Calculate the `K_(alpha)` x-ray wavelength for a gold atom `(Z=79)`.

Options

A. `5.13xx10^(-10)m`
B. `2.00xx10^(-11)m`
C. `8.54xx10^(-10)m`
D. `3.60xx10^(-11)m`

Answer

To calculate the K-alpha x-ray wavelength for a gold atom (Z=79), we will use the formula for the wavelength of the K-alpha line:

\[ \lambda_{\alpha} = \frac{4}{3} R (Z - 1)^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)^{-1} \]

### Step-by-step Solution:

1. **Identify the constants and values:**
- For K-alpha transition:
- \( n_1 = 1 \)
- \( n_2 = 2 \)
- The atomic number \( Z = 79 \) for gold.
- The Rydberg constant \( R = 1.1 \times 10^7 \, \text{m}^{-1} \).

2. **Calculate the term \( (Z - 1) \):**
\[ Z - 1 = 79 - 1 = 78 \]

3. **Calculate \( (Z - 1)^2 \):**
\[ (Z - 1)^2 = 78^2 = 6084 \]

4. **Calculate the difference of squares:**
\[ \frac{1}{n_1^2} - \frac{1}{n_2^2} = \frac{1}{1^2} - \frac{1}{2^2} = 1 - \frac{1}{4} = \frac{3}{4} \]

5. **Substituting values into the formula:**
\[ \lambda_{\alpha} = \frac{4}{3} R (Z - 1)^2 \left( \frac{3}{4} \right)^{-1} \]
\[ \lambda_{\alpha} = \frac{4}{3} \times 1.1 \times 10^7 \times 6084 \times \frac{4}{3} \]

6. **Simplifying the expression:**
\[ \lambda_{\alpha} = \frac{4 \times 1.1 \times 10^7 \times 6084}{3} \times \frac{4}{3} \]
\[ \lambda_{\alpha} = \frac{4 \times 1.1 \times 10^7 \times 6084 \times 4}{9} \]

7. **Calculating the final value:**
\[ \lambda_{\alpha} = \frac{4 \times 1.1 \times 10^7 \times 6084 \times 4}{9} \approx 1.9923 \times 10^{-11} \text{ meters} \]

8. **Convert to scientific notation:**
\[ \lambda_{\alpha} \approx 2.00 \times 10^{-11} \text{ meters} \]

### Final Answer:
The wavelength of the K-alpha x-ray for a gold atom is approximately \( 2.00 \times 10^{-11} \) meters.