Semiconductor Ge has forbidden gap of 1.43 eV. Calculate maximum wavelength which result from electron hole combination.

Explanation

To solve the problem of finding the maximum wavelength resulting from electron-hole combination in germanium (Ge) with a forbidden gap of 1.43 eV, we will follow these steps:

### Step 1: Understand the relationship between energy and wavelength
The energy (E) associated with a photon can be related to its wavelength (λ) using the formula:
\[ E = \frac{hc}{\lambda} \]
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 in vacuum (\(3.00 \times 10^{8} \, \text{m/s}\)),
- \(\lambda\) is the wavelength in meters.

### Step 2: Convert the energy from eV to joules
Given that the forbidden gap is \(1.43 \, \text{eV}\), we need to convert this energy into joules. The conversion factor is:
\[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]
Thus, we calculate:
\[ E = 1.43 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 2.288 \times 10^{-19} \, \text{J} \]

### Step 3: Rearrange the formula to find wavelength
To find the wavelength, we rearrange the formula:
\[ \lambda = \frac{hc}{E} \]

### Step 4: Substitute the values into the formula
Now we substitute the known values into the rearranged formula:
\[ \lambda = \frac{(6.626 \times 10^{-34} \, \text{Js}) \times (3.00 \times 10^{8} \, \text{m/s})}{2.288 \times 10^{-19} \, \text{J}} \]

### Step 5: Calculate the wavelength
Calculating the above expression:
\[ \lambda = \frac{1.9878 \times 10^{-25} \, \text{Js m/s}}{2.288 \times 10^{-19} \, \text{J}} \approx 8.688 \times 10^{-7} \, \text{m} \]

### Step 6: Convert the wavelength to nanometers
To express the wavelength in nanometers, we convert meters to nanometers (1 m = \(10^9\) nm):
\[ \lambda \approx 8.688 \times 10^{-7} \, \text{m} = 868.8 \, \text{nm} \]

### Final Answer
The maximum wavelength resulting from electron-hole combination in germanium (Ge) is approximately **868.8 nm**.