Forbidden energy gap of Ge is `0.75eV`, maximum wave length of incident radiation for producing electron-gole pair in germanium semiconductor is

Options

A. `4200A^(@)`
B. `16500A^(@)`
C. `4700A^(@)`
D. `4000A^(@)`

Answer

To find the maximum wavelength of incident radiation that can produce an electron-hole pair in germanium (Ge) with a forbidden energy gap of 0.75 eV, we can use the relationship between energy and wavelength.

### Step-by-Step Solution:

1. **Understand the relationship between energy and wavelength**:
The energy (E) of a photon is related to its wavelength (λ) by the equation:
\[ E = \frac{hc}{\lambda} \]
where:
- \(E\) is the energy in joules,
- \(h\) is Planck's constant (\(6.63 \times 10^{-34} \, \text{m}^2 \text{kg/s}\)),
- \(c\) is the speed of light (\(3 \times 10^8 \, \text{m/s}\)),
- \(\lambda\) is the wavelength in meters.

2. **Convert the energy gap from eV to joules**:
The energy gap given is 0.75 eV. To convert this to joules, we use the conversion factor \(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\):
\[ E = 0.75 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 1.2 \times 10^{-19} \, \text{J} \]

3. **Rearrange the equation to solve for wavelength**:
We need to find the maximum wavelength (\(\lambda_{\text{max}}\)):
\[ \lambda_{\text{max}} = \frac{hc}{E} \]

4. **Substitute the values into the equation**:
Substitute \(h\), \(c\), and \(E\) into the equation:
\[ \lambda_{\text{max}} = \frac{(6.63 \times 10^{-34} \, \text{m}^2 \text{kg/s}) \times (3 \times 10^8 \, \text{m/s})}{1.2 \times 10^{-19} \, \text{J}} \]

5. **Calculate the wavelength**:
\[ \lambda_{\text{max}} = \frac{1.989 \times 10^{-25} \, \text{m}^2 \text{kg/s}}{1.2 \times 10^{-19} \, \text{J}} \approx 1.6575 \times 10^{-6} \, \text{m} \]
Converting this to nanometers (1 m = \(10^9\) nm):
\[ \lambda_{\text{max}} \approx 16575 \, \text{nm} = 16500 \, \text{Å} \, (\text{since } 1 \, \text{nm} = 10 \, \text{Å}) \]

6. **Final Answer**:
The maximum wavelength of incident radiation for producing an electron-hole pair in germanium is approximately \(16500 \, \text{Å}\).