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The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than 2480 nm is incident on it. The band gap in (eV) for the semiconductor is
- a)2.5 eV
- b)1.1 eV
- c)0.7 eV
- d)0.5 eV
Correct answer is option 'D'. Can you explain this answer?
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The electrical conductivity of a semiconductor increases when electrom...
Band gap = energy of photon of wavelength 2480 nm.
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The electrical conductivity of a semiconductor increases when electrom...
The conductivity of a semiconductor is determined by the presence of charge carriers, which can be either electrons or holes. The band gap is the energy difference between the valence band (where electrons are tightly bound to atoms) and the conduction band (where electrons are free to move and conduct electricity). When photons of electromagnetic radiation are incident on a semiconductor, they can excite electrons from the valence band to the conduction band, creating electron-hole pairs and increasing the conductivity of the material.
In this question, it is stated that the conductivity of the semiconductor increases when electromagnetic radiation of wavelength shorter than 2480 nm is incident on it. To understand why the band gap is 0.5 eV, we need to consider the relationship between energy and wavelength.
- Relationship between Energy and Wavelength:
- The energy of a photon is given by the equation E = hc/λ, where E is the energy, h is Planck's constant (approximately 6.63 x 10^-34 J·s), c is the speed of light (approximately 3 x 10^8 m/s), and λ is the wavelength of the radiation.
- As the wavelength decreases, the energy of the photon increases.
- Effect of Shorter Wavelength Radiation:
- The question states that the conductivity increases when the semiconductor is exposed to radiation with a wavelength shorter than 2480 nm.
- Since shorter wavelength radiation has higher energy, it implies that the energy of the photons in this range is sufficient to excite electrons from the valence band to the conduction band.
- This indicates that the band gap of the semiconductor must be smaller than the energy of these photons.
- Determining the Band Gap:
- Given that the wavelength threshold for increased conductivity is 2480 nm, we can calculate the corresponding energy of the photons.
- Using the equation E = hc/λ, we can substitute the values of h, c, and λ to find E.
- E = (6.63 x 10^-34 J·s)(3 x 10^8 m/s)/(2480 x 10^-9 m)
- E ≈ 7.98 x 10^-20 J
- Converting this energy to electron volts (eV) by dividing by the charge of an electron (approximately 1.6 x 10^-19 C), we get:
- E ≈ 7.98 x 10^-20 J / 1.6 x 10^-19 C ≈ 0.49875 eV
- Rounded to the nearest tenth, the band gap is approximately 0.5 eV.
Therefore, the correct answer is option D) 0.5 eV.
In this question, it is stated that the conductivity of the semiconductor increases when electromagnetic radiation of wavelength shorter than 2480 nm is incident on it. To understand why the band gap is 0.5 eV, we need to consider the relationship between energy and wavelength.
- Relationship between Energy and Wavelength:
- The energy of a photon is given by the equation E = hc/λ, where E is the energy, h is Planck's constant (approximately 6.63 x 10^-34 J·s), c is the speed of light (approximately 3 x 10^8 m/s), and λ is the wavelength of the radiation.
- As the wavelength decreases, the energy of the photon increases.
- Effect of Shorter Wavelength Radiation:
- The question states that the conductivity increases when the semiconductor is exposed to radiation with a wavelength shorter than 2480 nm.
- Since shorter wavelength radiation has higher energy, it implies that the energy of the photons in this range is sufficient to excite electrons from the valence band to the conduction band.
- This indicates that the band gap of the semiconductor must be smaller than the energy of these photons.
- Determining the Band Gap:
- Given that the wavelength threshold for increased conductivity is 2480 nm, we can calculate the corresponding energy of the photons.
- Using the equation E = hc/λ, we can substitute the values of h, c, and λ to find E.
- E = (6.63 x 10^-34 J·s)(3 x 10^8 m/s)/(2480 x 10^-9 m)
- E ≈ 7.98 x 10^-20 J
- Converting this energy to electron volts (eV) by dividing by the charge of an electron (approximately 1.6 x 10^-19 C), we get:
- E ≈ 7.98 x 10^-20 J / 1.6 x 10^-19 C ≈ 0.49875 eV
- Rounded to the nearest tenth, the band gap is approximately 0.5 eV.
Therefore, the correct answer is option D) 0.5 eV.
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The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than 2480 nm is incident on it. The band gap in (eV) for the semiconductor isa)2.5 eVb)1.1 eVc)0.7 eVd)0.5 eVCorrect answer is option 'D'. Can you explain this answer?
Question Description
The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than 2480 nm is incident on it. The band gap in (eV) for the semiconductor isa)2.5 eVb)1.1 eVc)0.7 eVd)0.5 eVCorrect answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than 2480 nm is incident on it. The band gap in (eV) for the semiconductor isa)2.5 eVb)1.1 eVc)0.7 eVd)0.5 eVCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than 2480 nm is incident on it. The band gap in (eV) for the semiconductor isa)2.5 eVb)1.1 eVc)0.7 eVd)0.5 eVCorrect answer is option 'D'. Can you explain this answer?.
The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than 2480 nm is incident on it. The band gap in (eV) for the semiconductor isa)2.5 eVb)1.1 eVc)0.7 eVd)0.5 eVCorrect answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than 2480 nm is incident on it. The band gap in (eV) for the semiconductor isa)2.5 eVb)1.1 eVc)0.7 eVd)0.5 eVCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than 2480 nm is incident on it. The band gap in (eV) for the semiconductor isa)2.5 eVb)1.1 eVc)0.7 eVd)0.5 eVCorrect answer is option 'D'. Can you explain this answer?.
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