Q.Derive an expression for the distance of closest approach $r_0$ when an alpha particle is fired head-on at a nucleus of atomic number $Z$.

Why Distance of Closest Approach Depends on Kinetic Energy

When an alpha-particle ($\alpha$, carrying charge $+2e$) is fired head-on at a nucleus (charge $+Ze$), it slows down as electrostatic repulsion does negative work. At the distance of closest approach, the particle momentarily stops: all its initial kinetic energy has been converted into electrostatic potential energy.

The key insight is that the distance of closest approach is determined entirely by energy conservation. The greater the initial kinetic energy, the closer the alpha-particle can get before being turned back.

Step-by-Step Solution

1. Write the energy conservation equation at closest approach.

   Initially, the alpha-particle has kinetic energy $KE = \frac{1}{2}m v^2$ and is far from the nucleus (so $PE \approx 0$). At closest approach (distance $d$), it has zero velocity and maximum potential energy:

where $k = \frac{1}{4\pi\epsilon_0}$ is Coulomb's constant.

2. Solve for the distance of closest approach $d$.

   Rearranging:

This shows that $d \propto \frac{1}{v^2}$.

3. Find the new distance when velocity is halved.

   If the new velocity is $v' = \frac{v}{2}$, the new distance $d'$ is:

Impact parameter in Rutherford scattering

The impact parameter $b$ (perpendicular distance between the incident alpha particle's initial path and the nucleus) is related to the scattering angle $\theta$ by

where $Z$ is the atomic number of the target nucleus and $E$ is the alpha particle's kinetic energy.

Checking Assertion (A). As $\theta\to180^\circ$ (a head-on collision, bouncing straight back), $\cot(\theta/2)=\cot(90^\circ)=0$, so $b\to0$ -- the impact parameter is minimum, not maximum, at $\theta=180^\circ$. A large scattering angle corresponds to a small impact parameter (the particle aims almost directly at the nucleus); a large impact parameter (particle passing far from the nucleus) gives only a small deflection. So A is false.

Checking Reason (R). The formula above shows $b$ is directly proportional to $Z$ -- a larger nuclear charge produces a stronger Coulomb repulsion, deflecting the particle more strongly for the same aiming distance, so a given scattering angle corresponds to a larger $b$ for larger $Z$. The impact parameter clearly does depend on the atomic number, so R is also false.

The Geiger-Marsden experiment revealed the nuclear structure of the atom through alpha-particle scattering. When an alpha particle approaches a gold nucleus head-on, it experiences a repulsive Coulomb force that slows it down. At the distance of closest approach, the particle momentarily stops before reversing direction. This is a pure energy-conversion problem: kinetic energy transforms entirely into electrostatic potential energy.

The key insight is conservation of energy. Initially, the alpha particle has kinetic energy $K$ and negligible potential energy (it starts far away). At closest approach distance $d$, the particle has zero kinetic energy and maximum potential energy.

1. Write the initial energy state. Far from the nucleus, the alpha particle has kinetic energy $K$ and potential energy $U_i \approx 0$ (taking $U = 0$ at infinity).

2. Write the final energy state at closest approach. At distance $d$, the particle stops momentarily, so kinetic energy is zero. The potential energy between the alpha particle (charge $q_\alpha = 2e$) and gold nucleus (charge $q_{\text{Au}} = Ze$, where $Z = 79$ for gold) is:

3. Apply conservation of energy. Since total mechanical energy is conserved:

4. Solve for $d$ in terms of $K$. Rearranging:

5. Identify the proportionality. All factors except $K$ are constants (Coulomb constant $k$, atomic number $Z$, elementary charge $e$). Therefore:

At the distance of closest approach, the $\alpha$-particle momentarily stops (its velocity along the line of approach becomes zero) as it is repelled by the gold nucleus. All of its initial kinetic energy has been converted into electrostatic potential energy of the $\alpha$-nucleus system. Thus $\tfrac{1}{2}mv^2 = \dfrac{1}{4\pi\varepsilon_0}\dfrac{(2e)(Ze)}{r_0}$.