Question

An alpha particle moves towards a fixed nucleus carrying charge \(Ze\), with initial speed \(v_0\) and impact parameter \(b\). Starting from a large distance from the nucleus, its distance of closest approach is \(r_m\) and its speed there is \(v_m\). Then which of the following options is correct?

\( \left(k = \dfrac{1}{4\pi\epsilon_0} \text{ and } r_0 = k\dfrac{Ze^2}{mv_0^2}\right) \)

• A. \( v_0 b = v_m r_m \)
• B. \( v_0 b = 2 v_m r_0 \)
• C. For \( \dfrac{b}{r_0} \ll 1 \), \( r_m = 4r_0 + \dfrac{b^2}{2r_0} \), ignoring higher order corrections in \( \dfrac{b}{r_0} \)
• D. For \( \dfrac{b}{r_0} \ll 1 \), \( r_m = 4r_0 + \dfrac{b^2}{8r_0} \), ignoring higher order corrections in \( \dfrac{b}{r_0} \)

Hint:

Hint:
The Coulomb force on the alpha particle always points along the line to the nucleus, so it produces zero torque. Use that to relate \(v_0\), \(b\), \(v_m\), \(r_m\), then bring in energy conservation and expand for small \(b/r_0\).

Answer 1

The Correct Option is A

Step 1: Set up the orbit as a hyperbola.
For a repulsive inverse square force, the alpha particle path around the nucleus is one branch of a hyperbola with the nucleus at the outer focus. Two constants fix this hyperbola: the semi latus rectum $l$, set by the angular momentum, and the eccentricity $e$, set by the energy.

Step 2: Find the semi latus rectum $l$.
For a force $F = k'/r^2$ with $k' = 2kZe^2$ (the alpha particle carries charge $2e$), the orbit relation gives $l = L^2/(mk')$, where $L = mv_0b$ is the conserved angular momentum. Since $r_0 = kZe^2/(mv_0^2)$ gives $k' = 2r_0mv_0^2$,
\[ l = \frac{(mv_0b)^2}{m(2r_0mv_0^2)} = \frac{b^2}{2r_0} \]

Step 3: Find the eccentricity $e$.
The eccentricity of this scattering hyperbola is $e = \sqrt{1 + \dfrac{2EL^2}{mk'^2}}$, with total energy $E = \frac{1}{2}mv_0^2$. Substituting $L = mv_0b$ and $k' = 2r_0mv_0^2$,
\[ e = \sqrt{1 + \frac{2\left(\frac{1}{2}mv_0^2\right)(mv_0b)^2}{m(2r_0mv_0^2)^2}} = \sqrt{1 + \frac{b^2}{4r_0^2}} \]

Step 4: Write the closest approach distance and check $b=0$.
For the repulsive branch, $r_m = l/(e-1)$. At $b=0$ this is $0/0$, so take the small $b$ limit: $e - 1 \approx b^2/(8r_0^2)$, giving
\[ r_m \to \frac{b^2/(2r_0)}{b^2/(8r_0^2)} = 4r_0 \quad \text{as } b \to 0 \]
This already fixes the head on distance at $4r_0$, matching the constant term in options (C) and (D).

Step 5: Expand $r_m$ to order $b^2$.
Carrying the expansion of $e-1$ one order further and dividing,
\[ r_m = \frac{l}{e-1} = 4r_0 + \frac{b^2}{4r_0} + O(b^4) \]
The coefficient of $b^2$ is $1/(4r_0)$. Option (C) claims $1/(2r_0)$ and option (D) claims $1/(8r_0)$, so both are FALSE.

Step 6: Check the momentum relation directly.
The Coulomb force always points along the radius vector to the nucleus, so it exerts zero torque and angular momentum never changes: $mv_0b = mv_mr_m$ at every point of the path, in particular at closest approach where the velocity is purely tangential. So $v_0b = v_mr_m$ holds exactly for any $b$, confirming option (A). Option (B), $v_0b = 2v_mr_0$, would need $r_m$ to always equal $2r_0$, which Step 5 already rules out in general, so (B) is FALSE.

Final Answer:
Only option (A), the angular momentum relation, is an identity that holds for every impact parameter. \[ \boxed{\text{(A)}} \]