Question

Rutherford discovered the nucleus of the atom by firing a particles at gold foil. An a particle has a charge of q = +2e and a mass of m = 6.64-10-27 kg. A gold nucleus has charge of Q-+79e. You may ignore the motion of the gold nucleus in this problem +2e +79e Suppose an α particles is traveling directly toward a gold nucleus. If the speed of the a particle is u = 1.9. 107 m/s when it is 1 m from the gold nucleus, how close to the gold nucleus will the a particle come before it stops and reverses direction?

Answer 1

Answer:$r_0=3.037\times 10^{-14}m$

Explanation:

Given

charge on alpha particle=+2e

mass of alpha particle=$6.64\times 10^{-27}$ kg

Charge on gold nucleus=+79e

Velocity at r=1m is $1.9\times 10^{7}$

Using Energy conservation

Kinetic energy of particle will be converting to Potential energy as it approaches to nucleus

therefore

$\frac{1}{2}mv^2+U_{r=1m}=U_{closest\ to\ nucleus}$

$\frac{1}{2}\left ( 6.64\times 10^{-27}\right )\left ( 1.9\times 10^{7}^2\right )+\frac{K\left ( 2e\right )\left ( 79e\right )}{1}=\frac{K\left ( 2e\right )\left ( 79e\right )}{r_0}$

$\frac{1}{2}\left ( 6.64\times 10^{-27}\right )\left ( 1.9\times 10^{7}^2\right )=\frac{9\times 10^9\times 158\times \left ( 1.6\times 10^{-19}\right )}{y}\left [\frac{1}{r_0}-\frac{1}{1}\right ]$

on solving we get

$\frac{1}{r_0}=3.292\times 10^{13}$

$r_0=3.037\times 10^{-14}m$