The photon energy, in eV, for electromagnetic waves of wavelength 40 m is

Options

A. `3.1 xx 10^(-5)`
B. `4.8 xx 10^(-27)`
C. `6.2 xx 10^(-8)`
D. `9.6 xx 10^(-27)`

Explanation

To find the photon energy for electromagnetic waves of wavelength 40 m, we can use the formula for photon energy:

\[ E = \frac{hc}{\lambda} \]

Where:
- \( E \) is the photon 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.

### Step-by-step Solution:

1. **Identify the given values**:
- Wavelength (\( \lambda \)) = 40 m
- Planck's constant (\( h \)) = \( 6.626 \times 10^{-34} \, \text{J s} \)
- Speed of light (\( c \)) = \( 3 \times 10^8 \, \text{m/s} \)

2. **Substitute the values into the energy formula**:
\[ E = \frac{(6.626 \times 10^{-34} \, \text{J s}) \times (3 \times 10^8 \, \text{m/s})}{40 \, \text{m}} \]

3. **Calculate the numerator**:
\[ (6.626 \times 10^{-34}) \times (3 \times 10^8) = 1.9878 \times 10^{-25} \, \text{J m} \]

4. **Divide by the wavelength**:
\[ E = \frac{1.9878 \times 10^{-25} \, \text{J m}}{40 \, \text{m}} = 4.9695 \times 10^{-27} \, \text{J} \]

5. **Convert energy from joules to electron volts**:
- The charge of an electron (\( e \)) is approximately \( 1.602 \times 10^{-19} \, \text{C} \).
\[ E \, (\text{in eV}) = \frac{E \, (\text{in J})}{1.602 \times 10^{-19} \, \text{C}} = \frac{4.9695 \times 10^{-27}}{1.602 \times 10^{-19}} \approx 3.1 \times 10^{-8} \, \text{eV} \]

### Final Answer:
The photon energy for electromagnetic waves of wavelength 40 m is approximately \( 3.1 \times 10^{-8} \, \text{eV} \).

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