JEE Main 2024 · PhysicsNumerical answerNumerical valueMediumCalculation

JEE Main 8 April 2024, Shift 1, Physics Q58

Question 58 of 90 in this shift, Physics question 28 of 30, Section B.

In an alpha particle scattering experiment distance of closest approach for the

\alpha

particle is

4.5\times10^{-14}

m. If target nucleus has atomic number 80, then maximum velocity of

\alpha

- particle is ____________

\times10^5

m/s approximately. (

\frac{1}{4\pi\epsilon_0}=9\times10^9

SI unit, mass of

\alpha

particle

=6.72\times10^{-27}

kg)

Official answer

156

NTA final key.

Chapter: Atoms
Topic: Coulomb's Law
Idea tested: Distance of Closest Approach

Same topic in other shifts

All Coulomb's Law questions

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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 42 of the 174 JEE Main shifts held from 2017 to 2026.