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

MCQ_SINGLE
Two capacitors $C$ and $2C$ charged to $V$ and $2V$ respectively are connected in parallel with opposite polarity. The common potential is:
NUMERICAL
In the given situation force at center on $1~kg$ mass is $F_{1}$. Now if $4~m$ and $3~m$ is interchanged the force is $F_{2}$ Given: $\frac{F_{1}}{F_{2}}=\frac{2}{\sqrt{\alpha}}$ Find $\alpha$ . (Diagram shows masses $4m$, $3m$, $m$, $2m$ arranged around a center point) .
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).
MCQ_SINGLE
Two disc having same moment of inertia about their axis. Thickness is $t_{1}$ and $t_{2}$ and they have same density. If $R_{1}/R_{2}=1/2$, then find $t_{1}/t_{2}$ :
MCQ_SINGLE
Find dimensions of $\frac{A}{B}$ if $(P+\frac{A^{2}}{B})+\frac{1}{2}\rho V^{2}=$ constant, where $P\rightarrow$ pressure, $\rho\rightarrow$ density, $V$ speed.
View All Questions