For head-on collision, entire kinetic energy converts into electrostatic potential energy at the closest approach.
\[ K = U \]
\[ 7.7 \text{ MeV} = \frac{1}{4\pi\epsilon_0} \frac{(2e)(79e)}{r} \]
Convert 7.7 MeV to joules:
\[ 7.7 \times 10^6 \times 1.6 \times 10^{-19} \]
\[ = 1.232 \times 10^{-12} \text{ J} \]
Now,
\[ 1.232 \times 10^{-12} = 9 \times 10^9 \times \frac{158 e^2}{r} \]
Using \( e = 1.6 \times 10^{-19} \),
\[ r = 2.95 \times 10^{-14} \text{ m} \]
Therefore correct answer is (C).
The concept of distance of closest approach originates from Rutherford’s alpha scattering experiment, which fundamentally changed our understanding of atomic structure. In this experiment, alpha particles were directed towards a thin metal foil, typically gold. Most alpha particles passed straight through, but a small fraction were deflected at large angles, and a few even rebounded. This observation led to the conclusion that positive charge and most of the atomic mass are concentrated in a very small central nucleus.
When an alpha particle approaches a positively charged nucleus, it experiences Coulomb repulsion. If the collision is head-on, the particle momentarily comes to rest at the closest point before reversing direction. At this instant, kinetic energy becomes zero and entire initial kinetic energy converts into electrostatic potential energy.
The electrostatic potential energy between two point charges is given by:
\[ U = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r} \]
For alpha particle, charge = +2e. For gold nucleus, charge = +79e.
Applying energy conservation:
\[ K = \frac{1}{4\pi\epsilon_0} \frac{2e \cdot 79e}{r} \]
This formula is extremely important for JEE Main and Advanced. Students must remember that this applies only for head-on collision.
Important points:
• Alpha particle charge = 2e
• Atomic number gives nuclear charge
• MeV must be converted into joules in SI problems
• Closest approach distance is typically of order 10⁻¹⁴ m
In many JEE problems, they directly give formula:
\[ r = \frac{1}{4\pi\epsilon_0} \frac{2Ze^2}{K} \]
This result shows distance inversely proportional to kinetic energy. Higher energy means closer penetration.
Common mistakes:
• Forgetting to multiply alpha charge by 2
• Not converting MeV to joules
• Using atomic mass instead of atomic number
• Arithmetic mistakes in powers of 10
This concept also gives insight into nuclear size. If calculated closest approach becomes comparable to nuclear radius (~10⁻¹⁵ m), nuclear forces become significant and pure Coulomb law no longer applies.
Rutherford scattering formula for angular distribution is another advanced topic in JEE Advanced. However, JEE Main generally asks distance of closest approach or qualitative conceptual questions.
Understanding this topic builds bridge between electrostatics and nuclear physics. It demonstrates real-life application of Coulomb’s law at subatomic scale.
This concept is important not only for JEE but also for understanding particle accelerators and nuclear reactions.