SELECT * FROM question_mgmt as q WHERE id=2762 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=3 AND subId=8 AND chapterId=128 and ex_no='1' AND status=1 ORDER BY CAST(question_no AS UNSIGNED)

Question:

A p-n photodiode is fabricated from a semiconductor with band gap of 2.8 eV. Can it detect a wavelength of 6000 nm?

Answer:

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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