What is the distance of closest approach where the


Rutherford observed that an alpha particle (Q1 = 2 x 1.6 x 10^-19 coulomb) having a kinetic energy of 7.68 x 10 ^6 electron volts (7.68 x 10^6 x 1.6 x 10^-19 joule) rebounds backward in a head on collision with a gold nucleus (Q2 = 79 x 1.6 x 10^-19 coulomb).

a. What is the distance of closest approach where the electrostatic potential energy is equal to the initial kinetic energy? Express results in femtometers (10^-15 meter).

b. What is the maximum force of repulsion?

c. What is the maximum acceleration in g's? The mass of the alpha particle is about 4 times that of a proton, or 4 x 1.7 x 10^-27 kilogram.

Solution Preview :

Prepared by a verified Expert
Physics: What is the distance of closest approach where the
Reference No:- TGS01263740

Now Priced at $15 (50% Discount)

Recommended (98%)

Rated (4.3/5)