Let `lambda_(alpha), lambda_(beta),and lambda'_(alpha)` denote the wavelength of the X-ray of the `K_(alpha), K_(beta), and L_(alpha)` lines in the characteristic X-rays for a metal. Then.

Options

A. `lambda_(alpha) gt lambda_(alpha) gt lambda_(beta)`
B. `lambda_(alpha) gt lambda_(beta) gt lambda_(alpha)`
C. `(1)/(lambda_(beta)) = (1)/(lambda_(alpha)) + (1)/(lambda'_(alpha))`
D. `(1)/(lambda_(alpha)) = (1)/(lambda_(beta)) + (1)/(lambda_(alpha))`

Explanation

To solve the problem, we need to analyze the relationships between the wavelengths of the characteristic X-rays emitted by a metal. We are given the wavelengths for the K_alpha, K_beta, and L_alpha lines, denoted as \( \lambda_{\alpha} \), \( \lambda_{\beta} \), and \( \lambda'_{\alpha} \) respectively.

### Step-by-Step Solution:

1. **Understanding the Energy Levels**:
- The K shell (n=1) is the closest shell to the nucleus, followed by the L shell (n=2), and then the M shell (n=3).
- The K_alpha line corresponds to the transition from the L shell to the K shell, while the K_beta line corresponds to the transition from the M shell to the K shell. The L_alpha line corresponds to the transition from the M shell to the L shell.

2. **Energy Difference and Wavelength Relation**:
- The energy difference between two energy levels can be expressed using the equation:
\[ E = \frac{hc}{\lambda} \]
- Here, \( E \) is the energy difference, \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength of the emitted photon.

3. **Setting Up the Equations**:
- For the K_beta line (transition from M to K):
\[ E_{M \to K} = \frac{hc}{\lambda_{\beta}} \]
- For the L_alpha line (transition from M to L):
\[ E_{M \to L} = \frac{hc}{\lambda'_{\alpha}} \]
- For the K_alpha line (transition from L to K):
\[ E_{L \to K} = \frac{hc}{\lambda_{\alpha}} \]

4. **Applying the Energy Conservation Principle**:
- The total energy difference from M to K can be expressed as the sum of the energy differences from M to L and L to K:
\[ E_{M \to K} = E_{M \to L} + E_{L \to K} \]
- Substituting the equations from above, we get:
\[ \frac{hc}{\lambda_{\beta}} = \frac{hc}{\lambda'_{\alpha}} + \frac{hc}{\lambda_{\alpha}} \]

5. **Cancelling \( hc \)**:
- Since \( hc \) appears in all terms, we can cancel it out:
\[ \frac{1}{\lambda_{\beta}} = \frac{1}{\lambda'_{\alpha}} + \frac{1}{\lambda_{\alpha}} \]

6. **Final Result**:
- This equation shows the relationship between the wavelengths of the K_alpha, K_beta, and L_alpha lines:
\[ \frac{1}{\lambda_{\beta}} = \frac{1}{\lambda'_{\alpha}} + \frac{1}{\lambda_{\alpha}} \]

### Conclusion:
The correct relationship among the wavelengths is given by:
\[ \frac{1}{\lambda_{\beta}} = \frac{1}{\lambda'_{\alpha}} + \frac{1}{\lambda_{\alpha}} \]
Thus, the answer to the question is option C.