Question

An $\alpha$- particle of KE $5.4MeV$ is projected towards $Cr$- nucleus $(Z=24)$. What is its distance of closest approach? ($e=1.6\times {10}^{-19}C$)

Answer 1

First convert the kinetic energy to joules, $5.4\times {10}^6\times 1.6\times {10}^{-19}=8.64\times 10^{-13}$. The distance closest approach will be when all the kinetic energy has been converted to electrostatic potential energy.

The equation for electrostatic potential energy is $E=\frac{(Q_{gold}\times q_{alpha})}{4\Pi {ϵ}_{o}r}$ where r is the distance of closest approach. Equate the kinetic and potential energies and rearrange for $r=\frac{(Q_{gold}\times q_{alpha})}{4\pi {ϵ}_{o}E}$.

Substitute in the values of the charges, constants and energy to find

$r=\frac{(+79e)(+2e)}{(4\pi \times 8.9\times 10^{-12}\times 8.64\times {10}^{-13})}$

$r=\frac{(79\times 2)\times (1.6\times {10}^{-19}{)}^2}{(4\pi \times 8.9\times 10^{-12}\times 8.64\times {10}^{-13})}$

$r=4.188\times {10}^{-14}m$