If the energy of a photon of sodium light ( `lambda`=589 nm) equals the band gap of semiconductor, the minimum energy required to create hole electron pair

Options

A. 1.1 eV
B. 2.1 eV
C. 3.2 eV
D. 1.5 eV

Explanation

To find the minimum energy required to create an electron-hole pair in a semiconductor when the energy of a photon of sodium light (with a wavelength of 589 nm) equals the band gap of the semiconductor, we can follow these steps:

### Step 1: Understand the relationship between energy, wavelength, and the constants
The energy \(E\) of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \(h\) 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 of the light.

### Step 2: Convert the wavelength from nanometers to meters
Given that the wavelength \(\lambda\) is 589 nm, we need to convert this to meters:
\[ \lambda = 589 \, \text{nm} = 589 \times 10^{-9} \, \text{m} \]

### Step 3: Substitute the values into the energy formula
Now, substitute the values of \(h\), \(c\), and \(\lambda\) into the energy formula:
\[ E = \frac{(6.626 \times 10^{-34} \, \text{Js}) \times (3 \times 10^8 \, \text{m/s})}{589 \times 10^{-9} \, \text{m}} \]

### Step 4: Calculate the energy
Now, perform the calculation:
\[ E = \frac{(6.626 \times 3) \times 10^{-34 + 8}}{589 \times 10^{-9}} \]
\[ E = \frac{19.878 \times 10^{-26}}{589 \times 10^{-9}} = \frac{19.878}{589} \times 10^{-26 + 9} \]
\[ E \approx 0.0338 \times 10^{-17} \, \text{J} \]

### Step 5: Convert the energy from Joules to electronvolts
To convert Joules to electronvolts, we use the conversion factor \(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\):
\[ E \text{(in eV)} = \frac{0.0338 \times 10^{-17} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \]
\[ E \approx 0.211 \, \text{eV} \]

### Final Answer
The minimum energy required to create an electron-hole pair in the semiconductor is approximately **0.211 eV**.

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