JEE Main 2020 · PhysicsMultiple choiceSingle correctEasyCalculation

JEE Main 3 September 2020, Shift 2, Physics Q19

Question 19 of 75 in this shift, Physics question 19 of 25.

If a semiconductor photodiode can detect a photon with a maximum wavelength of 400 nm, then its band gap energy is : Planck's constant

h = 6.63 \times 10^{-34}

J.s. Speed of light

c = 3 \times 10^{8}

m/s

(1)3.1 eVOfficial answer
(2)1.1 eV
(3)2.0 eV
(4)1.5 eV

Official answer

Option 1

NTA final key (Sep 2020).

Chapter: Semiconductor Electronics: Materials, Devices and Simple Circuits
Topic: Semiconductor Diodes
Idea tested: LED Bandgap and Emission Wavelength

Same idea in other shifts

Asked 3× in all

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Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.