The `K_(alpha)` line from molybdenum (atomic number=42) has a wavelength off .7078Å. The wavelength off `K_(alpha)` line of zinc (atomic number=30) will be

Options

A. 1Å
B. 1.3872Å
C. 0.3541Å
D. 0.5Å

Explanation

To find the wavelength of the K-alpha line for zinc (atomic number = 30) given the wavelength of the K-alpha line for molybdenum (atomic number = 42) is 0.7078 Å, we can use the relationship derived from Moseley's law.

### Step-by-Step Solution:

1. **Understanding Moseley's Law**:
Moseley's law states that the frequency (ν) of the emitted X-ray is proportional to the square of the atomic number (Z) of the element minus a constant (b). For K-alpha lines, b is typically taken as 1.
\[ \nu \propto (Z - b)^2 \]

2. **Relating Wavelength and Frequency**:
The relationship between wavelength (λ) and frequency (ν) is given by:
\[ c = \nu \cdot \lambda \]
where \( c \) is the speed of light. This implies that:
\[ \nu = \frac{c}{\lambda} \]

3. **Setting Up the Proportionality**:
Since frequency is inversely proportional to wavelength, we can express this as:
\[ \lambda \propto \frac{1}{(Z - b)^2} \]

4. **Finding the Ratio of Wavelengths**:
Let \( \lambda_M \) be the wavelength for molybdenum and \( \lambda_Z \) be the wavelength for zinc. Then, we can write:
\[ \frac{\lambda_Z}{\lambda_M} = \frac{(Z_M - b)^2}{(Z_Z - b)^2} \]
where \( Z_M = 42 \) (for molybdenum) and \( Z_Z = 30 \) (for zinc).

5. **Substituting Values**:
For molybdenum:
\[ Z_M - b = 42 - 1 = 41 \]
For zinc:
\[ Z_Z - b = 30 - 1 = 29 \]
Thus, the ratio becomes:
\[ \frac{\lambda_Z}{0.7078} = \frac{41^2}{29^2} \]

6. **Calculating the Squares**:
Calculate \( 41^2 \) and \( 29^2 \):
\[ 41^2 = 1681, \quad 29^2 = 841 \]

7. **Finding the Ratio**:
Now substituting these values into the ratio:
\[ \frac{\lambda_Z}{0.7078} = \frac{1681}{841} \]

8. **Solving for \( \lambda_Z \)**:
Rearranging gives:
\[ \lambda_Z = 0.7078 \cdot \frac{1681}{841} \]
Calculate the right side:
\[ \lambda_Z = 0.7078 \cdot 2 = 1.4156 \, \text{Å} \]

9. **Final Answer**:
The wavelength of the K-alpha line for zinc is approximately:
\[ \lambda_Z \approx 1.4142 \, \text{Å} \]