The `K_(alpha)andK_(beta)` lines of charateristic X-ray spectrum of molybdenum are `0.76 overset(@)A and 0.64overset(@)A`, respectively. The wavelength of `L_(alpha)` line is

Options

A. `1.4overset(@)A`
B. `2.4overset(@)A`
C. `4.1overset(@)A`
D. `3.6overset(@)A`

Explanation

To find the wavelength of the L-alpha line for molybdenum based on the given K-alpha and K-beta lines, we can follow these steps:

### Step 1: Understand the transitions
The K-alpha line corresponds to the transition from the second energy level (E2) to the first energy level (E1), while the K-beta line corresponds to the transition from the third energy level (E3) to the first energy level (E1). The L-alpha line corresponds to the transition from the third energy level (E3) to the second energy level (E2).

### Step 2: Write the energy equations
The energy of the emitted X-ray can be related to its wavelength using the equation:
\[ E = \frac{hc}{\lambda} \]
Where:
- \( E \) is the energy,
- \( h \) is Planck's constant,
- \( c \) is the speed of light,
- \( \lambda \) is the wavelength.

For K-alpha and K-beta, we have:
- \( E_{K\alpha} = E_2 - E_1 = \frac{hc}{\lambda_{K\alpha}} \)
- \( E_{K\beta} = E_3 - E_1 = \frac{hc}{\lambda_{K\beta}} \)

### Step 3: Substitute the given wavelengths
From the problem, we know:
- \( \lambda_{K\alpha} = 0.76 \, \text{Å} \)
- \( \lambda_{K\beta} = 0.64 \, \text{Å} \)

### Step 4: Calculate the energy difference for L-alpha
The energy for the L-alpha transition can be expressed as:
\[ E_{L\alpha} = E_3 - E_2 = E_{K\beta} - E_{K\alpha} \]

Using the energy equations:
\[ E_{L\alpha} = \frac{hc}{\lambda_{K\beta}} - \frac{hc}{\lambda_{K\alpha}} \]

### Step 5: Factor out \( hc \)
This can be simplified to:
\[ E_{L\alpha} = hc \left( \frac{1}{\lambda_{K\beta}} - \frac{1}{\lambda_{K\alpha}} \right) \]

### Step 6: Find the wavelength of L-alpha
The wavelength of the L-alpha line can then be calculated as:
\[ \lambda_{L\alpha} = \frac{hc}{E_{L\alpha}} \]

Substituting the expression for \( E_{L\alpha} \):
\[ \lambda_{L\alpha} = \frac{hc}{hc \left( \frac{1}{\lambda_{K\beta}} - \frac{1}{\lambda_{K\alpha}} \right)} \]
This simplifies to:
\[ \lambda_{L\alpha} = \frac{1}{\left( \frac{1}{\lambda_{K\beta}} - \frac{1}{\lambda_{K\alpha}} \right)} \]

### Step 7: Substitute the values
Now substituting the wavelengths:
\[ \lambda_{L\alpha} = \frac{1}{\left( \frac{1}{0.64} - \frac{1}{0.76} \right)} \]

Calculating the individual terms:
- \( \frac{1}{0.64} = 1.5625 \)
- \( \frac{1}{0.76} = 1.3158 \)

Thus:
\[ \lambda_{L\alpha} = \frac{1}{1.5625 - 1.3158} = \frac{1}{0.2467} \approx 4.05 \, \text{Å} \]

### Final Answer
Therefore, the wavelength of the L-alpha line is approximately:
\[ \lambda_{L\alpha} \approx 4.1 \, \text{Å} \]