The `K_(alpha)` X-rays of moylydenum has a wavelength of `71xx10^(-12)m`. If the energy of molybdenum atom with molybdenum atom when an L-electron removed is `(hc=12.42xx10^(-7)eV)`

Options

A. 17.5 keV
B. 40.82 keV
C. 23.32 keV
D. 5.28 keV

Explanation

To solve the problem, we need to find the energy of the molybdenum atom when an L-electron is removed, given the wavelength of the K-alpha X-rays and the value of \( hc \).

### Step-by-step Solution:

1. **Identify Given Data:**
- Wavelength of K-alpha X-rays, \( \lambda = 71 \times 10^{-12} \, \text{m} \)
- Value of \( hc = 12.42 \times 10^{-7} \, \text{eV} \)

2. **Calculate the Energy of K-alpha X-rays:**
The energy \( E \) of the X-rays can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
Substituting the values:
\[ E = \frac{12.42 \times 10^{-7} \, \text{eV}}{71 \times 10^{-12} \, \text{m}} \]

3. **Perform the Calculation:**
\[ E = \frac{12.42 \times 10^{-7}}{71 \times 10^{-12}} = \frac{12.42}{71} \times 10^{5} \, \text{eV} \]
\[ E \approx 0.175 \times 10^{5} \, \text{eV} = 17.5 \, \text{keV} \]

4. **Determine the Energy of the L-electron:**
The energy of the K-alpha X-ray can be expressed as:
\[ E_{K\alpha} = E_L + \Delta E \]
where \( \Delta E \) is the energy difference between the K-alpha and L-level.

Given that \( E_{K\alpha} = 23.32 \, \text{keV} \):
\[ \Delta E = E_{K\alpha} - E \]
Substituting the values:
\[ \Delta E = 23.32 \, \text{keV} - 17.5 \, \text{keV} = 5.82 \, \text{keV} \]

5. **Final Calculation for \( E_L \):**
\[ E_L = E_{K\alpha} - \Delta E \]
\[ E_L = 23.32 \, \text{keV} - 5.82 \, \text{keV} = 17.5 \, \text{keV} \]

### Conclusion:
The energy of the molybdenum atom when an L-electron is removed is approximately \( 17.5 \, \text{keV} \).