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

Numerical

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

#### Solution

Energy band gap of the given photodiode, Eg = 2.8 eV

Wavelength, λ = 6000 nm = 6000 × 10−9 m

The energy of a signal is given by the relation:

"E" = ("hc")/lambda

Where

h = Planck’s constant

= 6.626 × 10−34 Js

c = Speed of light

= 3 × 108 m/s

"E" = (6.626 xx 10^(-34) xx 3 xx 10^8)/(6000 xx 10^(-9))

= 3.313 × 10−20 J

But 1.6 × 10−19 J = 1 eV

∴ E = 3.313 × 10−20 J

= (3.313 xx 10^(-20))/(1.6 xx 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.

Concept: Special Purpose P-n Junction Diodes
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#### APPEARS IN

NCERT Physics Part 1 and 2 Class 12
Chapter 14 Semiconductor Electronics: Materials, Devices and Simple Circuits
Exercise | Q 14.7 | Page 497
NCERT Class 12 Physics Textbook
Chapter 14 Semiconductor Electronics: Materials, Devices and Simple Circuits
Exercise | Q 11 | Page 510

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