The wavelength of `k_(alpha) ` X- rays produced by an X - rays tube is `0.76 Å` . The atomic number of the anode material of the tube is …….

Options

A. 20
B. 60
C. 41
D. 80

Explanation

To find the atomic number of the anode material of the X-ray tube that produces K-alpha X-rays with a wavelength of 0.76 Å, we can use the Rydberg formula for the K-alpha transition. Here’s the step-by-step solution:

### Step 1: Convert Wavelength to Meters
The given wavelength of K-alpha X-rays is 0.76 Å. We need to convert this to meters for calculations.

\[ \text{Wavelength} (\lambda) = 0.76 \, \text{Å} = 0.76 \times 10^{-10} \, \text{m} \]

### Step 2: Use the Rydberg Formula
The Rydberg formula for the wavelength of emitted X-rays can be expressed as:

\[ \frac{1}{\lambda} = \frac{3}{4} R (Z - 1)^2 \]

Where:
- \( \lambda \) is the wavelength,
- \( R \) is the Rydberg constant, approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \),
- \( Z \) is the atomic number of the anode material.

### Step 3: Substitute the Values into the Formula
Substituting the known values into the equation:

\[ \frac{1}{0.76 \times 10^{-10}} = \frac{3}{4} \times 1.097 \times 10^7 \times (Z - 1)^2 \]

### Step 4: Calculate \( \frac{1}{\lambda} \)
Calculating \( \frac{1}{\lambda} \):

\[ \frac{1}{0.76 \times 10^{-10}} \approx 1.3158 \times 10^{10} \, \text{m}^{-1} \]

### Step 5: Rearranging the Equation
Now we can rearrange the equation to solve for \( (Z - 1)^2 \):

\[ 1.3158 \times 10^{10} = \frac{3}{4} \times 1.097 \times 10^7 \times (Z - 1)^2 \]

### Step 6: Calculate \( (Z - 1)^2 \)
Calculating the right side:

\[ \frac{3}{4} \times 1.097 \times 10^7 \approx 8.2275 \times 10^6 \]

Now, substituting this back into the equation:

\[ 1.3158 \times 10^{10} = 8.2275 \times 10^6 \times (Z - 1)^2 \]

Dividing both sides by \( 8.2275 \times 10^6 \):

\[ (Z - 1)^2 = \frac{1.3158 \times 10^{10}}{8.2275 \times 10^6} \approx 1600 \]

### Step 7: Solve for \( Z - 1 \)
Taking the square root of both sides:

\[ Z - 1 = \sqrt{1600} = 40 \]

### Step 8: Find the Atomic Number \( Z \)
Adding 1 to both sides gives:

\[ Z = 40 + 1 = 41 \]

### Conclusion
The atomic number of the anode material of the tube is \( Z = 41 \).

### Answer
Thus, the correct option is **C) 41**.