If `lamda_(K_(alpha)),lamda_(K_(beta))` and `lamda_(L_(alpha))` are the wavelengths of `K_(alpha),K_(beta)` and `L_(alpha)` lines, then

Options

A. `lamda_(K_(beta))=(lamda_(K_(alpha))lamda_(L_(alpha)))/(lamda_(K_(alpha))+lamda_(L_(alpha)))`
B. `lamda_(L_(alpha))=(lamda_(K_(alpha))lamda_(K_(beta)))/(lamda_(K_(alpha))+lamda_(K_(beta)))`
C. `lamda_(L_(alpha))=(lamda_(K_(alpha))lamda_(K_(beta)))/(lamda_(K_(beta))-lamda_(K_(alpha)))`
D. `lamda_(K_(beta))=(lamda_(K_(alpha))lamda_(L_(alpha)))/(lamda_(K_(alpha))-lamda_(L_(alpha)))`

Answer

To solve the problem regarding the relationship between the wavelengths of the K_alpha, K_beta, and L_alpha lines, we will follow these steps:

### Step 1: Understand the Energy Levels
We start by identifying the energy levels involved in the transitions:
- Let \( n = 1 \) correspond to the K shell,
- \( n = 2 \) correspond to the L shell,
- \( n = 3 \) correspond to the M shell, and
- \( n = 4 \) correspond to the N shell.

### Step 2: Define the Transitions
We define the transitions that lead to the emission of the different spectral lines:
- The transition from the L shell (n=2) to the K shell (n=1) corresponds to the K_alpha line.
- The transition from the M shell (n=3) to the K shell (n=1) corresponds to the K_beta line.
- The transition from the M shell (n=3) to the L shell (n=2) corresponds to the L_alpha line.

### Step 3: Write the Wavelength Relationships
From the transitions, we can express the wavelengths in terms of energy:
- The energy of a photon emitted during a transition is given by \( E = \frac{hc}{\lambda} \), where \( h \) is Planck's constant and \( c \) is the speed of light.
- Thus, we can write the relationships:
- For K_alpha: \( \frac{1}{\lambda_{K_\alpha}} = E_{K_\alpha} \)
- For K_beta: \( \frac{1}{\lambda_{K_\beta}} = E_{K_\beta} \)
- For L_alpha: \( \frac{1}{\lambda_{L_\alpha}} = E_{L_\alpha} \)

### Step 4: Establish the Energy Relationship
From the energy conservation principle, we know that:
- The energy for the K_beta transition can be expressed as:
\[ E_{K_\beta} = E_{K_\alpha} + E_{L_\alpha} \]
This implies:
\[ \frac{1}{\lambda_{K_\beta}} = \frac{1}{\lambda_{K_\alpha}} + \frac{1}{\lambda_{L_\alpha}} \]

### Step 5: Rearranging the Equation
We can rearrange the equation to find the relationship between the wavelengths:
\[ \frac{1}{\lambda_{K_\beta}} = \frac{\lambda_{L_\alpha} + \lambda_{K_\alpha}}{\lambda_{K_\alpha} \cdot \lambda_{L_\alpha}} \]
Taking the reciprocal gives us:
\[ \lambda_{K_\beta} = \frac{\lambda_{K_\alpha} \cdot \lambda_{L_\alpha}}{\lambda_{K_\alpha} + \lambda_{L_\alpha}} \]

### Conclusion
Thus, we have derived the relationship:
\[ \lambda_{K_\beta} = \frac{\lambda_{K_\alpha} \cdot \lambda_{L_\alpha}}{\lambda_{K_\alpha} + \lambda_{L_\alpha}} \]