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Question #43639

Difficulty: 8/10 | 🟢 Free

The Question

An alpha particle of mass $m$ and charge $+2e$ is projected from infinity with an initial speed $v_0$ towards a massive, fixed nucleus of charge $+Ze$ . The impact parameter of the alpha particle is $b = \frac{2kZe^2}{mv_0^2}$ , where $k = \frac{1}{4\pi\epsilon_0}$ . What is the distance of closest approach of the alpha particle to the nucleus?

Options

A.
$b(1+\sqrt{2})$
B.
$2b$
C.
$b(\sqrt{2}-1)$
D.
$b\sqrt{2}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Distance of Closest Approach|Maximum Height Trap|Variable Force Integration

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to transition from 1D head-on collision models to 2D central force kinematics.

A common trap is assuming that the velocity at the distance of closest approach is zero, which is only true for a head-on collision ( $b=0$ ).

By forcing a non-zero impact parameter, students must recognize that the particle retains tangential kinetic energy at the periapsis.

Failing to apply Conservation of Angular Momentum to account for this residual kinetic energy leads directly to the incorrect 1D result ( $r_m = 2b$ ).

Additionally, a sign error in the potential energy term (assuming an attractive rather than repulsive force) leads to another trap option.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Atoms - Class 12
Concept	Distance of Closest Approach
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free