Question

A proton and an alpha particle (q=+2e,m=4u) are fired directly toward each other from far away, each with an initial speed

of 0.014 c.

What is their distance of closest approach, as measured between their centers?

Answer 1

k =9x109 N.m2/C2

mp = 1.67x10-27 kg , m\alpha =4mp

m =mp +m\alpha=5mp

qp =e =1.6x10-19 C, q\alpha = 2e

Relative velocity v =0.014 c- (-0.014c) =0.028 c

c =3x108 m/s

U =kqpq\alpha /r, K.E =1/2(mv2)

At closest approach potential energy is equal to kineticenergy.

kqpq\alpha /r,=1/2(mv2)

r = 2 x kqpq\alpha/mv2)

r = [2x 9x109x 1.6x10-19x2x1.6x10-19] / [5x1.67x10-27x(0.028x3x108)2]

r = 1.6x10-15m