A p-n photodiode is made of a material with a band gap of 1 e V. The minimum frequency of the radiation that can be absorbed by the material is nearly
(hc= 1240 eV nm)

Options

A. `1 xx 10^(14) Hz`
B. `2.4 xx 10^(14) Hz`
C. `1.5 xx 10^(14) Hz`
D. `5xx 10^(14) Hz`

Answer

To find the minimum frequency of the radiation that can be absorbed by a p-n photodiode made of a material with a band gap of 1 eV, we can use the relationship between energy and frequency given by the equation:

\[ E = h \nu \]

where:
- \( E \) is the energy in joules,
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \) J·s),
- \( \nu \) is the frequency in Hz.

1. **Convert the band gap energy from eV to joules**:
The energy in electron volts (eV) can be converted to joules (J) using the conversion factor \( 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} \).

\[ E = 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} \]

2. **Rearranging the formula to find frequency**:
From the equation \( E = h \nu \), we can rearrange it to solve for frequency \( \nu \):

\[ \nu = \frac{E}{h} \]

3. **Substituting the values**:
Now, substitute the values of \( E \) and \( h \) into the equation:

\[ \nu = \frac{1.6 \times 10^{-19} \text{ J}}{6.626 \times 10^{-34} \text{ J·s}} \]

4. **Calculating the frequency**:
Performing the calculation:

\[ \nu = \frac{1.6 \times 10^{-19}}{6.626 \times 10^{-34}} \approx 2.41 \times 10^{14} \text{ Hz} \]

Rounding this to two significant figures gives:

\[ \nu \approx 2.4 \times 10^{14} \text{ Hz} \]

5. **Conclusion**:
The minimum frequency of the radiation that can be absorbed by the material is approximately \( 2.4 \times 10^{14} \text{ Hz} \). Therefore, the correct option is Option 2.