Question

A deuterium nucleus and an alpha particle approach a target nucleus separately in head-on position. Find the ratio of their distance of closest approach to the nucleus, when both have the same momentum.

Hint:

Whenever a question presents a "same [quantity]" scenario, always immediately manipulate the core formula to explicitly include that constant quantity (like converting $v$ to $p$ here) before taking any ratios. It prevents critical substitution errors.

Answer 1

Concept:
• As established, the distance of closest approach occurs when initial kinetic energy entirely equals electrostatic potential energy.
• Kinetic energy $K$ can be expressed in terms of momentum $p$ using the relation $K = \frac{p^2}{2m}$.
• Substituting this into the energy conservation equation provides a different proportional relationship suitable for constant momentum conditions.

Step 1: Derive $r_0$ in terms of momentum
Equate kinetic energy (in terms of momentum) to potential energy at closest approach:
\[ K = U \]
\[ \frac{p^2}{2m} = \frac{1}{4\pi\varepsilon_0} \frac{(Z e) (q)}{r_0} \]
Rearranging to solve for the distance of closest approach, $r_0$:
\[ r_0 = \frac{1}{4\pi\varepsilon_0} \frac{2 m Z e \cdot q}{p^2} \]
This indicates that for a given target ($Z$) and a strictly constant momentum ($p$), the distance of closest approach is directly proportional to the product of the particle's mass and its charge:
\[ r_0 \propto m \cdot q \]

Step 2: Identify properties of the interacting particles
As detailed previously:
For a deuterium nucleus (d): Charge $q_d = e$, Mass $m_d = 2m_p$.
For an alpha particle ($\alpha$): Charge $q_\alpha = 2e$, Mass $m_\alpha = 4m_p$.

Step 3: Calculate the ratio for the case of same momentum
Since both particles share identical initial momentum $p$, we utilize the proportionality $r_0 \propto m \cdot q$.
The ratio of their closest approach distances will be:
\[ \frac{r_{0d}}{r_{0\alpha}} = \frac{m_d \cdot q_d}{m_\alpha \cdot q_\alpha} \]
Substitute the known masses and charges into the ratio:
\[ \frac{r_{0d}}{r_{0\alpha}} = \frac{(2m_p) \cdot (e)}{(4m_p) \cdot (2e)} \]
Simplify the numerator and denominator:
\[ \frac{r_{0d}}{r_{0\alpha}} = \frac{2 m_p e}{8 m_p e} \]
Cancel the common terms ($m_p e$) and reduce the fraction:
\[ \frac{r_{0d}}{r_{0\alpha}} = \frac{2}{8} = \frac{1}{4} \]

Step 4: Conclusion
When approaching with the same initial momentum, the ratio of their distances of closest approach is $1 : 4$.