The electrical conductivity of a semiconductor increases when a radiation of wavelength shorter than 1000 nm is incident on it. The band gap of the semiconductor is A- 2.4 eV B- 2.2 eV C- 1.2 eV D- 0.8 eV LO
Question
The electrical conductivity of a semiconductor increases when a radiation of wavelength shorter than 1000 nm is incident on it. The band gap of the semiconductor is A- 2.4 eV B- 2.2 eV C- 1.2 eV D- 0.8 eV LO
Solution
The energy of the radiation can be calculated using the formula E = hc/λ, where h is Planck's constant (6.626 x 10^-34 Js), c is the speed of light (3 x 10^8 m/s), and λ is the wavelength.
First, convert the wavelength from nm to m: 1000 nm = 1 x 10^-6 m.
Then, substitute these values into the formula to find the energy:
E = (6.626 x 10^-34 Js)(3 x 10^8 m/s) / 1 x 10^-6 m = 1.987 x 10^-19 J
To convert this energy to electron volts (eV), use the conversion factor 1 eV = 1.6 x 10^-19 J:
E = 1.987 x 10^-19 J * (1 eV / 1.6 x 10^-19 J) = 1.24 eV
Therefore, the band gap of the semiconductor is closest to 1.2 eV, so the answer is C- 1.2 eV.
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