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Academic session 2026-27, Chapter 14 in the current curriculum.
Welcome to the NCERT Solutions for Class 12 Physics - Chapter Semiconductor Electronics. This page offers a step-by-step solution to the specific question from Exercise 1, Question 11:
A p-n photodiode is fabricated from a semiconductor with band gap of 2.8 eV. Can it detect a wavelength of 6000 nm?
. With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.A p-n photodiode is fabricated from a semiconductor with band gap of 2.8 eV. Can it detect a wavelength of 6000 nm?
Given that,
Energy band gap of the given photodiode, E g = 2.8 eV
Wavelength, λ = 6000 nm = 6000 × 10 −9 m
The energy of a signal is given by the relation: E = hc/λ
Where, h = Planck’s constant = 6.626 × 10 −34 Js
c = Speed of light = 3 × 10 8 m/s
E = 6.626 x 10-34 x 3 x 108 / 6000 x 10-9 = 3.313 x 10-20 J
But 1.6 × 10 −19 J = 1 eV
E = 3.313 × 10 −20 J
∴E = 3.313 × 10 −20 J = 3.313 x 10-20 / 1.6 x 10-19 = 0.207 eV
The energy of a signal of wavelength 6000 nm is 0.207 eV, which is less than 2.8 eV − the energy band gap of a photodiode. Hence, the photodiode cannot detect the signal.
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Welcome to the NCERT Solutions for Class 12 Physics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 11: A p-n photodiode is fabricated from a semiconductor with band gap of 2.8 eV. Can it detect a wavelen....
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