The forbidden energy app in Ge is 0.75 eV, given hc=`11250eVÅ`. The maximum wavelength of radiation that will generate electron hole pair is

Options

A. `1500Å`
B. `15000Å`
C. `150.52Å`
D. `150Å`

Explanation

To solve the problem of finding the maximum wavelength of radiation that will generate an electron-hole pair in germanium (Ge), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Information**:
- The forbidden energy gap (EG) in germanium (Ge) is given as 0.75 eV.
- The value of \( hc \) is given as 11250 eV·Å.

2. **Use the Formula for Energy and Wavelength**:
- The relationship between energy (E), wavelength (λ), and the constants \( h \) (Planck's constant) and \( c \) (speed of light) is given by the formula:
\[ E = \frac{hc}{\lambda} \]
- Rearranging this formula to find the wavelength (λ) gives us:
\[ \lambda = \frac{hc}{E} \]

3. **Substitute the Known Values**:
- Substitute \( hc = 11250 \, \text{eV·Å} \) and \( E = 0.75 \, \text{eV} \) into the rearranged formula:
\[ \lambda = \frac{11250 \, \text{eV·Å}}{0.75 \, \text{eV}} \]

4. **Calculate the Wavelength**:
- Performing the calculation:
\[ \lambda = \frac{11250}{0.75} = 15000 \, \text{Å} \]

5. **Final Result**:
- The maximum wavelength of radiation that will generate an electron-hole pair in germanium is:
\[ \lambda = 15000 \, \text{Å} \]

### Conclusion:
The answer to the question is that the maximum wavelength of radiation that will generate an electron-hole pair in germanium is **15000 Å**.