Question

An alpha particle (mass = 4 u) has a head-on, elastic collision with a nucleus (mass = 158 u) that is initially at rest. What percentage of the kinetic energy of the alpha particle is transferred to the nucleus in the collision? %

Answer 1

m mass of alpha and M mass of platinum nucleus.
u the initial velocity of alpha and v and vp are the final velocities of alpha and platinum respectively
BY conservation of momentum, mu = m v + M vp----------> (1)
Because of elastic collision, KE before and after would be the same.
That is, mu^2 = m v^2 + M vp^2------------>(2) cancelling 1/2 through out
From (1), m(u-v) = M vp--------------->(3)
From (2), m(u-v)(u+v) = M vp^2------------>(4)
(4)/(3)===> vp = u+v------------> (5)
From (3) u-v = (M/m)vp =39.5vp

(5)+(6) ===> 2u = 40.5vp

vp/u = 1/20.25

The kinetic energy lost by alpha will be with the platinum nucleus
ie 1/2 M vp^2
Hence percetage loss of KE is100* [1/2 M vp^2]/[1/2 mu^2]
Hence, required % loss = 100* (M/m) (vp/u)^2

= 100*39.5*(1/20.25)*(1/20.25) = 9.63 %

Percentage loss in KE of the alpha particle will be 9.63%