The `K_(alpha)` X-ray of molybdenum has wavelength 0.071 nm. If the energy of a molybdenum atoms with a K electron knocked out is 27.5 keV, the energy of this atom when an L electron is knocked out will be ____________ keV. (Round off to the nearest integer)
`["h"=4.14xx10^(-15)" eVs",c=3xx10^(8)" ms"^(-1)]`

Answer

To solve the problem, we need to find the energy required to knock out an electron from the L shell of a molybdenum atom, given that the energy required to knock out a K electron is 27.5 keV and the K-alpha X-ray has a wavelength of 0.071 nm.

### Step-by-step Solution:

1. **Identify the Energy of the K-alpha X-ray:**
The energy of the K-alpha X-ray can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h = 4.14 \times 10^{-15} \, \text{eV s} \) (Planck's constant),
- \( c = 3 \times 10^8 \, \text{m/s} \) (speed of light),
- \( \lambda = 0.071 \, \text{nm} = 0.071 \times 10^{-9} \, \text{m} \).

2. **Calculate the Energy:**
Substituting the values into the formula:
\[ E = \frac{(4.14 \times 10^{-15}) \times (3 \times 10^8)}{0.071 \times 10^{-9}} \]
Performing the calculation:
\[ E \approx \frac{(1.242 \times 10^{-6})}{0.071 \times 10^{-9}} \approx 17.5 \, \text{keV} \]

3. **Determine the Energy Gap Between K and L Shells:**
The energy gap between the K and L shells can be identified as the energy of the K-alpha photon:
\[ E_{K \to L} = 17.5 \, \text{keV} \]

4. **Calculate the Energy Required to Knock Out an L Electron:**
The energy required to knock out an L electron can be found using the energy required to knock out a K electron and the energy gap:
\[ E_L = E_K - E_{K \to L} \]
where:
- \( E_K = 27.5 \, \text{keV} \) (energy to knock out K electron),
- \( E_{K \to L} = 17.5 \, \text{keV} \).

Therefore:
\[ E_L = 27.5 \, \text{keV} - 17.5 \, \text{keV} = 10 \, \text{keV} \]

5. **Final Answer:**
The energy required to knock out an L electron from a molybdenum atom is approximately:
\[ \boxed{10} \, \text{keV} \]