Chapter 3, Problem 3.1P

To determine

The closest approach that a 5.3 MeV alpha particle can make to a gold nucleus.

Answer to Problem 3.1P

  $d=4.29\times {10}^{-14}m$

Explanation of Solution

Introduction:

When an alpha particle having kinetic energy ‘E_a’and charge ‘q’ is ejected on a gold nucleus, it undergoes repulsion. The force of repulsion increases with the decreasing distance between the two. On reaching the closest distance of approach ( d ), the total kinetic energy converts into potential energy hence, it comes to rest.

Formula:

  $d=(\frac{1}{4\pi {\epsilon }_0})\frac{qQ}{Ea}$

Consider an alpha particle of energy $Ea=5.3\text{MeV=}(5.3\times {10}^6\times 1.6\times {10}^{-19})$

Using the formula, we get:

  $d=(\frac{1}{4\pi {\epsilon }_0})\frac{qQ}{Ea}$

  $d=9\times {10}^9\left(\frac{2\times 1.6\times {10}^{-19}\times 79\times 1.6\times {10}^{-19}}{5.3\times 10\times 1.6\times {10}^{-19}}\right)$

  $d=4.29\times {10}^{-14}m$

Conclusion:

Therefore, the closest distance of approach ( d ) for a 5.3 MeV alpha particle is: $d=4.29\times {10}^{-14}m$ .