Question

An α particle is a helium nucleus and has mass 4 u, which is "4 atomic units." suppose it collides head-on in an elastic collision with a stationary gold nucleus inside a block of material. the mass of a gold nucleus is 197 u. what percentage of the α's kinetic energy is lost in this collision

Answer 1

velocity of helium nuclei after it collide with stationary gold nuclei will be calculated by momentum conservation and coefficient of restitution

here we have

`m_1 u = m_1 v_1 + m_2 v_2`

also we know that

`v_2 - v_1 = u`

by solving above two equations

`v = (m_1 - m_2)/(m_1 + m_2) u`

`v = (4u - 197u)/(4u + 197u) u`

`v = 0.96u`

now loss in kinetic energy of alpha particle will be

`Loss = (0.5mu^2 - 0.5mv^2)/(0.5mu^2)`

`Loss =1 - (v^2)/(u^2)`

`Loss =1 - 0.96^2`

`Loss =0.078 = 7.8%`