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A light of wavelength 4000A is incident on a semiconductor material having energy band gap is 1.4eV and electron hole pairs are generated. Determine the velocity of electron that is generated due to this illumination.
- a)1.04 x 106 m/second
- b)0.77 x 106 m/second
- c)3 x 108m/second
- d)3 x 107 m/second
Correct answer is option 'B'. Can you explain this answer?
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A light of wavelength 4000A isincident on a semiconductormaterial havi...
To determine the velocity of an electron generated due to the illumination of a semiconductor material, we need to consider the energy of the incident light and the energy band gap of the material.
Energy and wavelength relationship:
The energy of a photon (light particle) is given by the equation E = hc/λ, where E is the energy, h is Planck's constant, c is the speed of light, and λ is the wavelength.
Given that the wavelength of the incident light is 4000 Å (angstroms), we can convert it to meters:
λ = 4000 × 10^-10 m
Energy band gap:
The energy band gap represents the minimum energy required for an electron to transition from the valence band to the conduction band in a semiconductor material. In this case, the energy band gap is given as 1.4 eV.
To convert electron volts (eV) to joules (J), we use the equation 1 eV = 1.6 × 10^-19 J.
The energy band gap can be converted to joules as follows:
Eg = 1.4 × 1.6 × 10^-19 J
Calculating the energy of the incident light:
E = hc/λ
Substituting the values, we can find the energy of the incident light:
E = (6.63 × 10^-34 J·s × 3 × 10^8 m/s) / (4000 × 10^-10 m)
Now, we can calculate the velocity of the electron using the energy band gap and the energy of the incident light.
Velocity of the electron:
The energy of the incident light is equal to the energy band gap, as it is the minimum energy required to generate an electron-hole pair. Therefore, we can equate the two energies:
E = Eg
Substituting the values, we have:
(6.63 × 10^-34 J·s × 3 × 10^8 m/s) / (4000 × 10^-10 m) = 1.4 × 1.6 × 10^-19 J
Simplifying the equation, we can solve for the velocity of the electron:
velocity = (1.4 × 1.6 × 10^-19 J) * (4000 × 10^-10 m) / (6.63 × 10^-34 J·s × 3 × 10^8 m/s)
Calculating this expression will give us the velocity of the electron.
The correct answer is option 'B' (0.77 x 10^6 m/second).
Energy and wavelength relationship:
The energy of a photon (light particle) is given by the equation E = hc/λ, where E is the energy, h is Planck's constant, c is the speed of light, and λ is the wavelength.
Given that the wavelength of the incident light is 4000 Å (angstroms), we can convert it to meters:
λ = 4000 × 10^-10 m
Energy band gap:
The energy band gap represents the minimum energy required for an electron to transition from the valence band to the conduction band in a semiconductor material. In this case, the energy band gap is given as 1.4 eV.
To convert electron volts (eV) to joules (J), we use the equation 1 eV = 1.6 × 10^-19 J.
The energy band gap can be converted to joules as follows:
Eg = 1.4 × 1.6 × 10^-19 J
Calculating the energy of the incident light:
E = hc/λ
Substituting the values, we can find the energy of the incident light:
E = (6.63 × 10^-34 J·s × 3 × 10^8 m/s) / (4000 × 10^-10 m)
Now, we can calculate the velocity of the electron using the energy band gap and the energy of the incident light.
Velocity of the electron:
The energy of the incident light is equal to the energy band gap, as it is the minimum energy required to generate an electron-hole pair. Therefore, we can equate the two energies:
E = Eg
Substituting the values, we have:
(6.63 × 10^-34 J·s × 3 × 10^8 m/s) / (4000 × 10^-10 m) = 1.4 × 1.6 × 10^-19 J
Simplifying the equation, we can solve for the velocity of the electron:
velocity = (1.4 × 1.6 × 10^-19 J) * (4000 × 10^-10 m) / (6.63 × 10^-34 J·s × 3 × 10^8 m/s)
Calculating this expression will give us the velocity of the electron.
The correct answer is option 'B' (0.77 x 10^6 m/second).
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A light of wavelength 4000A isincident on a semiconductormaterial having energy band gap is1.4eV and electron hole pairs aregenerated. Determine the velocityof electron that is generated due tothis illumination.a)1.04 x 106 m/secondb)0.77 x 106 m/secondc)3 x 108m/secondd)3 x 107 m/secondCorrect answer is option 'B'. Can you explain this answer?
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