From what meterial is the anod of an X-ray tube made if the `K_alpha` line wavelength of thej characteristic spectrum is `0.76 Å`?

Explanation

To determine the material of the anode of an X-ray tube given the K-alpha line wavelength of the characteristic spectrum is 0.76 Å, we can follow these steps:

### Step 1: Understand the relationship between wavelength and atomic number
The K-alpha line wavelength can be related to the atomic number (Z) of the material using the formula:
\[ \frac{c}{\lambda_{K\alpha}} = 2.48 \times 10^{15} (Z - 1)^2 \]
where:
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \))
- \( \lambda_{K\alpha} \) is the wavelength of the K-alpha line (in meters)

### Step 2: Convert the wavelength from angstroms to meters
Given:
\[ \lambda_{K\alpha} = 0.76 \, \text{Å} = 0.76 \times 10^{-10} \, \text{m} \]

### Step 3: Substitute the values into the equation
Substituting the values into the equation:
\[ \frac{3 \times 10^8}{0.76 \times 10^{-10}} = 2.48 \times 10^{15} (Z - 1)^2 \]

### Step 4: Calculate the left-hand side
Calculating the left-hand side:
\[ \frac{3 \times 10^8}{0.76 \times 10^{-10}} = \frac{3 \times 10^8}{7.6 \times 10^{-11}} \approx 3.947 \times 10^{20} \]

### Step 5: Set up the equation
Now we can set up the equation:
\[ 3.947 \times 10^{20} = 2.48 \times 10^{15} (Z - 1)^2 \]

### Step 6: Solve for \( (Z - 1)^2 \)
Rearranging gives:
\[ (Z - 1)^2 = \frac{3.947 \times 10^{20}}{2.48 \times 10^{15}} \]
Calculating the right-hand side:
\[ (Z - 1)^2 \approx 1.591 \times 10^{5} \]

### Step 7: Take the square root
Taking the square root:
\[ Z - 1 \approx \sqrt{1.591 \times 10^{5}} \approx 399 \]
Thus:
\[ Z \approx 400 \]

### Step 8: Identify the material
The atomic number \( Z \) corresponds to the element Niobium (Nb), which has an atomic number of 41.

### Final Answer
The material of the anode of the X-ray tube is Niobium (Nb).

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