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

$v = \frac{m_1 - m_2}{m_1 + m_2} u$

$v = \frac{4u - 197u}{4u + 197u} u$

$v = 0.96u$

now loss in kinetic energy of alpha particle will be

$Loss = \frac{0.5mu^2 - 0.5mv^2}{0.5mu^2}$

$Loss =1 - \frac{v^2}{u^2}$

$Loss =1 - 0.96^2$

$Loss =0.078 = 7.8%$