Question

If the band gap of a semiconductor is equal to the energy of a photon of wavelength \(620\) nm, then the minimum thermal energy required for the generation of \(8\) electron-hole pairs is nearly

• A. \(8\) eV
• B. \(4\) eV
• C. \(16\) eV
• D. \(2\) eV

Hint:

The energy of a photon is \[ \boxed{ E(\text{eV})=\frac{1240}{\lambda(\text{nm})}. } \] The minimum energy required to generate \(n\) electron-hole pairs is \[ \boxed{nE_g.} \]

Answer 1

The Correct Option is C

Step 1: Calculate the band gap energy. The band gap energy is equal to the energy of a photon of wavelength \[ \lambda=620\text{ nm}. \] Using \[ E=\frac{1240}{\lambda(\text{nm})}\text{ eV}, \] we get \[ E_g = \frac{1240}{620} = 2\text{ eV}. \]

Step 2: Find the energy required for \(8\) electron-hole pairs. Each electron-hole pair requires energy equal to the band gap. Therefore, \[ E = 8E_g = 8\times2 = 16\text{ eV}. \]

Step 3: Write the answer. Hence, \[ \boxed{16\text{ eV}}. \] Thus, \[ \boxed{(C)} \] is the correct answer.