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Question #42663

Difficulty: 8/10 | 🟢 Free

The Question

A particle of mass $m$ and initial speed $v_0$ approaches a fixed repulsive scattering center from infinity. The potential energy of the interaction is given by $V(r) = \frac{\alpha}{r^2}$ , where $\alpha > 0$ is a constant and $r$ is the radial distance from the center. If the impact parameter of the particle is $b$ , what is the scattering angle $\theta$ of the particle?

Options

A.
$\pi \left( 1 - \frac{b}{\sqrt{b^2 + \frac{2\alpha}{m v_0^2}}} \right)$
B.
$\pi \left( \frac{b}{\sqrt{b^2 + \frac{2\alpha}{m v_0^2}}} \right)$
C.
$\pi \left( 1 - \frac{b}{\sqrt{b^2 + \frac{\alpha}{m v_0^2}}} \right)$
D.
$\frac{\pi}{2} \left( 1 - \frac{b}{\sqrt{b^2 + \frac{2\alpha}{m v_0^2}}} \right)$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Potential Gradient Direction|Impact Parameter Geometry|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to derive scattering angles for non-Coulombic central potentials using the impact parameter integral, bypassing memorized Rutherford formulas.

The primary trap lies in the 'Non-Central_Force_Constraint' misconception, where students either attempt to use a small-angle impulse approximation (assuming constant force direction) or fail to correctly incorporate the centrifugal barrier $L^2/(2mr^2)$ into the effective potential.

Furthermore, students frequently confuse the total swept angle $2\phi_0$ with the actual scattering deviation angle $\theta = \pi - 2\phi_0$ , leading them to select the trap option representing the asymptote angle.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Electrostatic Potential and Capacitance - Class 12
Concept	Potential Gradient Direction
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free