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 the equation:

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

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

Given:
- The forbidden energy gap \( E = 0.72 \, \text{eV} \)
- \( hc = 12440 \, \text{eV} \cdot \text{Å} \)

### Step 1: Rearranging the formula
We need to rearrange the formula to solve for \( \lambda \):

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

### Step 2: Substituting the values
Now, we substitute the values of \( hc \) and \( E \) into the equation:

\[ \lambda = \frac{12440 \, \text{eV} \cdot \text{Å}}{0.72 \, \text{eV}} \]

### Step 3: Performing the calculation
Now we perform the calculation:

\[ \lambda = \frac{12440}{0.72} \approx 17277.78 \, \text{Å} \]

### Step 4: Rounding the result
Rounding this value gives us:

\[ \lambda \approx 17278 \, \text{Å} \]

### Conclusion
Thus, the maximum wavelength of radiation that will generate an electron-hole pair in Ge is approximately **17278 Å**.

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