula:
\[ \frac{1}{\lambda} = R \left( Z - 1 \right)^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( n_1 = 1 \) and \( n_2 = 2 \) for K-alpha transition.

2. **Substituting Values**: Given:
- Wavelength \( \lambda = 1.785 \, \text{Å} = 1.785 \times 10^{-8} \, \text{cm} \)
- Rydberg constant \( R = 109737 \, \text{cm}^{-1} \)
- Transition states \( n_1 = 1 \) and \( n_2 = 2 \)

Substitute these values into the equation:
\[ \frac{1}{1.785 \times 10^{-8}} = 109737 \left( Z - 1 \right)^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]

3. **Calculate the Difference of Squares**: The term \( \frac{1}{1^2} - \frac{1}{2^2} \) simplifies to:
\[ 1 - \frac{1}{4} = \frac{3}{4} \]

4. **Rearranging the Equation**: Now, we can rewrite the equation:
\[ \frac{1}{1.785 \times 10^{-8}} = 109737 \left( Z - 1 \right)^2 \cdot \frac{3}{4} \]

5. **Isolate \( (Z - 1)^2 \)**: Rearranging gives:
\[ (Z - 1)^2 = \frac{4}{3} \cdot \frac{1}{1.785 \times 10^{-8}} \cdot \frac{1}{109737} \]

6. **Calculate the Right Side**: First, calculate \( \frac{1}{1.785 \times 10^{-8}} \):
\[ \frac{1}{1.785 \times 10^{-8}} \approx 56000000 \, \text{cm}^{-1} \]
Then calculate:
\[ (Z - 1)^2 = \frac{4}{3} \cdot 56000000 \cdot \frac{1}{109737} \]

7. **Final Calculation**: This simplifies to:
\[ (Z - 1)^2 \approx 680.686 \]

8. **Taking the Square Root**: Now take the square root:
\[ Z - 1 \approx \sqrt{680.686} \approx 26.12 \]

9. **Finding \( Z \)**: Finally, add 1 to find \( Z \):
\[ Z \approx 26.12 + 1 \approx 27.12 \]

10. **Conclusion**: The atomic number of the anode material is approximately 27.

### Final Answer:
The atomic number of the anode material is approximately **27**.