JEE Main 2022 · PhysicsMultiple choiceSingle correctEasyRecall

JEE Main 28 July 2022, Shift 1, Physics Q47

Question 47 of 90 in this shift, Physics question 17 of 30, Section A.

The equation

\lambda=\frac{1.227}{x}

nm can be used to find the de-Brogli wavelength of an electron. In this equation

x

stands for : Where m = mass of electron P = momentum of electron K = Kinetic energy of electron V = Accelerating potential in volts for electron

(1) $\sqrt{mK}$
(2) $\sqrt{P}$
(3) $\sqrt{K}$
(4) $\sqrt{V}$ Official answer

Official answer

Option 4

NTA final key (2022 Session 2).

Chapter: Dual Nature of Radiation and Matter
Topic: Matter Waves
Idea tested: Wavelength from Accelerating Potential

Same idea in other shifts

Asked 5× in all

6 Apr 2026, Shift 2 · Q47The de Broglie wavelength for an electron accelerated through the potential difference of $V_1$ volt is $\lambda_1$ . When the potential…EasyNumerical value
12 Apr 2023, Shift 1 · Q47A proton and an $\alpha$ -particle are accelerated from rest by 2 V and 4 V potentials, respectively. The ratio of their de-Broglie…MediumSingle correct
30 Jul 2022, Shift 2 · Q19The de Broglie wavelength associated with an electron, accelerated through a potential difference of 15056 V, is:EasySingle correct
22 Jul 2021, Shift 2 · Q10An electron of mass $\mathrm{m_e}$ and a proton of mass $\mathrm{m_p}$ are accelerated through the same potential difference. The ratio of…EasySingle correct

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.