JEE Main 2026 · PhysicsNumerical answerNumerical valueEasyCalculation

JEE Main 6 April 2026, Shift 2, Physics Q47

Question 47 of 75 in this shift, Physics question 22 of 25, Section B.

The de Broglie wavelength for an electron accelerated through the potential difference of

V_1

volt is

\lambda_1

. When the potential difference is changed to

V_2

volt, the associated de Broglie wavelength is increased by 50%. If

(V_1/V_2)=(9/\alpha)

, then the value of

\alpha

is __________.

Official answer

NTA final key.

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

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
28 Jul 2022, Shift 1 · Q47The 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 :…EasySingle 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.