A proton and an alpha particle$(q=+2e,m=4\mathrm{u})$are fired directly toward each other from far away, each with an initial speed of $0.010c$. What is their distance of closest approach, as measured between their centers?

Given:

Mass of proton, $m_p=1.67\times {10}^{-27}\mathrm{kg}$

Mass of an alpha particle, $m_a=4\times 1.67\times {10}^{-27}\mathrm{kg}$

Charge of proton, $q=1.6\times {10}^{-19}\mathrm{C}$

Charge of alpha particle, $q_{\alpha }=2\times 1.6\times {10}^{-19}$

Initial velocity of both proton and alpha particle,$v=0.010c$

Formulae

We understand that the system's potential energy is

$\mathrm{PE}=PE=\frac{1}{4\pi {\epsilon }_0}\times \frac{q_1{q}_2}{r}$

Here

$q_1{q}_2$are charges

${\epsilon }_o$is the permittivity

$\mathit{r}$is the separation distance

The system's initial energy is kinetic energy.

$E_i=\frac{1}{2}{m}_p{v}_p^2+\frac{1}{2}{m}_{\alpha }{v}_{\alpha }^2$

The kinetic energy of the system is calculated as

$\begin{array}{rcl}KE& =& \frac{1}{2}\times 1.67\times {10}^{-27}\times {\left(0.01\times 3\times {10}^8\right)}^2+\frac{1}{2}\times 4\times 1.67\times {10}^{-27}\times {\left(0.01\times 3\times {10}^8\right)}^2\\ & =& 3.75\times {10}^{-14}\mathrm{J}\end{array}$

The particle will be closest when the entire kinetic energy will convert into potential energy

$KE=PE$

Plugging the values in the above equation

width="378">3.75×10-14=14πeo×1.6×10-19×4×1.6×10-19r3.75×10-14=9×109×1.6×10-19×4×1.6×10-19rr=9×109×1.6×10-19×4×1.6×10-193.75×10-14=2.45×10-14m=24.5fm

Distance of closest approach between the center of proton and an alpha particle is $24.5\mathrm{fm}$