Atomic number of anti cathode material in an X- ray tube is 41. Wavelength of `K_(a)` X-ray produced in the tube is

Options

A. 0.66 A
B. 0.76 A
C. 0.82 A
D. 0.88 A

Answer

To find the wavelength of the K-alpha X-ray produced in an X-ray tube with an anti-cathode material of atomic number 41, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Atomic Number (Z)**:
The atomic number of the anti-cathode material is given as \( Z = 41 \).

2. **Use Moseley's Law for Wavelength**:
According to Moseley's law, the relationship for the wavelength \( \lambda \) of the emitted X-ray can be expressed as:
\[ \frac{1}{\lambda} = R \cdot (Z - b)^2 \cdot \left( \frac{1}{n_L^2} - \frac{1}{n_H^2} \right) \]
where:
- \( R \) is the Rydberg constant, approximately \( 1.09 \times 10^7 \, \text{m}^{-1} \).
- \( b \) is a constant that accounts for screening effects (for K-alpha transition, \( b = 1 \)).
- \( n_L \) and \( n_H \) are the principal quantum numbers of the lower and higher energy levels, respectively.

3. **Determine Quantum Numbers for K-alpha Transition**:
For K-alpha transition:
- The lower energy level \( n_L = 1 \) (K shell).
- The higher energy level \( n_H = 2 \) (L shell).

4. **Calculate the Difference in Quantum Numbers**:
Substitute \( n_L \) and \( n_H \) into the equation:
\[ \frac{1}{n_L^2} - \frac{1}{n_H^2} = \frac{1}{1^2} - \frac{1}{2^2} = 1 - \frac{1}{4} = \frac{3}{4} \]

5. **Substitute Values into Moseley's Law**:
Now substitute \( Z = 41 \), \( b = 1 \), and the calculated value into the equation:
\[ \frac{1}{\lambda} = R \cdot (41 - 1)^2 \cdot \frac{3}{4} \]
\[ \frac{1}{\lambda} = 1.09 \times 10^7 \cdot (40)^2 \cdot \frac{3}{4} \]
\[ = 1.09 \times 10^7 \cdot 1600 \cdot \frac{3}{4} \]

6. **Calculate the Value**:
\[ = 1.09 \times 10^7 \cdot 1200 \]
\[ = 1.308 \times 10^{10} \, \text{m}^{-1} \]

7. **Find Wavelength \( \lambda \)**:
Now, take the reciprocal to find \( \lambda \):
\[ \lambda = \frac{1}{1.308 \times 10^{10}} \approx 7.64 \times 10^{-11} \, \text{m} \]

8. **Convert to Angstroms**:
Since \( 1 \, \text{Angstrom} = 10^{-10} \, \text{m} \):
\[ \lambda \approx 0.764 \, \text{Angstroms} \]

### Final Answer:
The wavelength of the K-alpha X-ray produced in the tube is approximately \( 0.764 \, \text{Angstroms} \).