Question

In one of the classic nuclear physics experiments at the beginning of the 20th century, an alpha particle was accelerated toward a gold nucleus, and its path was substantially deflected by the Coulomb interaction. If the energy of the doubly charged alpha nucleus was 2.25 MeV, how close (in m) to the gold nucleus (79 protons) could it come before being deflected

Answer 1

The answer is "
`1.01 * 10^(-13)`"

Step-by-step explanation:

Using the law of conservation for energy. Equating the kinetic energy to the potential energy.

`KE=U=(kqq')/(r)\\\\`

Calculating the closest distance:

`\to r=(kqq')/(KE)\\\\`

`=(k(2e)(79e))/(KE)\\\\=(k(2)(79)e^2)/(KE)\\\\=(9.0* 10^9 \ N \cdot (m^2)/(c)(2)(79)(1.6 *10^(-19) \ C)^2)/((2.25\ meV) ((1.6 * 10^(-13) \ J)/(1 \ MeV)))\\\\`

`=(9.0* 10^9 * 2* 79* 1.6 *10^(-19)* 1.6 *10^(-19) )/((2.25 * 1.6 * 10^(-13)) )\\\\=(3,640.32* 10^(-29))/(3.6 * 10^(-13) )\\\\=(3,640.32)/(3.6) * 10^(-16)\\\\=1011.2 * 10^(-16)\\\\=1.01 * 10^(-13)`