The `K_(alpha)` line obtained for molybdenum `(Z = 42)` is 0.71 Å. Then, the wavelength of the `K_(alpha)` line of copper (Z = 29) is

Options

A. 2.14 Å
B. 1.52 Å
C. 1.04 Å
D. 1.71 Å

Answer

To find the wavelength of the K_alpha line for copper (Z = 29) given the K_alpha line for molybdenum (Z = 42) is 0.71 Å, we can use the formula that relates the wavelengths of K_alpha lines for different elements:

\[ \frac{\lambda_2}{\lambda_1} = \frac{(Z_1 - b)^2}{(Z_2 - b)^2} \]

Where:
- \(\lambda_1\) is the wavelength of the K_alpha line for molybdenum (0.71 Å).
- \(Z_1\) is the atomic number of molybdenum (42).
- \(Z_2\) is the atomic number of copper (29).
- \(b\) is a constant, which is typically taken as 1 for K_alpha transitions.

### Step-by-Step Solution:

1. **Identify the known values:**
- \(\lambda_1 = 0.71 \, \text{Å}\)
- \(Z_1 = 42\) (for molybdenum)
- \(Z_2 = 29\) (for copper)
- \(b = 1\)

2. **Substitute the values into the formula:**
\[ \frac{\lambda_2}{0.71} = \frac{(42 - 1)^2}{(29 - 1)^2} \]

3. **Calculate \(Z_1 - b\) and \(Z_2 - b\):**
- \(Z_1 - b = 42 - 1 = 41\)
- \(Z_2 - b = 29 - 1 = 28\)

4. **Substitute these values into the equation:**
\[ \frac{\lambda_2}{0.71} = \frac{41^2}{28^2} \]

5. **Calculate \(41^2\) and \(28^2\):**
- \(41^2 = 1681\)
- \(28^2 = 784\)

6. **Now substitute these squared values back into the equation:**
\[ \frac{\lambda_2}{0.71} = \frac{1681}{784} \]

7. **Cross-multiply to solve for \(\lambda_2\):**
\[ \lambda_2 = 0.71 \times \frac{1681}{784} \]

8. **Calculate the right-hand side:**
\[ \lambda_2 = 0.71 \times 2.141 \]
(where \( \frac{1681}{784} \approx 2.141\))

9. **Final calculation:**
\[ \lambda_2 \approx 1.52 \, \text{Å} \]

### Final Answer:
The wavelength of the K_alpha line of copper is approximately \(1.52 \, \text{Å}\).