X-ray emitted from a copper target and a molybdenum target are found to contains a line of wavelength `22.85 n m` attributed to the `K_(alpha)` line of an impurity element . The `K_(alpha)` line of copper `(Z = 29)` and molybdenum `(Z = 42)` have wavelength `15.42 nm and 7.12 nm`, respectively. The atomic number of the impurity element is

Options

A. 22
B. 23
C. 24
D. 25

Explanation

To find the atomic number of the impurity element based on the given information about the K-alpha lines of copper and molybdenum, we can follow these steps:

### Step 1: Calculate the frequencies of the K-alpha lines for copper, molybdenum, and the impurity element.

The frequency (ν) can be calculated using the formula:
\[ \nu = \frac{c}{\lambda} \]
where \(c\) is the speed of light (\(3 \times 10^8 \, \text{m/s}\)) and \(\lambda\) is the wavelength.

1. **For Copper (Cu)**:
\[ \lambda_{\text{Cu}} = 15.42 \, \text{nm} = 15.42 \times 10^{-9} \, \text{m} \]
\[ \nu_{\text{Cu}} = \frac{3 \times 10^8}{15.42 \times 10^{-9}} \approx 1.95 \times 10^{16} \, \text{Hz} \]

2. **For Molybdenum (Mo)**:
\[ \lambda_{\text{Mo}} = 7.12 \, \text{nm} = 7.12 \times 10^{-9} \, \text{m} \]
\[ \nu_{\text{Mo}} = \frac{3 \times 10^8}{7.12 \times 10^{-9}} \approx 4.21 \times 10^{16} \, \text{Hz} \]

3. **For the Impurity Element (X)**:
\[ \lambda_{X} = 22.85 \, \text{nm} = 22.85 \times 10^{-9} \, \text{m} \]
\[ \nu_{X} = \frac{3 \times 10^8}{22.85 \times 10^{-9}} \approx 1.31 \times 10^{16} \, \text{Hz} \]

### Step 2: Apply Moseley's Law

Moseley's Law states:
\[ \sqrt{\nu} = A(Z - B) \]
where \(A\) and \(B\) are constants, and \(Z\) is the atomic number.

1. **For Copper**:
\[ \sqrt{1.95 \times 10^{16}} = A(29 - B) \quad \text{(Equation 1)} \]

2. **For Molybdenum**:
\[ \sqrt{4.21 \times 10^{16}} = A(42 - B) \quad \text{(Equation 2)} \]

### Step 3: Solve the equations

1. **Divide Equation 1 by Equation 2**:
\[ \frac{\sqrt{1.95 \times 10^{16}}}{\sqrt{4.21 \times 10^{16}}} = \frac{29 - B}{42 - B} \]

Simplifying gives:
\[ \frac{1.95}{4.21} = \left(\frac{29 - B}{42 - B}\right)^2 \]

2. **Calculate B**:
After solving the above equation, we find \(B \approx 1.3\).

### Step 4: Find A using one of the equations

Substituting \(B\) back into Equation 1:
\[ \sqrt{1.95 \times 10^{16}} = A(29 - 1.3) \]
Solving for \(A\) gives \(A \approx 5.13 \times 10^6\).

### Step 5: Use A and B to find the atomic number of the impurity element

Using the frequency of the impurity element:
\[ \sqrt{1.31 \times 10^{16}} = A(Z - 1.3) \]
Substituting \(A\):
\[ \sqrt{1.31 \times 10^{16}} = 5.13 \times 10^6(Z - 1.3) \]
Solving for \(Z\) gives \(Z \approx 24\).

### Conclusion
The atomic number of the impurity element is **24**.

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