JEE Exam > JEE Questions > The electrical conductivity of a semiconducto...
Start Learning for Free
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?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
The electrical conductivity of a semiconductor increases when electrom...
Band gap = energy of photon of wavelength 2480 nm.
So,
So,
Most Upvoted Answer
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.
Attention JEE Students!
To make sure you are not studying endlessly, EduRev has designed JEE study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in JEE.
|
Explore Courses for JEE exam
|
|
Similar JEE Doubts
Top Courses for JEEView all
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?.
Solutions 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? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of 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? defined & explained in the simplest way possible. Besides giving the explanation of
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?, a detailed solution 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? has been provided alongside types of 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? theory, EduRev gives you an
ample number of questions to practice 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? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup