When an alpha particle collides elastically with a nucleus, the nucleus recoils. Suppose an 5.00MeV alpha particle has a head-on elastic collision with a gold nucleus that is initially at rest. What is the kinetic energy of (a) the recoiling nucleus and (b) the rebounding alpha particle?

• Energy of an alpha particle,${\mathrm{K}}_{\mathrm{\alpha i}}=5.00\mathrm{MeV}$
• A head-on elastic collision of alpha particle with a gold nucleus that is initially at rest.
• Atomic mass of gold,$m_{Au}=197u$
• Atomic mass of an alpha particle,${\mathrm{m}}_{\mathrm{\alpha }}=4.00\mathrm{u}$

Suppose in the Rutherford scattering experiment, after an elastic collision of an alpha particle with the nucleus, the nucleus recoils back with a recoiling velocity calculated using conservation of linear momentum. Then the initial velocity of the alpha particle changes to a different velocity, which can be found using the conservation of linear momentum. This velocity is called the rebounding velocity of the alpha particle. Using these velocities in the kinetic energy formula, we can get the energy by which both the particles move.

Formulae:

The kinetic energy of a body in motion,$\mathrm{K}=\frac{1}{2}{\mathrm{mv}}^2.......\left(1\right)$

According to conservation of linear momentum, $m_1{u}_1+m_2{u}_2+m_1{v}_1+m_2{v}_2..........\left(2\right)$

Where, are the masses of the two bodies in collision.

$u_1,u_2$are initial velocities of the bodies.

$v_1,v_2$are the final velocities of the bodies.

The conservation laws of (classical kinetic) energy and (linear) momentum determine the outcome of the collision. The final speed of the particle is given using equation (2) as follows:

$v_{\alpha f}=\frac{m_{\alpha }-m_{Au}}{m_{\alpha }+m_{Au}}{v}_{\alpha i}............\left(3\right)$

And that of the recoiling gold nucleus is given using same equation (2) as follows:

$v_{Au,f}=\frac{2m_{\alpha }}{m_{\alpha }+m_{Au}}{v}_{\alpha i}............\left(4\right)$

Thus, the kinetic energy of the recoiling gold nucleus is given using equation (4) in equation (1) as follows:

$$\begin{gathered} {\mathrm{K}}_{\mathrm{Au},\mathrm{f}}=\frac{1}{2}{\mathrm{m}}_{\mathrm{Au}}{\left(\frac{2{\mathrm{m}}_{\mathrm{\alpha }}}{{\mathrm{m}}_{\mathrm{\alpha }}+{\mathrm{m}}_{\mathrm{Au}}}{\mathrm{v}}_{\mathrm{\alpha i}}\right)}^2 \\ ={\mathrm{K}}_{\mathrm{\alpha i}}\frac{4{\mathrm{m}}_{\mathrm{Au}}{\mathrm{m}}_{\mathrm{\alpha }}}{{\left({\mathrm{m}}_{\mathrm{\alpha }}+{\mathrm{m}}_{\mathrm{Au}}\right)}^2} \\ =\left(5.00\mathrm{MeV}\right)\frac{4(197\mathrm{u})(4.00\mathrm{u})}{{(4.00\mathrm{u}+197\mathrm{u})}^2} \\ =0.390\mathrm{MeV} \end{gathered}$$

Hence, the value of the recoiling velocity is 0.390 MeV .

The rebounding velocity of the alpha particle is none other than the final velocity of the alpha particle. Thus, it is given using equation (3) in equation (2) as follows:

$$\begin{gathered} {\mathrm{K}}_{\mathrm{Au},\mathrm{f}}=\frac{1}{2}{\mathrm{m}}_{\mathrm{\alpha }}{\left(\frac{{\mathrm{m}}_{\mathrm{\alpha }}-{\mathrm{m}}_{\mathrm{Au}}}{{\mathrm{m}}_{\mathrm{\alpha }}+{\mathrm{m}}_{\mathrm{Au}}}{\mathrm{v}}_{\mathrm{\alpha i}}\right)}^2 \\ ={\mathrm{K}}_{\mathrm{\alpha i}}{\left(\frac{{\mathrm{m}}_{\mathrm{\alpha }}-{\mathrm{m}}_{\mathrm{Au}}}{{\mathrm{m}}_{\mathrm{\alpha }}+{\mathrm{m}}_{\mathrm{Au}}}\right)}^2\left(\because {\mathrm{K}}_{\mathrm{\alpha i}}=\frac{1}{2}{\mathrm{m}}_{\mathrm{\alpha }}{{\mathrm{v}}_{\mathrm{\alpha i}}}^2,\mathrm{from}\mathrm{equation}\left(1\right)\right) \\ =\left(5.00\mathrm{MeV}\right)\frac{{\left(4.00\mathrm{u}-197\mathrm{u}\right)}^2}{{\left(4.00\mathrm{u}+197\mathrm{u}\right)}^2} \\ =4.61\mathrm{MeV} \end{gathered}$$

Hence, the value of the rebounding velocity is 4.61MeV .