To determine the kinetic energy of the alpha particle when it makes a head-on collision with a gold nucleus, we can use the concept of electrostatic potential energy at the closest approach. The kinetic energy of the alpha particle will be completely converted into electrostatic potential energy at the point of closest approach.
The formula for the electrostatic potential energy between two charges is given by:
U=‌‌where:
q1 and
q2 are the charges of the alpha particle and gold nucleus, respectively
r is the distance of closest approach
E0 is the permittivity of free space, which is approximately
8.85×10−12F∕mFor an alpha particle,
q1 is the charge of 2 protons, so:
q1=2e=2×1.6×10−19C For a gold nucleus (with atomic number 79 ),
q2 is the charge of 79 protons, so:
q2=79e=79×1.6×10−19C The distance of closest approach,
r, is given as:
r=10×10−14m Now, substituting these values into the formula for electrostatic potential energy:
Simplifying the equation:
U=‌(2×79×1.62×10−38) Calculating each component:
U=‌(2×79×2.56×10−38)Or:
U=9×1022×404.48×10−38Simplifying further:
U=3.64×10−13JTherefore, the kinetic energy of the alpha particle is:
Option D:
3.64×10−13J