91Ó°ÊÓ

The mass of the alpha particle is

\({m_1} = 4.0\;{\rm{u}}\).

The total mass of the nucleus is \(m = 222\;{\rm{u}}\).

The mass of the remaining nucleus is \({m_2} = \left( {222\;{\rm{u}} - 4.0\;{\rm{u}}} \right) = 218\;{\rm{u}}\)

The initial speed of the nucleus is \(v = 320\;{\rm{m/s}}\).

The seed of the remaining nucleus is \({v_2} = 280\;{\rm{m/s}}\).

Let \({v_1}\) be the alpha particle after emitted from the nucleus.

The total momentum of the alpha particle and the remaining nucleus is equal to the total momentum of the nucleus before emitting the alpha particle.

The total momentum of the nucleus before emitting the alpha particle is \(mv\).

The total momentum after emitting the alpha particle is \(\left( {{m_1}{v_1} + {m_2}{v_2}} \right)\).

From the concept of momentum conservation,

\(\begin{array}{c}{m_1}{v_1} + {m_2}{v_2} = mv\\{m_1}{v_1} = mv - {m_2}{v_2}\\{v_1} = \frac{{mv - {m_2}{v_2}}}{{{m_1}}}\end{array}\)

Now, after further calculation,

\(\begin{array}{c}{v_1} = \frac{{\left[ {\left( {222\;{\rm{u}}} \right) \times \left( {320\;{\rm{m/s}}} \right)} \right] - \left[ {\left( {218\;{\rm{u}}} \right) \times \left( {280\;{\rm{m/s}}} \right)} \right]}}{{4.0\;{\rm{u}}}}\\ = 2500\;{\rm{m/s}}\end{array}\)

Hence, the speed of the alpha particle is \(2500\;{\rm{m/s}}\) when emitted from the nucleus.