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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

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

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Solution

The energy of the radiation can be calculated using the formula:

E = hc/λ

where: E is the energy, 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 = 1000 x 10^-9 m = 1 x 10^-6 m

Then, substitute the values into the formula:

E = (6.626 x 10^-34 Js)(3 x 10^8 m/s) / 1 x 10^-6 m E = 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) E = 1.24 eV

Therefore, the band gap of the semiconductor is closest to 1.2 eV, so the answer is C- 1.2 eV.

This problem has been solved

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