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Question

In Rutherford's experiment, alpha particles with kinetic energy 'K' were scattered by a gold foil. If the kinetic energy of the alpha particles is quadrupled, how does the distance of closest approach ($r_{min}$) change?

ABDUL JILANI BASHA · Accepted Answer

## CORE INFORMATION
✓ **Answer**: (2) becomes one-fourth
## CONCEPT FRAMEWORK
 **Primary Concept**: Rutherford scattering and conservation of energy
 **Related Topics**: Coulomb's law, potential energy
 **Prerequisites**: Understanding of kinetic and potential energy, Coulomb's law.
## SOLUTION PATHWAY
**Given Information**:
  Initial kinetic energy of alpha particle = K
  Final kinetic energy of alpha particle = 4K
  The alpha particle is scattered by a gold foil (containing a positive nucleus).
**Method & Steps**:
 1. At the distance of closest approach ($r_{min}$), the kinetic energy of the alpha particle is converted into potential energy due to the Coulomb repulsion between the alpha particle and the gold nucleus.
  Reasoning: Conservation of energy principle.  The initial kinetic energy is entirely converted into potential energy at the point of closest approach.
  Formula/Rule: $K = \frac{1}{4\pi\epsilon_0} \frac{2Ze^2}{r_{min}}$, where K is the kinetic energy, Z is the atomic number of the gold nucleus, e is the elementary charge, and $r_{min}$ is the distance of closest approach.
 2. Let's denote the initial distance of closest approach as $r_{min,1}$ and the final distance of closest approach as $r_{min,2}$.  We have:
  $K = \frac{1}{4\pi\epsilon_0} \frac{2Ze^2}{r_{min,1}}$ (1)
  $4K = \frac{1}{4\pi\epsilon_0} \frac{2Ze^2}{r_{min,2}}$ (2)
  Working: Divide equation (2) by equation (1):
  $\frac{4K}{K} = \frac{\frac{1}{4\pi\epsilon_0} \frac{2Ze^2}{r_{min,2}}}{\frac{1}{4\pi\epsilon_0} \frac{2Ze^2}{r_{min,1}}} = \frac{r_{min,1}}{r_{min,2}}$
  $4 = \frac{r_{min,1}}{r_{min,2}}$
  $r_{min,2} = \frac{r_{min,1}}{4}$
 3. The distance of closest approach becomes one-fourth of its original value.
  Result: $r_{min}$ becomes one-fourth.
  Verification: The relationship is inversely proportional, thus quadrupling the kinetic energy reduces the distance of closest approach by a factor of 4.
 ## OPTION ANALYSIS
[A] :x: Doubling the kinetic energy would halve the distance of closest approach.
[B] ✓:  Correct. Quadrupling the kinetic energy reduces the distance of closest approach to one-fourth.
[C] :x: Halving the distance is incorrect. The relationship is inversely proportional.
[D] :x: The distance of closest approach depends on the kinetic energy; it does not remain the same.
 ## LEARNING AIDS
 **Common Mistakes**:
  1.Incorrectly applying the conservation of energy principle: Ensure that all kinetic energy is converted into potential energy at the point of closest approach.
  2.Misinterpreting the inverse square relationship between kinetic energy and distance of closest approach.
 **Quick Tips**:
  1.Remember the inverse square relationship: $K ∝ \frac{1}{r_{min}}$
  2.Use the formula directly and solve for the ratio of the distances.

ABDUL JILANI BASHA · Answer

(1) doubles

ABDUL JILANI BASHA · Answer

The correct Option is "(2) becomes one-fourth"

ABDUL JILANI BASHA · Answer

(3) becomes one-half

ABDUL JILANI BASHA · Answer

(4) remains the same

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