Question

if 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 4.5 MeV. How close the gold nucleus (79 proton) could it come before being deflected?

Answer 1

Answer:

r=4.55 x 10^-14 m

Explanation:

Identify the unknown:  

How close to the gold nucleus could alpha particles come before being deflected?

List the Knowns:  

Charge of alpha particle: Q_1 = 2 x 1.6 x 10^-19 C

Energy of alpha particle: K.E = 4.5 x 10^6 eV = 5 x 10^6 x 1.6 x 10^-19 J  

Charge of gold nucleus: Q_2 = 79 x 1.6 x 10^-19 C

Electric force constant: k = 9 x 10^9 N • m^2/c^2  

Set Up the Problem:  

The alpha particles will be repelled by the gold nucleus and it will approaches until all its kinetic energy transformed into electrical potential energy. so at this distance from the gold nucleus, potential energy equals KE  

Potential energy of a two-charge system:

 U(r)=k*Q_1*Q_2/r

     r=k*Q_1*Q_2/ U

Solve the Problem:  

     r=4.55 x 10^-14 m