A P-N junction becomes active as photons of wavelength, `lambda=400 nm` falls on it. Find the energy band gap?

Options

A. `3.09 eV`
B. `4.51eV`
C. `2.45eV`
D. `5.34eV`

Answer

To find the energy band gap of a P-N junction when photons of wavelength \( \lambda = 400 \, \text{nm} \) fall on it, we can use the relationship between the energy of a photon and its wavelength. The energy \( E \) of a photon is given by 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 of the photon.

### Step-by-Step Solution:

1. **Convert Wavelength to Meters**:
\[ \lambda = 400 \, \text{nm} = 400 \times 10^{-9} \, \text{m} \]

2. **Substitute Values into the Energy Formula**:
\[ E = \frac{(6.626 \times 10^{-34} \, \text{J s}) \times (3 \times 10^{8} \, \text{m/s})}{400 \times 10^{-9} \, \text{m}} \]

3. **Calculate the Energy**:
\[ E = \frac{1.9878 \times 10^{-25} \, \text{J m}}{400 \times 10^{-9} \, \text{m}} = 4.9695 \times 10^{-19} \, \text{J} \]

4. **Convert Energy from Joules to Electron Volts**:
To convert energy from Joules to electron volts, use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ E = \frac{4.9695 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \approx 3.1059 \, \text{eV} \]

5. **Round the Result**:
The energy band gap \( E_g \) is approximately \( 3.1 \, \text{eV} \).

### Final Answer:
The energy band gap of the P-N junction is approximately \( 3.1 \, \text{eV} \).

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