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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

Respuesta :

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

here we have

[tex]m_1 u = m_1 v_1 + m_2 v_2[/tex]

also we know that

[tex]v_2 - v_1 = u[/tex]

by solving above two equations

[tex]v = \frac{m_1 - m_2}{m_1 + m_2} u[/tex]

[tex]v = \frac{4u - 197u}{4u + 197u} u[/tex]

[tex]v = 0.96u[/tex]

now loss in kinetic energy of alpha particle will be

[tex] Loss = \frac{0.5mu^2 - 0.5mv^2}{0.5mu^2}[/tex]

[tex] Loss =1 - \frac{v^2}{u^2}[/tex]

[tex] Loss =1 - 0.96^2[/tex]

[tex] Loss =0.078 = 7.8%[/tex]

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