The photon energy in units of eV for electromagnetic waves of wavelength 2 cm is

Options

A. `2.5xx10^(-19)`
B. `5.2xx10^(16)`
C. `3.2xx10^(-16)`
D. `6.2xx10^(-5)`

Explanation

To find the photon energy in electron volts (eV) for electromagnetic waves of wavelength 2 cm, we can follow these steps:

### Step-by-Step Solution:

1. **Convert Wavelength to Meters**:
The wavelength given is 2 cm. We need to convert this to meters:
\[ \text{Wavelength} (\lambda) = 2 \text{ cm} = 0.02 \text{ m} \]

2. **Use the Photon Energy Formula**:
The energy of a photon (E) can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \(h\) 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.

3. **Substitute the Values**:
Now, substitute the values into the formula:
\[ E = \frac{(6.626 \times 10^{-34} \text{ J s})(3 \times 10^8 \text{ m/s})}{0.02 \text{ m}} \]

4. **Calculate the Energy in Joules**:
Performing the calculation:
\[ E = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{0.02} \]
\[ E = \frac{1.9878 \times 10^{-25}}{0.02} = 9.939 \times 10^{-24} \text{ J} \]

5. **Convert Joules to Electron Volts**:
To convert the energy from joules to electron volts, we use the conversion factor \(1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}\):
\[ E (\text{eV}) = \frac{9.939 \times 10^{-24} \text{ J}}{1.6 \times 10^{-19} \text{ J/eV}} \]
\[ E (\text{eV}) = 6.21125 \times 10^{-5} \text{ eV} \]

6. **Final Result**:
Rounding the result, we find:
\[ E \approx 6.2 \times 10^{-5} \text{ eV} \]

### Conclusion:
The photon energy for electromagnetic waves of wavelength 2 cm is approximately \(6.2 \times 10^{-5} \text{ eV}\).

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