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 atomic number of the anti-cathode material as 41, we can use the modified Rydberg formula for X-rays. Here’s a step-by-step solution:

### Step 1: Understand the Rydberg Formula for X-rays
The formula for the wavelength of X-rays can be expressed as:

\[ \frac{1}{\lambda} = R \cdot (Z - b)^2 \cdot \left( \frac{1}{n^2} - \frac{1}{m^2} \right) \]

Where:
- \( \lambda \) is the wavelength of the X-ray.
- \( R \) is the Rydberg constant, approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \).
- \( Z \) is the atomic number of the anti-cathode material.
- \( b \) is a constant, which for K-alpha X-rays is equal to 1.
- \( n \) and \( m \) are the principal quantum numbers of the energy levels involved in the transition.

### Step 2: Identify the Values
For K-alpha X-rays:
- The transition is from \( n = 2 \) to \( n = 1 \), so \( n = 1 \) and \( m = 2 \).
- The atomic number \( Z \) is given as 41.
- The constant \( b \) is 1.

### Step 3: Substitute the Values into the Formula
Substituting the values into the formula:

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

\[ = 1.097 \times 10^7 \cdot (41 - 1)^2 \cdot \left( 1 - \frac{1}{4} \right) \]

### Step 4: Simplify the Equation
Calculating \( Z - b \):

\[ Z - b = 41 - 1 = 40 \]

Calculating \( (Z - b)^2 \):

\[ (40)^2 = 1600 \]

Calculating \( \left( 1 - \frac{1}{4} \right) \):

\[ 1 - \frac{1}{4} = \frac{3}{4} \]

Now substituting these values back into the equation:

\[ \frac{1}{\lambda} = 1.097 \times 10^7 \cdot 1600 \cdot \frac{3}{4} \]

### Step 5: Calculate the Result
Calculating the right-hand side:

\[ \frac{1}{\lambda} = 1.097 \times 10^7 \cdot 1600 \cdot 0.75 \]

Calculating \( 1.097 \times 10^7 \cdot 1600 \):

\[ = 1.7552 \times 10^{10} \]

Now multiplying by \( 0.75 \):

\[ \frac{1}{\lambda} = 1.7552 \times 10^{10} \cdot 0.75 = 1.3164 \times 10^{10} \]

### Step 6: Find \( \lambda \)
Taking the reciprocal to find \( \lambda \):

\[ \lambda = \frac{1}{1.3164 \times 10^{10}} \approx 0.76 \times 10^{-10} \, \text{m} \]

### Step 7: Convert to Angstroms
Since \( 1 \, \text{Å} = 10^{-10} \, \text{m} \):

\[ \lambda \approx 0.76 \, \text{Å} \]

### Final Answer
The wavelength of the K-alpha X-ray produced in the tube is approximately **0.76 Å**.

---