The X-rays of wavelength 0.5 A are scattered by a target. What will be the energy of incident X-rays, if these are scattered at an angle of `72^@` ?

Options

A. 12.41
B. 6.2 KeV
C. 18.6 keV
D. 24.82 keV

Explanation

To solve the problem of finding the energy of incident X-rays with a given wavelength of 0.5 Å and scattered at an angle of 72 degrees, we will follow these steps:

### Step-by-Step Solution:

1. **Convert Wavelength to Meters**:
The given wavelength \( \lambda \) is 0.5 Å. We need to convert this to meters.
\[ \lambda = 0.5 \, \text{Å} = 0.5 \times 10^{-10} \, \text{m} \]

2. **Use the Energy-Wavelength Relation**:
The energy \( E \) of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h = 6.626 \times 10^{-34} \, \text{J s} \) (Planck's constant)
- \( c = 3 \times 10^8 \, \text{m/s} \) (speed of light)

3. **Substituting Values**:
Plugging in the values into the energy equation:
\[ E = \frac{(6.626 \times 10^{-34} \, \text{J s}) \times (3 \times 10^8 \, \text{m/s})}{0.5 \times 10^{-10} \, \text{m}} \]

4. **Calculating the Energy**:
First, calculate the numerator:
\[ 6.626 \times 3 = 19.878 \quad \text{(approximately 19.88)} \]
Now, substituting back:
\[ E = \frac{19.878 \times 10^{-34 + 8}}{0.5} = \frac{19.878 \times 10^{-26}}{0.5} = 39.756 \times 10^{-26} \, \text{J} \]

5. **Convert Energy to Electron Volts**:
To convert Joules to electron volts, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ E = \frac{39.756 \times 10^{-26} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} = 24.85 \times 10^3 \, \text{eV} = 24.85 \, \text{keV} \]

6. **Final Answer**:
Therefore, the energy of the incident X-rays is approximately:
\[ E \approx 24.82 \, \text{keV} \]

### Conclusion:
The energy of the incident X-rays is **24.82 keV**, which corresponds to option D.