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 velocity.

Hint:

An alpha particle is essentially twice as heavy and has twice the charge of a deuteron. Because kinetic energy depends on mass linearly, and potential energy depends on charge linearly, the doubling effects cancel each other out entirely when velocity is constant.

Answer 1

Concept:
• When a positively charged particle is fired directly at a target nucleus, it experiences strong electrostatic repulsion.
• As it approaches, its initial kinetic energy is gradually converted into electrostatic potential energy.
• At the "distance of closest approach" ($r_0$), the particle momentarily stops before rebounding. Here, its entire initial kinetic energy is completely transformed into electrostatic potential energy.
• By conservation of energy: $\frac{1}{2} m v^2 = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r_0}$.

Step 1: Establish the general formula for distance of closest approach
Let the target nucleus have a charge $Z e$.
Let the approaching particle have mass $m$, charge $q$, and initial velocity $v$.
Equating initial kinetic energy to potential energy at closest approach:
\[ \frac{1}{2} m v^2 = \frac{1}{4\pi\varepsilon_0} \frac{(Z e) (q)}{r_0} \]
Rearranging to solve for the distance of closest approach, $r_0$:
\[ r_0 = \frac{1}{4\pi\varepsilon_0} \frac{2 Z e \cdot q}{m v^2} \]
This indicates that for a given target ($Z$) and a constant velocity ($v$), the distance of closest approach is directly proportional to the charge-to-mass ratio of the incident particle:
\[ r_0 \propto \frac{q}{m} \]

Step 2: Identify properties of the interacting particles
For a deuterium nucleus (Deuteron, d): It consists of one proton and one neutron. Its charge is $q_d = +e$. Its mass is approximately $m_d = 2m_p$ (where $m_p$ is the proton mass).
For an alpha particle ($\alpha$): It consists of two protons and two neutrons. Its charge is $q_\alpha = +2e$. Its mass is approximately $m_\alpha = 4m_p$.

Step 3: Calculate the ratio for the case of same velocity
Since both particles have the identical initial velocity $v$, we utilize the proportionality $r_0 \propto \frac{q}{m}$.
The ratio of their closest approach distances will be:
\[ \frac{r_{0d}}{r_{0\alpha}} = \frac{\left( \frac{q_d}{m_d} \right)}{\left( \frac{q_\alpha}{m_\alpha} \right)} \]
Substitute the known charges and masses:
\[ \frac{r_{0d}}{r_{0\alpha}} = \frac{\left( \frac{e}{2m_p} \right)}{\left( \frac{2e}{4m_p} \right)} \]
Simplify the denominator fraction ($\frac{2e}{4m_p} = \frac{e}{2m_p}$):
\[ \frac{r_{0d}}{r_{0\alpha}} = \frac{\left( \frac{e}{2m_p} \right)}{\left( \frac{e}{2m_p} \right)} \]
\[ \frac{r_{0d}}{r_{0\alpha}} = 1 \]

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