The `X`- ray wavelength of `L_(alpha)` line of platinum `(Z = 78)` is `1.30 Å`. The `X` - ray wavelength of `L_(alpha)` line of Molybdenum `(Z = 42)` is

Options

A. `5.41 Å`
B. `4.20 Å`
C. `2.70 Å`
D. `1.35 Å`

Answer

To find the X-ray wavelength of the L-alpha line of Molybdenum (Mo, Z = 42), we can use the relationship between the wavelengths of the L-alpha lines of two elements based on their effective nuclear charge.

### Step-by-step Solution:

1. **Understand the Formula**:
The formula for the wavelength of the emitted photon is given by:
\[ \frac{1}{\lambda} = R \cdot Z_{\text{effective}}^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]
For the L-alpha transition, the initial state \( n_i = 3 \) and the final state \( n_f = 2 \). Thus, we can rewrite this as:
\[ \frac{1}{\lambda} = R \cdot Z_{\text{effective}}^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]

2. **Calculate the Difference in Terms**:
\[ \frac{1}{2^2} - \frac{1}{3^2} = \frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36} \]
Therefore, we can express the equation as:
\[ \frac{1}{\lambda} = R \cdot Z_{\text{effective}}^2 \cdot \frac{5}{36} \]

3. **Effective Nuclear Charge**:
The effective nuclear charge \( Z_{\text{effective}} \) is given by:
\[ Z_{\text{effective}} = Z - 7.4 \]
For Platinum (Z = 78):
\[ Z_{\text{effective}}(\text{Pt}) = 78 - 7.4 = 70.6 \]
For Molybdenum (Z = 42):
\[ Z_{\text{effective}}(\text{Mo}) = 42 - 7.4 = 34.6 \]

4. **Relate Wavelengths**:
From the relationship of wavelengths:
\[ \frac{\lambda_1}{\lambda_2} = \frac{Z_{\text{effective, Mo}}^2}{Z_{\text{effective, Pt}}^2} \]
Rearranging gives:
\[ \lambda_2 = \lambda_1 \cdot \frac{Z_{\text{effective, Pt}}^2}{Z_{\text{effective, Mo}}^2} \]

5. **Substituting Values**:
Given \( \lambda_1 = 1.30 \, \text{Å} \):
\[ \lambda_2 = 1.30 \cdot \left(\frac{70.6^2}{34.6^2}\right) \]

6. **Calculating the Squares**:
\[ 70.6^2 = 4984.36 \quad \text{and} \quad 34.6^2 = 1197.16 \]
Therefore:
\[ \frac{70.6^2}{34.6^2} = \frac{4984.36}{1197.16} \approx 4.16 \]

7. **Final Calculation**:
\[ \lambda_2 = 1.30 \cdot 4.16 \approx 5.41 \, \text{Å} \]

### Conclusion:
The X-ray wavelength of the L-alpha line of Molybdenum is approximately **5.41 Å**.