Light of an electromagnetic radiation has an energy 2.06 eV each. To which region of electromagnetic spectrum does it belong?

Options

A. X - ray region
B. Ultra violet region
C. Infrared region
D. Visible region

Explanation

To determine the region of the electromagnetic spectrum to which light with an energy of 2.06 eV belongs, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Relationship Between Energy and Wavelength**:
The energy (E) of a photon can be related to its wavelength (λ) using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \(E\) is the energy in electron volts (eV),
- \(h\) 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 Energy to Wavelength**:
Rearranging the formula gives:
\[ \lambda = \frac{hc}{E} \]
To simplify calculations, we can use a convenient value for \(hc\) in electron volts and nanometers:
\[ hc \approx 1240 \text{ eV nm} \]
Thus, we can write:
\[ \lambda = \frac{1240 \text{ eV nm}}{E} \]

3. **Substitute the Given Energy**:
Plugging in the given energy \(E = 2.06 \text{ eV}\):
\[ \lambda = \frac{1240 \text{ eV nm}}{2.06 \text{ eV}} \approx 601.94 \text{ nm} \]

4. **Determine the Region of the Electromagnetic Spectrum**:
The wavelength of approximately 601.94 nm falls within the range of visible light, which is approximately from 380 nm to 750 nm.

5. **Conclusion**:
Therefore, the light with an energy of 2.06 eV belongs to the visible region of the electromagnetic spectrum.

### Final Answer:
The light belongs to the **visible region** of the electromagnetic spectrum.

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