# When an alpha particle collides elastically with a nucleus, the nucleus recoils. Suppose a...

## Question:

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

## Kinetic energy:

The form of energy that gets produced in a body due to the state of motion is termed as kinetic energy. The faster the object moves, the higher it will have the kinetic energy, and it decreases as the object attains height.

## Answer and Explanation:

- The energy of alpha particle is {eq}{E_{i,\alpha }} = 5.00\;{\rm{MeV}} {/eq}.

The mass of gold nucleus is {eq}{m_{Au}} = 197u {/eq}.

The mass of alpha particle is {eq}{m_\alpha } = 4u {/eq}.

The expression that can be written using energy conservation law is,

{eq}\dfrac{1}{2}{m_a}v_{i,\alpha }^2 = \dfrac{1}{2}{m_a}v_{f,\alpha }^2 + \dfrac{1}{2}{m_{Au}}v_{f,Au}^2......\left( 1 \right) {/eq}

The expression from conservation of momentum,

{eq}{m_\alpha }{v_{i,\alpha }} = {m_\alpha }{v_{f,\alpha }} + {m_{Au}}{v_{f,Au}}......\left( 2 \right) {/eq}

(a)

The expression for the kinetic energy of recoiling nucleus,

{eq}{K_{f,Au}} = \dfrac{1}{2}{m_{Au}}v_{f,Au}^2......\left( 3 \right) {/eq}

From equation 1, velocity of gold partial is,

{eq}\begin{align*} {m_{Au}}v_{f,Au}^2 &= {m_a}v_{i,\alpha }^2 - {m_a}v_{f,\alpha }^2\\ v_{f,Au}^2 &= \dfrac{{{m_a}\left( {v_{i,\alpha }^2 - v_{f,\alpha }^2} \right)}}{{{m_{Au}}}}\\ {v_{f,Au}} &= \sqrt {\dfrac{{{m_a}\left( {v_{i,\alpha }^2 - v_{f,\alpha }^2} \right)}}{{{m_{Au}}}}} ......\left( 4 \right) \end{align*} {/eq}

Substituting equation 3 in 2,

{eq}\begin{align*} {m_\alpha }{v_{i,\alpha }} &= {m_\alpha }{v_{f,\alpha }} + {m_{Au}}\sqrt {\dfrac{{{m_a}\left( {v_{i,\alpha }^2 - v_{f,\alpha }^2} \right)}}{{{m_{Au}}}}} \\ {m_\alpha }{v_{i,\alpha }} - {m_\alpha }{v_{f,\alpha }} &= {m_{Au}}\sqrt {\dfrac{{{m_a}\left( {v_{i,\alpha }^2 - v_{f,\alpha }^2} \right)}}{{{m_{Au}}}}} \end{align*} {/eq}

Squaring the above expression,

{eq}\begin{align*} {\left( {{m_\alpha }{v_{i,\alpha }} - {m_\alpha }{v_{f,\alpha }}} \right)^2} &= m_{Av}^2{\left( {\sqrt {\dfrac{{{m_a}\left( {v_{i,\alpha }^2 - v_{f,\alpha }^2} \right)}}{{{m_{Au}}}}} } \right)^2}\\ m_\alpha ^2{\left( {{v_{i,\alpha }} - {v_{f,\alpha }}} \right)^2} &= m_{Au}^2\dfrac{{{m_a}\left( {v_{i,\alpha }^2 - v_{f,\alpha }^2} \right)}}{{{m_{Au}}}}\\ {m_\alpha }{\left( {{v_{i,\alpha }} - {v_{f,\alpha }}} \right)^2} &= {m_{Au}}\left( {v_{i,\alpha }^2 - v_{f,\alpha }^2} \right)\\ {m_\alpha }{\left( {{v_{i,\alpha }} - {v_{f,\alpha }}} \right)^2} &= {m_{Au}}\left( {{v_{i,\alpha }} - {v_{f,\alpha }}} \right)\left( {{v_{i,\alpha }} + {v_{f,\alpha }}} \right)\\ {m_\alpha }\left( {{v_{i,\alpha }} - {v_{f,\alpha }}} \right) &= {m_{Au}}\left( {{v_{i,\alpha }} + {v_{f,\alpha }}} \right) \end{align*} {/eq}

The above expression can be rewritten as,

{eq}\begin{align*} \left( {{m_\alpha } - {m_{Au}}} \right){v_{i,\alpha }} &= \left( {{m_\alpha } + {m_{Au}}} \right){v_{f,\alpha }}\\ {v_{f,\alpha }} &= {v_{i,\alpha }}\dfrac{{\left( {{m_\alpha } - {m_{Au}}} \right)}}{{\left( {{m_\alpha } + {m_{Au}}} \right)}}......\left( 5 \right) \end{align*} {/eq}

Substituting the values in equation 4,

{eq}{v_{f,Au}} = \sqrt {\dfrac{{{m_a}\left( {v_{i,\alpha }^2 - {{\left( {{v_{i,\alpha }}\dfrac{{\left( {{m_\alpha } - {m_{Au}}} \right)}}{{\left( {{m_\alpha } + {m_{Au}}} \right)}}} \right)}^2}} \right)}}{{{m_{Au}}}}} {/eq}

Further solving the above expression, the velocity can be written as,

{eq}{v_{f,Au}} = \left( {\dfrac{{2{m_\alpha }}}{{{m_\alpha } + {m_{Au}}}}} \right){v_{i,\alpha }}......\left( 6 \right) {/eq}

Substituting the values in equation 3,

{eq}\begin{align*} {K_{f,Au}} &= \dfrac{1}{2}{m_{Au}}{\left( {\left( {\dfrac{{2{m_\alpha }}}{{{m_\alpha } + {m_{Au}}}}} \right){v_{i,\alpha }}} \right)^2}\\ {K_{f,Au}} &= \dfrac{1}{2}{m_\alpha }v_{i,\alpha }^2 \times \dfrac{{{m_{Au}}}}{{{m_\alpha }}}{\left( {\dfrac{{2{m_\alpha }}}{{{m_\alpha } + {m_{Au}}}}} \right)^2} \end{align*} {/eq}

As, the kinetic energy of alpha particle is,

{eq}{K_{i,\alpha }} = \dfrac{1}{2}{m_\alpha }v_{i,\alpha }^2 {/eq}

So, above expression can be written as,

{eq}{K_{f,Au}} = {K_{i,\alpha }} \times \dfrac{{{m_{Au}}}}{{{m_\alpha }}}{\left( {\dfrac{{2{m_\alpha }}}{{{m_\alpha } + {m_{Au}}}}} \right)^2} {/eq}

Substituting the values in above expression,

{eq}$\begin{align*} {K_{f,Au}}& = 5.00 \times \dfrac{{197u}}{{4u}} \times {\left( {\dfrac{{2 \times 4u}}{{4u + 197u}}} \right)^2}\\ {K_{f,Au}} &= 0.390\;{\rm{MeV}} \end{align*}$ {/eq}

Thus, the kinetic energy of recoiling nucleus is {eq}0.390\;{\rm{MeV}} {/eq}.

(b)

The expression for the energy of alpha particle is given as,

{eq}{K_{f,\alpha }} = \dfrac{1}{2}{m_\alpha }v_{f,\alpha }^2 {/eq}

{eq}\begin{align*} {K_{f,\alpha }} &= \dfrac{1}{2}{m_\alpha }v_{i,\alpha }^2{\left( {\dfrac{{{m_\alpha } - {m_{Au}}}}{{{m_\alpha } + {m_{Au}}}}} \right)^2}\\ {K_{f,\alpha }} &= {K_{i,\alpha }}{\left( {\dfrac{{{m_\alpha } - {m_{Au}}}}{{{m_\alpha } + {m_{Au}}}}} \right)^2} \end{align*} {/eq}

Substituting the values,

{eq}\begin{align*} {K_{f,\alpha }} &= \dfrac{1}{2}{m_\alpha }v_{i,\alpha }^2{\left( {\dfrac{{{m_\alpha } - {m_{Au}}}}{{{m_\alpha } + {m_{Au}}}}} \right)^2}\\ {K_{f,\alpha }} &= 5.00 \times {\left( {\dfrac{{4u - 197u}}{{4u + 197u}}} \right)^2}\\ {K_{f,\alpha }} &= 4.60\;{\rm{MeV}} \end{align*} {/eq}

Thus, the kinetic energy of rebounding alpha particle is {eq}4.60\;{\rm{MeV}} {/eq}.

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Chapter 4 / Lesson 14