Question

What is the closest distance of approach for an alpha particle with velocity 10^5 m/s striking the gold foil

Answer 1

Given:

An alpha particle was projected with Velocity 10⁵/s towards a gold foil.

To find:

The closest distance of approach

Calculation:

At the closest distance between alpha particle and an gold atom in the foil , the whole Kinetic Energy of the alpha particle will be converted to Electrostatic Potential Energy ;

Let the closest distance be r ;

$\therefore \: \dfrac{1}{2} m {v}^{2} = \dfrac{k(ze)(2e)}{r}$

$= > \: r = \dfrac{k(ze)(2e)}{ \frac{1}{2} m {v}^{2} }$

For gold atom , z = 79

$= > \: r = \dfrac{4k \times 79 {e}^{2} }{ m {v}^{2} }$

$= > \: r = \dfrac{4\times 79 {e}^{2} }{4\pi \epsilon_{0} m {v}^{2} }$

$= > \: r = \dfrac{79 {e}^{2} }{\pi \epsilon_{0} m {v}^{2} }$

$= > \: r = \dfrac{79 {e}^{2} }{\pi \epsilon_{0} m {( {10}^{5} )}^{2} }$

$= > \: r = \dfrac{79 {e}^{2} }{\pi \epsilon_{0} m {10}^{10} }$

$= > \: r = \dfrac{79 {e}^{2} \times {10}^{ - 10} }{\pi \epsilon_{0} m }$

So, final answer is :

$\boxed{ \bold{ \large{ \: r = \dfrac{79 {e}^{2} \times {10}^{ - 10} }{\pi \epsilon_{0} m } }}}$