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 0.007c, where c is the speed of light. What is their distance of closest approach?

Hint: There are two conserved quantities. Make use of both of them. (c = 3.00 × 10^8 m/s, k = 1/40 = 8.99 × 10^9 N m2/C2, e = 1.60 × 10^-19 C, mproton = 1.67 x 10^-27 kg) (Give your answer to the nearest 0.001 pico meters (10^-12 m)

The answer is 0.039, please try to break down the math as much as possible and don't just post a single equation line.

Explanation / Answer

At very far away electric potential energy of system, PEi = 0

KE = (4m + m) (0.007c)^2 /2

when they are at closest approach, (distance d )

they will be having same velocity.

using energy conservation,

4m(0.007c) - m(0.007c) = (4m + m)v

v= 0.0042c

KE = 5m (0.0042c)^2 / 2

PE = k(2q)(q) / d

Using energy conservation,

0 + (4m + m) (0.007c)^2 /2 = k(2q)(q) / d + 5m (0.0042c)^2 / 2

(5 x 1.67 x 10^-27 ) (0.007 x 3 x 10^8)^2 /2 = [(9 x 10^9 x 2 x (1.6 x 10^-19)^2) / d] + 5m (0.0042c)^2 / 2

1.842 x 10^-14 = 4.608 x 10^-28 / d + 6.628 x 10^-15

d = 3.91 x 10^-14 m = 0.039 x 10^-12 m = 0.039 pm

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