Class JEE Physics ALL Q #1161
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.
(A) $1.72 nm$
(B) $6.2 nm$
(C) $16.8 nm$
(D) $0.2 nm$
Prev Next

AI Tutor Explanation

Powered by Gemini

Step-by-Step Solution

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

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

Potential Energy (PE) = (1 / 4πε0) * (q1q2 / r)

Where:

  • q1 = charge of alpha particle = 2e = 2 * 1.6 * 10-19 C
  • q2 = charge of gold nucleus = 79e = 79 * 1.6 * 10-19 C
  • r = distance of closest approach
  • 1 / 4πε0 = 9 * 109 Nm2/C2

At the distance of closest approach, KE = PE

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

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

r = (9 * 2 * 79 * 1.6 * 10-29) / (7.7 * 10-13)

r = (9 * 2 * 79 * 1.6) * 10-16 / 7.7

r = 2275.2 * 10-16 / 7.7

r = 295.48 * 10-16 m

r = 2.9548 * 10-14 m

r = 2.9548 * 10-14 m = 0.29548 * 10-12 m = 0.29548 pm

r ≈ 3.0 * 10-14 m

r ≈ 0.03 nm

Correct Answer: 0.2 nm

APPLY|||COMPETENCY|||PROCEDURAL|||HARD|||
Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student needs to apply the concepts of conservation of energy and electrostatic potential energy to find the distance of closest approach.
Knowledge Dimension: PROCEDURAL
Justification: The solution involves a series of steps: converting units, equating kinetic and potential energies, and solving for the distance. This requires a procedural understanding of the concepts.
Syllabus Audit: In the context of JEE, this is classified as COMPETENCY. It requires the application of knowledge to solve a problem, rather than just recalling facts or definitions.

Step-by-Step Solution

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

Kinetic Energy (KE) = 7.7 MeV = 7.7 * 1.602 * 10-13 J

Potential Energy (PE) = (1 / 4πε0) * (q1q2 / r)

Where:

  • q1 = charge of alpha particle = 2e = 2 * 1.602 * 10-19 C
  • q2 = charge of gold nucleus = 79e = 79 * 1.602 * 10-19 C
  • r = distance of closest approach
  • 1 / 4πε0 = 9 * 109 Nm2/C2

At the distance of closest approach, KE = PE

7.7 * 1.602 * 10-13 = (9 * 109) * (2 * 1.602 * 10-19) * (79 * 1.602 * 10-19) / r

r = (9 * 109 * 2 * 1.602 * 10-19 * 79 * 1.602 * 10-19) / (7.7 * 1.602 * 10-13)

r = (9 * 2 * 79 * 1.602 * 1.602 * 10-29) / (7.7 * 1.602 * 10-13)

r = (3634.84 * 10-29) / (12.3354 * 10-13)

r = 294.67 * 10-16 m

r = 0.29467 * 10-14 m

r = 0.29467 * 10-14 m = 0.029467 * 10-12 m = 0.29467 pm

r ≈ 0.2 * 10-13 m

r ≈ 0.2 nm

Correct Answer: 0.2 nm

AI Suggestion: Option D

AI generated content. Review strictly for academic accuracy.

Pedagogical Audit
Bloom's Analysis: This is an APPLY question because it requires the student to apply the concepts of kinetic energy, potential energy, and conservation of energy to calculate the distance of closest approach.
Knowledge Dimension: PROCEDURAL
Justification: The question requires a series of steps to solve, including converting units, applying the conservation of energy principle, and using 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 just recalling definitions or formulas.

More from this Chapter

NUMERICAL
Given $\vec{E}=10x\hat{i}+5y\hat{j}$. Potential at $(10, 20)$ is $500~V$. Find potential at origin .
NUMERICAL
Six charges (four $+q$, two $-q$) are present at circle of radius $r$ and centred at origin as shown. Electric field at origin is .
MCQ_SINGLE
Ratio of de-Broglie wavelengths of a proton and an alpha particle accelerated through the same potential is:
MCQ_SINGLE
A wave propagates whose electric field is given by $\vec{E}=69~\sin(\omega t-kx)\hat{j}$. Find the direction of the magnetic field.
MCQ_SINGLE
Two rods of equal length 60 cm each are joined together end to end. The coefficients of linear expansion are $24\times10^{-6\circ}C^{-1}$ and $1.2\times10^{-5\circ}C^{-1}$. Initial temperature $30^{\circ}C$ is increased to $100^{\circ}C$. Find final length (in cm).
View All Questions