The wavelength of `K_(alpha)` line for an element of atomic number `43 is lambda`. Then the wavelength of `K_(alpha)` line for an element of atomic number `29` is

Options

A. `(43)/(29) lambda`
B. `(42)/(28) lambda`
C. `(9)/(4) lambda`
D. `(4)/(9) lambda`

Answer

To solve the problem, we will use Moseley's law, which relates the frequency of the X-ray emission lines to the atomic number of the element. The K-alpha line corresponds to transitions from the L shell to the K shell in an atom.

### Step-by-step solution:

1. **Understand Moseley's Law**:
Moseley's law states that the square root of the frequency (ν) of the emitted X-ray is proportional to the atomic number (Z) minus a constant (b). Mathematically, it can be expressed as:
\[ \sqrt{\nu} = a(Z - b) \]
where \(a\) is a constant and \(b\) is typically 1 for K-alpha transitions.

2. **Effective Atomic Number**:
For K-alpha transitions, the effective atomic number \(Z_{eff}\) is given by:
\[ Z_{eff} = Z - 1 \]
Therefore, we can rewrite the equation as:
\[ \sqrt{\nu} = a(Z - 1) \]

3. **Relate Frequency and Wavelength**:
The relationship between frequency (ν) and wavelength (λ) is given by:
\[ \nu = \frac{c}{\lambda} \]
where \(c\) is the speed of light. Substituting this into the previous equation gives:
\[ \sqrt{\frac{c}{\lambda}} = a(Z - 1) \]

4. **Rearranging the Equation**:
Rearranging the equation yields:
\[ \frac{1}{\sqrt{\lambda}}(Z - 1) = \text{constant} \]
This implies:
\[ \frac{1}{\sqrt{\lambda_1}}(Z_1 - 1) = \frac{1}{\sqrt{\lambda_2}}(Z_2 - 1) \]

5. **Substituting Values**:
Given:
- For the first element (atomic number \(Z_1 = 43\)), the wavelength is \(\lambda_1 = \lambda\).
- For the second element (atomic number \(Z_2 = 29\)), we need to find \(\lambda_2\).

Substitute these values into the equation:
\[ \frac{1}{\sqrt{\lambda}}(43 - 1) = \frac{1}{\sqrt{\lambda_2}}(29 - 1) \]
Simplifying gives:
\[ \frac{42}{\sqrt{\lambda}} = \frac{28}{\sqrt{\lambda_2}} \]

6. **Cross Multiplying**:
Cross multiplying yields:
\[ 42 \sqrt{\lambda_2} = 28 \sqrt{\lambda} \]

7. **Solving for \(\lambda_2\)**:
Dividing both sides by 28 gives:
\[ \sqrt{\lambda_2} = \frac{42}{28} \sqrt{\lambda} = \frac{3}{2} \sqrt{\lambda} \]
Squaring both sides results in:
\[ \lambda_2 = \left(\frac{3}{2}\right)^2 \lambda = \frac{9}{4} \lambda \]

### Final Result:
The wavelength of the K-alpha line for the element with atomic number 29 is:
\[ \lambda_2 = \frac{9}{4} \lambda \]