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?\( \left(k = \dfrac{1}{4\pi\epsilon_0} \text{ and } r_0 = k\dfrac{Ze^2}{mv_0^2}\right) \)
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?
\( \left(k = \dfrac{1}{4\pi\epsilon_0} \text{ and } r_0 = k\dfrac{Ze^2}{mv_0^2}\right) \)
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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 \(b/r_0\).
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 \(b/r_0\).
Updated On: Jul 28, 2026
- \( v_0 b = v_m r_m \)
- \( v_0 b = 2 v_m r_0 \)
- For \( \dfrac{b}{r_0} \ll 1 \), \( r_m = 4r_0 + \dfrac{b^2}{2r_0} \), ignoring higher order corrections in \( \dfrac{b}{r_0} \)
- For \( \dfrac{b}{r_0} \ll 1 \), \( r_m = 4r_0 + \dfrac{b^2}{8r_0} \), ignoring higher order corrections in \( \dfrac{b}{r_0} \)
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The Correct Option is A
Solution and Explanation
Step 1: Set up the two conservation laws.
The nucleus is fixed and its Coulomb force on the alpha particle always points along the line joining the two particles, so it exerts no torque about the nucleus. Angular momentum about the nucleus is conserved. The alpha particle carries charge \(2e\), so its potential energy at separation \(r\) is \(k(2e)(Ze)/r\), and total mechanical energy is also conserved.
Step 2: Apply angular momentum conservation.
Far from the nucleus the alpha particle moves in a straight line with speed \(v_0\) and impact parameter \(b\), so its angular momentum about the nucleus is \(mv_0b\). At the distance of closest approach the radial velocity is zero, so the whole speed \(v_m\) is tangential, giving angular momentum \(mv_mr_m\). Conservation gives
\[ v_0 b = v_m r_m \]
This is exactly option (A), and it holds at every value of \(b\), not only for small \(b\).
Step 3: Apply energy conservation.
Equating the energy at infinity to the energy at closest approach,
\[ \frac{1}{2}mv_0^2 = \frac{1}{2}mv_m^2 + \frac{2kZe^2}{r_m} \]
Using \(r_0 = kZe^2/(mv_0^2)\), the potential term becomes \(2kZe^2/r_m = 2r_0mv_0^2/r_m\). Dividing the whole equation by \(\frac{1}{2}mv_0^2\) gives
\[ 1 = \left(\frac{v_m}{v_0}\right)^2 + \frac{4r_0}{r_m} \]
Step 4: Eliminate \(v_m\) using Step 2.
From Step 2, \(v_m/v_0 = b/r_m\). Putting this in,
\[ 1 = \frac{b^2}{r_m^2} + \frac{4r_0}{r_m} \]
Multiplying through by \(r_m^2\) gives the quadratic
\[ r_m^2 - 4r_0 r_m - b^2 = 0 \]
Taking the physically sensible positive root,
\[ r_m = 2r_0 + \sqrt{4r_0^2 + b^2} \]
Step 5: Expand for \(b/r_0 \ll 1\) and check options (B), (C), (D).
Write \(\sqrt{4r_0^2+b^2} = 2r_0\sqrt{1 + b^2/(4r_0^2)} \approx 2r_0\left(1 + \frac{b^2}{8r_0^2}\right) = 2r_0 + \frac{b^2}{4r_0}\), keeping only the leading correction. So
\[ r_m \approx 4r_0 + \frac{b^2}{4r_0} \]
The coefficient of \(b^2\) is \(1/(4r_0)\), not \(1/(2r_0)\) as option (C) claims and not \(1/(8r_0)\) as option (D) claims, so both (C) and (D) are FALSE. For option (B): at \(b=0\), \(v_0b=0\) and \(v_m=0\) too, so this special case does not test it; for a general \(b\), Step 2 already fixes \(v_0b = v_mr_m\), and \(2v_mr_0\) equals this only if \(r_m = 2r_0\), which Step 5 shows is not true except possibly at one isolated \(b\). So (B) is FALSE in general.
Final Answer:
Only the angular momentum relation \(v_0b = v_mr_m\) is an identity holding for every \(b\); that is option (A). \[ \boxed{\text{(A)}} \]
The nucleus is fixed and its Coulomb force on the alpha particle always points along the line joining the two particles, so it exerts no torque about the nucleus. Angular momentum about the nucleus is conserved. The alpha particle carries charge \(2e\), so its potential energy at separation \(r\) is \(k(2e)(Ze)/r\), and total mechanical energy is also conserved.
Step 2: Apply angular momentum conservation.
Far from the nucleus the alpha particle moves in a straight line with speed \(v_0\) and impact parameter \(b\), so its angular momentum about the nucleus is \(mv_0b\). At the distance of closest approach the radial velocity is zero, so the whole speed \(v_m\) is tangential, giving angular momentum \(mv_mr_m\). Conservation gives
\[ v_0 b = v_m r_m \]
This is exactly option (A), and it holds at every value of \(b\), not only for small \(b\).
Step 3: Apply energy conservation.
Equating the energy at infinity to the energy at closest approach,
\[ \frac{1}{2}mv_0^2 = \frac{1}{2}mv_m^2 + \frac{2kZe^2}{r_m} \]
Using \(r_0 = kZe^2/(mv_0^2)\), the potential term becomes \(2kZe^2/r_m = 2r_0mv_0^2/r_m\). Dividing the whole equation by \(\frac{1}{2}mv_0^2\) gives
\[ 1 = \left(\frac{v_m}{v_0}\right)^2 + \frac{4r_0}{r_m} \]
Step 4: Eliminate \(v_m\) using Step 2.
From Step 2, \(v_m/v_0 = b/r_m\). Putting this in,
\[ 1 = \frac{b^2}{r_m^2} + \frac{4r_0}{r_m} \]
Multiplying through by \(r_m^2\) gives the quadratic
\[ r_m^2 - 4r_0 r_m - b^2 = 0 \]
Taking the physically sensible positive root,
\[ r_m = 2r_0 + \sqrt{4r_0^2 + b^2} \]
Step 5: Expand for \(b/r_0 \ll 1\) and check options (B), (C), (D).
Write \(\sqrt{4r_0^2+b^2} = 2r_0\sqrt{1 + b^2/(4r_0^2)} \approx 2r_0\left(1 + \frac{b^2}{8r_0^2}\right) = 2r_0 + \frac{b^2}{4r_0}\), keeping only the leading correction. So
\[ r_m \approx 4r_0 + \frac{b^2}{4r_0} \]
The coefficient of \(b^2\) is \(1/(4r_0)\), not \(1/(2r_0)\) as option (C) claims and not \(1/(8r_0)\) as option (D) claims, so both (C) and (D) are FALSE. For option (B): at \(b=0\), \(v_0b=0\) and \(v_m=0\) too, so this special case does not test it; for a general \(b\), Step 2 already fixes \(v_0b = v_mr_m\), and \(2v_mr_0\) equals this only if \(r_m = 2r_0\), which Step 5 shows is not true except possibly at one isolated \(b\). So (B) is FALSE in general.
Final Answer:
Only the angular momentum relation \(v_0b = v_mr_m\) is an identity holding for every \(b\); that is option (A). \[ \boxed{\text{(A)}} \]
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