The wavelength of the `K_(alpha)` line for an element of atomic number 57 is `lambda` . What is the wavelength of the `K_(alpha)` line for the element of atomic number29 ?

Options

A. `lambda`
B. `2lambda`
C. `4lambda`
D. `8lambda`

Answer

To solve the problem, we will use Moseley's law, which relates the frequency of X-ray emissions to the atomic number of the element. The steps are as follows:

### Step-by-Step Solution:

1. **Understand Moseley's Law**: According to Moseley's law, the square root of the frequency (ν) of the emitted X-ray is directly proportional to the difference between the atomic number (Z) and a constant (B). For K-series X-rays, B is typically taken as 1.
\[ \sqrt{\nu} \propto Z - B \]

2. **Relate Frequency and Wavelength**: The frequency (ν) is related to the wavelength (λ) by the equation:
\[ \nu = \frac{c}{\lambda} \]
where \( c \) is the speed of light. Thus, we can express the relationship in terms of wavelength:
\[ \frac{1}{\lambda} \propto Z - B \]

3. **Set Up the Proportionality**: For two different elements, we can write:
\[ \frac{1}{\lambda_1} \propto Z_1 - B \]
\[ \frac{1}{\lambda_2} \propto Z_2 - B \]
where \( Z_1 = 57 \) (for the first element) and \( Z_2 = 29 \) (for the second element).

4. **Substituting Values**: For the first element (atomic number 57):
\[ \frac{1}{\lambda_1} \propto 57 - 1 = 56 \]
For the second element (atomic number 29):
\[ \frac{1}{\lambda_2} \propto 29 - 1 = 28 \]

5. **Form the Ratio**: Now, we can form the ratio of the two wavelengths:
\[ \frac{\lambda_1}{\lambda_2} = \frac{Z_2 - B}{Z_1 - B} = \frac{28}{56} = \frac{1}{2} \]

6. **Express λ2 in Terms of λ1**: Rearranging gives:
\[ \lambda_2 = 2 \lambda_1 \]

7. **Substituting the Given Wavelength**: Since we know that \( \lambda_1 = \lambda \), we can substitute:
\[ \lambda_2 = 2 \lambda \]

### Conclusion:
The wavelength of the K-alpha line for the element with atomic number 29 is \( 2\lambda \).