JEE Main 2022 · PhysicsNumerical answerNumerical valueMediumCalculation

JEE Main 30 July 2022, Shift 2, Physics Q21

Question 21 of 90 in this shift, Physics question 21 of 30, Section B.

A nuclear physicist performed an experiment on the scattering of alpha particles by thin metal foil of gold. The alpha particles were accelerated up to kinetic energy 3.2 Mev. The estimation of the radius of the nucleus of gold element (Z = 79) by him is _______

\times

10^{-16}

m (Take

\frac{1}{4\pi \in_o} = 9\times10^9\ \frac{\mathrm{Nm^2}}{\mathrm{C^2}}

)

Official answer

711

NTA final key.

Chapter: Atoms
Topic: Electric Potential
Idea tested: Distance of Closest Approach

Same topic in other shifts

All Electric Potential questions

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2 Apr 2025, Shift 2 · Q42Two large plane parallel conducting plates are kept 10 cm apart as shown in figure. The potential difference between them is V. The…MediumDiagram basedHas a figure
3 Apr 2025, Shift 1 · Q39The electrostatic potential on the surface of uniformly charged spherical shell of radius R = 10 cm is 120 V. The potential at the centre…EasySingle correct
7 Apr 2025, Shift 1 · Q38Two charges $q_1$ and $q_2$ are separated by a distance of 30 cm. A third charge $q_3$ initially at 'C' as shown in the figure, is moved…MediumDiagram basedHas a figure
8 Apr 2025, Shift 2 · Q35Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R Assertion A : Work done in moving a…EasyAssertion–reason

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.