The `K, L` and `M` energy levels of platinum lie roughly at `78 keV, 12 keV` and `3 keV`, réspectively. The ratio of wavelength of `K_(alpha)` line to that of `K_(beta)` line in X-ray spee- trum is `(alpha)/(22) .` Find `alpha .`

Answer

To solve the problem, we need to find the value of \( \alpha \) in the ratio of the wavelengths of the K-alpha line to the K-beta line in the X-ray spectrum of platinum.

### Step-by-Step Solution:

1. **Identify Energy Levels**:
- The energy levels for platinum are given as:
- \( E_K = 78 \, \text{keV} \)
- \( E_L = 12 \, \text{keV} \)
- \( E_M = 3 \, \text{keV} \)

2. **Calculate Energy for K-alpha Transition**:
- The K-alpha transition occurs when an electron falls from the L-shell to the K-shell.
- The energy of the K-alpha line is calculated as:
\[ E_{K\alpha} = E_K - E_L = 78 \, \text{keV} - 12 \, \text{keV} = 66 \, \text{keV} \]

3. **Calculate Energy for K-beta Transition**:
- The K-beta transition occurs when an electron falls from the M-shell to the K-shell.
- The energy of the K-beta line is calculated as:
\[ E_{K\beta} = E_K - E_M = 78 \, \text{keV} - 3 \, \text{keV} = 75 \, \text{keV} \]

4. **Relate Wavelengths to Energies**:
- The wavelength \( \lambda \) is inversely proportional to the energy \( E \):
\[ \lambda \propto \frac{1}{E} \]
- Therefore, the ratio of the wavelengths is:
\[ \frac{\lambda_{K\alpha}}{\lambda_{K\beta}} = \frac{E_{K\beta}}{E_{K\alpha}} = \frac{75 \, \text{keV}}{66 \, \text{keV}} \]

5. **Set Up the Ratio**:
- According to the problem, we have:
\[ \frac{\lambda_{K\alpha}}{\lambda_{K\beta}} = \frac{\alpha}{22} \]
- Thus, we can equate the two expressions:
\[ \frac{75}{66} = \frac{\alpha}{22} \]

6. **Cross-Multiply to Solve for \( \alpha \)**:
- Cross-multiplying gives:
\[ 75 \cdot 22 = 66 \cdot \alpha \]
- Simplifying:
\[ 1650 = 66 \alpha \]
- Now, divide both sides by 66:
\[ \alpha = \frac{1650}{66} = 25 \]

### Final Answer:
Thus, the value of \( \alpha \) is \( \boxed{25} \).