The maximum wavelength of electromagnetic radiation, which can create a hole-electron pair in germanium. (Given that forbidden energy gap in germanium is 0.72 eV)

Options

A. `1.7xx10^(-6)` m
B. `1.5xx10^(-5)` m
C. `1.3xx10^(-4)` m
D. `1.9xx10^(-5)` m

Explanation

To find the maximum wavelength of electromagnetic radiation that can create a hole-electron pair in germanium, we can use the relationship between energy and wavelength given by the equation:

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

Where:
- \( E \) is the energy (in joules),
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \)),
- \( c \) is the speed of light (\( 3 \times 10^{8} \, \text{m/s} \)),
- \( \lambda \) is the wavelength (in meters).

Given that the forbidden energy gap in germanium is \( 0.72 \, \text{eV} \), we first need to convert this energy into joules. The conversion factor is:

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

So, we can calculate the energy in joules:

\[ E = 0.72 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} \]

Now, substituting the values:

1. Calculate the energy in joules:
\[ E = 0.72 \times 1.6 \times 10^{-19} = 1.152 \times 10^{-19} \, \text{J} \]

2. Rearranging the equation for wavelength:
\[ \lambda = \frac{hc}{E} \]

3. Substitute the values of \( h \), \( c \), and \( E \):
\[ \lambda = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^{8} \, \text{m/s})}{1.152 \times 10^{-19} \, \text{J}} \]

4. Calculate \( hc \):
\[ hc = 6.626 \times 10^{-34} \times 3 \times 10^{8} = 1.9878 \times 10^{-25} \, \text{Js m} \]

5. Now substitute \( hc \) into the wavelength equation:
\[ \lambda = \frac{1.9878 \times 10^{-25}}{1.152 \times 10^{-19}} \approx 1.726 \times 10^{-6} \, \text{m} \]

6. Rounding this value gives:
\[ \lambda \approx 1.7 \times 10^{-6} \, \text{m} \]

Thus, the maximum wavelength of electromagnetic radiation that can create a hole-electron pair in germanium is approximately:

\[ \lambda \approx 1.7 \times 10^{-6} \, \text{m} \]