An alpha particle is incident on a lead nucleus.Because of the Coulomb force between them,the alpha particle will only approach the nucleus to a minimum distance of 32.5 fm.Determine the kinetic energy initial kinetic energy of the alpha particle.
A)8.66 MeV
B)7.27 MeV
C)5.00 MeV
D)9.18 MeV

Question

Quizplus · Accepted Answer

We can use the Coulomb force equation to calculate the initial kinetic energy of the alpha particle. At the minimum distance of closest approach, the Coulomb force will equal the centripetal force needed for circular motion:

F_c = F_e

mv^2/r = kqQ/r^2

Solving for the initial kinetic energy:

K_i = 1/2mv^2 = (kqQ/r)(r/2)

Plugging in the values given in the problem:

K_i = (9x10^9 Nm^2/C^2)(2e)(82e)(1.6x10^-19 C)/(32.5x10^-15m)(2)

K_i = 7.27 MeV

Therefore, the correct answer is B) 7.27 MeV.

An Alpha Particle Is Incident on a Lead Nucleus