- #1

etotheipi

- Homework Statement
- Find the distance of closest approach of an alpha particle and nucleus (of arbitrary atomic number Z)

- Relevant Equations
- ##E_{k}##, relative velocities

I've found two methods for doing this problem and they give different answers.

$$m_{\alpha}v_{\alpha} = (m_{\alpha} + M_{nucleus})v_{common}$$

$$\frac{1}{2} m_{\alpha} v_{\alpha}^{2} = \frac{1}{2} (m_{\alpha} + M_{nucleus})v_{common}^{2} + \frac{qQ}{4\pi\epsilon_{0}r}$$It makes sense that the two methods give different answers for ##r##, however this is a very common exam question, and it is never usually stated whether we permit the larger nucleus to move or not. So, I was wondering which approach I should use for this and similar problems. Thank you!

**Method 1**: Assume the larger nucleus does not move, and simply equate energies before the collision and at the point of closest approach:$$\frac{1}{2} m v^{2} = \frac{qQ}{4\pi\epsilon_{0}r}$$**Method 2**: Assume the larger nucleus is capable of moving; the distance of closest approach occurs when the relative velocity of the two particles is zero. From this, we can use CLM and then do another energy calculation:*Conservation of Linear Momentum:*$$m_{\alpha}v_{\alpha} = (m_{\alpha} + M_{nucleus})v_{common}$$

*Energy:*$$\frac{1}{2} m_{\alpha} v_{\alpha}^{2} = \frac{1}{2} (m_{\alpha} + M_{nucleus})v_{common}^{2} + \frac{qQ}{4\pi\epsilon_{0}r}$$It makes sense that the two methods give different answers for ##r##, however this is a very common exam question, and it is never usually stated whether we permit the larger nucleus to move or not. So, I was wondering which approach I should use for this and similar problems. Thank you!