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The electricity conductivity of a semico...

The electricity conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than 2480nm is incident on it. The band gap (in eV) for the semiconductor is

A

0.9

B

0.7

C

0.5

D

0.1

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The correct Answer is:
To find the band gap of the semiconductor given that its electrical conductivity increases when electromagnetic radiation of wavelength shorter than 2480 nm is incident on it, we can use the formula relating the energy of a photon to its wavelength. ### Step-by-Step Solution: 1. **Identify the Given Wavelength**: - The wavelength (\( \lambda \)) is given as 2480 nm. Since we are interested in wavelengths shorter than this, we will use this value for our calculations. 2. **Convert Wavelength to Meters**: - Convert the wavelength from nanometers to meters: \[ \lambda = 2480 \, \text{nm} = 2480 \times 10^{-9} \, \text{m} \] 3. **Use the Energy-Wavelength Relation**: - The energy (\( E \)) of a photon can be calculated using the formula: \[ E = \frac{hc}{\lambda} \] - Where: - \( h \) = Planck's constant = \( 6.63 \times 10^{-34} \, \text{J s} \) - \( c \) = Speed of light = \( 3 \times 10^{8} \, \text{m/s} \) 4. **Substitute the Values into the Formula**: - Substitute \( h \), \( c \), and \( \lambda \) into the energy formula: \[ E = \frac{(6.63 \times 10^{-34} \, \text{J s}) \times (3 \times 10^{8} \, \text{m/s})}{2480 \times 10^{-9} \, \text{m}} \] 5. **Calculate the Energy**: - Calculate the numerator: \[ 6.63 \times 10^{-34} \times 3 \times 10^{8} = 1.989 \times 10^{-25} \, \text{J m} \] - Now divide by the wavelength in meters: \[ E = \frac{1.989 \times 10^{-25}}{2480 \times 10^{-9}} = 8.016 \times 10^{-19} \, \text{J} \] 6. **Convert Energy from Joules to Electron Volts**: - Use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \): \[ E \, (\text{in eV}) = \frac{8.016 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \approx 5.01 \, \text{eV} \] 7. **Final Result**: - The band gap of the semiconductor is approximately \( 5.01 \, \text{eV} \).
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