The forbidden energy gap in Ge is 0.72 eV , given `hc = 12440 eVÅ` - The maximum wavelength of radiation that will generate electron hole pair is

Options

A. `172220Å`
B. `172.2Å`
C. `17222Å`
D. `1722Å`

Answer

To find the maximum wavelength of radiation that will generate an electron-hole pair in germanium (Ge) with a forbidden energy gap of 0.72 eV, we can use the relationship between energy and wavelength given by Planck's equation:

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

Where:
- \( E \) is the energy (in electron volts),
- \( h \) is Planck's constant,
- \( c \) is the speed of light,
- \( \lambda \) is the wavelength (in angstroms).

### Step-by-Step Solution:

1. **Identify the given values**:
- Forbidden energy gap \( E = 0.72 \, \text{eV} \)
- \( hc = 12440 \, \text{eV} \cdot \text{Å} \)

2. **Rearrange the formula to find wavelength**:
We need to find \( \lambda \), so we rearrange the equation:
\[ \lambda = \frac{hc}{E} \]

3. **Substitute the known values into the equation**:
Substitute \( hc \) and \( E \) into the rearranged equation:
\[ \lambda = \frac{12440 \, \text{eV} \cdot \text{Å}}{0.72 \, \text{eV}} \]

4. **Calculate the wavelength**:
Perform the division:
\[ \lambda = \frac{12440}{0.72} \approx 17277.78 \, \text{Å} \]

5. **Conclusion**:
The maximum wavelength of radiation that will generate an electron-hole pair in germanium is approximately \( 17277.78 \, \text{Å} \).