An alpha particle is a nucleus of helium. It has twice the charge and four times

Question

An alpha particle is a nucleus of helium. It has twice the charge and four times the mass of the proton. When they were very far away from each other, but headed toward directly each other, a proton and an alpha particle each had an initial speed of 2×103c, where c is the speed of light. What is their distance of closest approach? There are two conserved quantities. Make use of both of them. (c = 3.00 × 108 m/s, k = 1/4 0 = 8.99 × 109 N · m2/C2, e = 1.60 × 10-19 C, m proton = 1.67 x 10-27kg)

Explanation / Answer

initial KE of system = Ke of proton + KE of alpha particle

      = [ 1.67 x 10^-27 x (0.002 x 3 x 10^8)^2 / 2 ] + [ 4 x 1.67 x 10^-27 x (0.002 x 3 x 10^8)^2 / 2 ]

= 1.503 10^-15 J

initially PE = 0

final PE when they are separated by distance d,

PE = kq1q2 /d = (9 x 10^9 x 1.6 x 10^-19 x 2 x 1.6 x 10^-19) / d

PE = 4.608 x 10^-28 / d

using momentum conservation,

4m v - mv = (4m + m) vf

vf = 0.6v = 1.2 x 10^-3 c

final KE = (5 x 1.67 x 10^-27 x (1.2 x 3 x 10^5)^2) / 2

       = 0.54 x 10^-15 J

Using energy conservation,

1.503 x 10^-15 + 0   = 4.608 x 10^-28 / d     +   0.54 x 10^-15

d = 4.8 x 10^-13 m

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