Question

Find the minimum kinetic energy in MeV necessary for an α particle to touch a 39 19 K nucleus that is initially at rest, assuming only Coulomb forces are present. Assume that the α particle has a radius of 2.0 fm, and the potassium nucleus of 3.7 fm, and take into account the masses of both particles.

Answer 1

Answer:

KE = 9.6MeV

Explanation:

Given the relationships we understand that,

$q_1 = 19e, q_2 = 2e$

The Potential at point p, is given by the following formula

$V_ {p} = \frac {Kq_1} {r}$

According to the graphic designed, you have,

$V_ {p} = \frac {Kq_1} {(r_1 + r_2)}$

$V_ {p} = \frac {(9 * 10 ^ 9) (19 * 1.6 * 10 ^ {- 15})} {(3.7 + 2) * 10 ^{- 15}}}$

$V_ {p} = 4.8 * 10 ^ 6V$

The kinetic energy of the particle would be given by

$KE = q_2 * V_ {p} = 2e * 4.8 * 10 ^ 6V$

$KE = 9.6MeV$