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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.
APPLY COMPETENCY 4 Marks
Concept Application
50%
Calculation / Logic
50%
Target Level
MEDIUM
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APPLY COMPETENCY MEDIUM

Q: 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.

Question Analysis & Solution

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
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