Question:
A p-type semiconductor has acceptor levels 57 meV above the valence band. The maximum wavelength of light required to create a hole is:
A p-type semiconductor has acceptor levels 57 meV above the valence band. The maximum wavelength of light required to create a hole is:
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To find the wavelength from the energy of a photon, use the relation \( \lambda = \frac{hc}{E} \). Remember to convert the energy to Joules before using this formula.
Updated On: Apr 22, 2026
- 57 Ã
- \( 57 \times 10^{-3} \, \text{Ã } \)
- 217105 Ã
- \( 11.61 \times 10^{-33} \, \text{Ã } \)
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The Correct Option is C
Solution and Explanation
Step 1: Use the energy-wavelength relation.
The energy \( E \) of a photon is related to its wavelength \( \lambda \) by the formula: \[ E = \frac{hc}{\lambda} \] where: - \( h \) is Planck's constant (\(6.626 \times 10^{-34} \, \text{J} \cdot \text{s}\)), - \( c \) is the speed of light (\(3 \times 10^8 \, \text{m/s}\)), - \( \lambda \) is the wavelength.
Step 2: Apply the given values.
We are given that the energy required to create a hole is \( 57 \, \text{meV} = 57 \times 10^{-3} \, \text{eV} \). We need to convert this to Joules: \[ 57 \times 10^{-3} \, \text{eV} = 57 \times 10^{-3} \times 1.602 \times 10^{-19} \, \text{J} = 9.13 \times 10^{-21} \, \text{J} \]
Step 3: Solve for the wavelength.
Rearranging the energy-wavelength formula: \[ \lambda = \frac{hc}{E} \] Substitute the values: \[ \lambda = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{9.13 \times 10^{-21}} = 217105 \, \text{Ã } \]
The energy \( E \) of a photon is related to its wavelength \( \lambda \) by the formula: \[ E = \frac{hc}{\lambda} \] where: - \( h \) is Planck's constant (\(6.626 \times 10^{-34} \, \text{J} \cdot \text{s}\)), - \( c \) is the speed of light (\(3 \times 10^8 \, \text{m/s}\)), - \( \lambda \) is the wavelength.
Step 2: Apply the given values.
We are given that the energy required to create a hole is \( 57 \, \text{meV} = 57 \times 10^{-3} \, \text{eV} \). We need to convert this to Joules: \[ 57 \times 10^{-3} \, \text{eV} = 57 \times 10^{-3} \times 1.602 \times 10^{-19} \, \text{J} = 9.13 \times 10^{-21} \, \text{J} \]
Step 3: Solve for the wavelength.
Rearranging the energy-wavelength formula: \[ \lambda = \frac{hc}{E} \] Substitute the values: \[ \lambda = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{9.13 \times 10^{-21}} = 217105 \, \text{Ã } \]
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