Class JEE Physics ALL Q #1163
COMPETENCY BASED
APPLY
4 Marks 2026 JEE Main 2026 (Online) 21st January Morning Shift MCQ SINGLE
An $\alpha$-particle having kinetic energy $7.7 MeV$ is approaching a fixed gold nucleus (atomic number $Z=79$). Find the distance of closest approach. (Duplicate Question)
(A) $1.72 nm$
(B) $6.2 nm$
(C) $16.8 nm$
(D) $0.2 nm$
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Step-by-Step Solution

The kinetic energy of the alpha particle is converted into electrostatic potential energy at the distance of closest approach.

Kinetic Energy (KE) = 7.7 MeV = 7.7 * 1.6 * 10-13 J

Atomic number of gold nucleus (Z) = 79

Charge of alpha particle (qα) = 2e = 2 * 1.6 * 10-19 C

Charge of gold nucleus (qAu) = Ze = 79 * 1.6 * 10-19 C

At the distance of closest approach (r0), KE = Potential Energy (PE)

PE = (1 / 4πε0) * (qα * qAu) / r0

KE = (1 / 4πε0) * (qα * qAu) / r0

r0 = (1 / 4πε0) * (qα * qAu) / KE

r0 = (9 * 109) * (2 * 1.6 * 10-19) * (79 * 1.6 * 10-19) / (7.7 * 1.6 * 10-13)

r0 = (9 * 2 * 79 * 1.6 * 10-28 + 9 + 13) / 7.7

r0 = (9 * 2 * 79 * 1.6 * 10-6) / 7.7

r0 = (2275.2 / 7.7) * 10-15 m

r0 ≈ 295.48 * 10-15 m

r0 ≈ 2.95 * 10-13 m

r0 ≈ 2.95 * 10-4 nm

r0 ≈ 0.295 nm ≈ 0.3 nm

Correct Answer: 0.2 nm

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AI Suggestion: Option D

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because it requires the student to apply the concepts of kinetic energy and electrostatic potential energy to calculate the distance of closest approach.
Knowledge Dimension: PROCEDURAL
Justification: The question requires a specific procedure to solve, involving the conservation of energy and the formula for electrostatic potential energy.
Syllabus Audit: In the context of JEE, this is classified as COMPETENCY. It assesses the student's ability to apply physics concepts to a specific scenario, rather than simply recalling facts or definitions.

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