Question:
The energy of a photon in a monochromatic light of wavelength \(621\ \text{nm}\) matches with the band gap of a semiconducting material. Then the minimum energy required to create an electron-hole pair from the semiconductor is
\[
[\text{Take } hc=1242\ \text{eV-nm}, \text{ where } h \text{ is Planck's constant and } c \text{ is speed of light in vacuum}]
\]
The energy of a photon in a monochromatic light of wavelength \(621\ \text{nm}\) matches with the band gap of a semiconducting material. Then the minimum energy required to create an electron-hole pair from the semiconductor is
\[ [\text{Take } hc=1242\ \text{eV-nm}, \text{ where } h \text{ is Planck's constant and } c \text{ is speed of light in vacuum}] \]
\[ [\text{Take } hc=1242\ \text{eV-nm}, \text{ where } h \text{ is Planck's constant and } c \text{ is speed of light in vacuum}] \]
Show Hint
For semiconductors, the minimum energy required to create an electron-hole pair is equal to the band gap energy:
\[
E_g=\frac{hc}{\lambda}
\]
When \(hc\) is given in eV-nm, use \(\lambda\) in nm.
Updated On: Jun 25, 2026
- \(3.4\ \text{eV}\)
- \(1.7\ \text{eV}\)
- \(2\ \text{eV}\)
- \(2.2\ \text{eV}\)
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Understand the relation between photon energy and band gap.
The minimum energy required to create an electron-hole pair in a semiconductor is equal to its band gap energy.
Given that the photon energy matches with the band gap of the semiconductor, we can write \[ E_g=E_{\text{photon}} \]
Step 2: Use the photon energy formula.
Photon energy is given by \[ E=\frac{hc}{\lambda} \] Given: \[ hc=1242\ \text{eV-nm} \] and \[ \lambda=621\ \text{nm} \] Therefore, \[ E=\frac{1242}{621} \] \[ E=2\ \text{eV} \]
Step 3: Relate this energy to electron-hole pair formation.
Since this photon energy matches the band gap, \[ E_g=2\ \text{eV} \] The minimum energy required to create an electron-hole pair is therefore \[ 2\ \text{eV} \]
Step 4: Final conclusion.
Hence, the minimum energy required is \[ \boxed{2\ \text{eV}} \]
The minimum energy required to create an electron-hole pair in a semiconductor is equal to its band gap energy.
Given that the photon energy matches with the band gap of the semiconductor, we can write \[ E_g=E_{\text{photon}} \]
Step 2: Use the photon energy formula.
Photon energy is given by \[ E=\frac{hc}{\lambda} \] Given: \[ hc=1242\ \text{eV-nm} \] and \[ \lambda=621\ \text{nm} \] Therefore, \[ E=\frac{1242}{621} \] \[ E=2\ \text{eV} \]
Step 3: Relate this energy to electron-hole pair formation.
Since this photon energy matches the band gap, \[ E_g=2\ \text{eV} \] The minimum energy required to create an electron-hole pair is therefore \[ 2\ \text{eV} \]
Step 4: Final conclusion.
Hence, the minimum energy required is \[ \boxed{2\ \text{eV}} \]
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