An electromagnetic radiation has an energy of 13.2 keV. Then the radiation belongs to region of

Options

A. visible light
B. ultraviolet
C. infrared
D. X-ray

Explanation

To determine the region of the electromagnetic spectrum to which an electromagnetic radiation with an energy of 13.2 keV belongs, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between energy, frequency, and wavelength:**
The energy (E) of electromagnetic radiation can be related to its wavelength (λ) using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \(E\) is the energy in joules,
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{J s}\)),
- \(c\) is the speed of light (\(3 \times 10^8 \, \text{m/s}\)),
- \(\lambda\) is the wavelength in meters.

2. **Convert the energy from keV to joules:**
The energy given is in kilo-electron volts (keV). We need to convert this to joules.
\[ 1 \, \text{keV} = 1.602 \times 10^{-16} \, \text{J} \]
Therefore,
\[ E = 13.2 \, \text{keV} = 13.2 \times 10^3 \times 1.602 \times 10^{-16} \, \text{J} \]
\[ E \approx 2.11 \times 10^{-12} \, \text{J} \]

3. **Calculate the wavelength (λ):**
Rearranging the formula for wavelength, we have:
\[ \lambda = \frac{hc}{E} \]
Substituting the values of \(h\), \(c\), and \(E\):
\[ \lambda = \frac{(6.626 \times 10^{-34} \, \text{J s})(3 \times 10^8 \, \text{m/s})}{2.11 \times 10^{-12} \, \text{J}} \]
\[ \lambda \approx 9.39 \times 10^{-10} \, \text{m} = 0.939 \, \text{Å} \]

4. **Identify the region of the electromagnetic spectrum:**
The wavelength of \(0.939 \, \text{Å}\) (or \(9.39 \times 10^{-10} \, \text{m}\)) falls within the range of X-rays, which is approximately from \(10^{-8} \, \text{m}\) to \(10^{-13} \, \text{m}\).

5. **Conclusion:**
Therefore, the electromagnetic radiation with an energy of 13.2 keV belongs to the region of X-rays.

### Final Answer:
The radiation belongs to the region of X-rays.