One of the waste products of a nuclear reactor is plutonium\( - 239\left( {^{239}Pu} \right)\). This nucleus is radioactive and decays by splitting into a helium-\(4\)nucleus and a uranium\( - 235\)nucleus\(\left( {^4He{ + ^{235}}U} \right)\), the latter of which is also radioactive and will itself decay some time later. The energy emitted in the plutonium decay is\(8.40 \times 1{0^{ - 13}}\;J\)and is entirely converted to kinetic energy of the helium and uranium nuclei. The mass of the helium nucleus is\(6.68 \times 1{0^{ - 27}}kg\),while that of the uranium is\(3.92 \times 1{0^{ - 25}}\;kg\)(note that the ratio of the masses is\(4\)to\(235\)). (a) Calculate the velocities of the two nuclei, assuming the plutonium nucleus is originally at rest. (b) How much kinetic energy does each nucleus carry away? Note that the data given here are accurate to three digits only.

Kinetic energy is the energy that an item or particle has as a result of its movement. When work is done on an object by exerting a net force, the object accelerates and gains kinetic energy as a result.

The mass of the helium nuclei is \({M_{He}} = 6.68 \times 1{0^{ - 27}}kg\)

The mass of the uranium nuclei is \({M_U} = 3.92 \times 1{0^{ - 25}}\;kg\)

The velocity of the helium and uranium nuclei before decay \({U_{He}} = {U_U} = 0 m/s\)

By putting all the value into the equation we get

\(\begin{aligned}{M_p}{V_p} &= {M_{He}}{V_{He}} + {M_U}{V_{Ua}}\\\cancel{{{M_p}{V_p}}} &= {M_{He}}{V_{He}} + {M_U}{V_{Ua}}\\ - {M_{He}}{V_{He}} &= {M_U}{V_{Ua}}\end{aligned}\)…………………(1)

The above equation we can use in kinetic energy equation.

\(\begin{aligned}K{E_{in}} &= K{E_f} - K{E_i}\\K{E_{in}} &= \left( {K{E_{He}} + K{E_U}} \right) - \cancel{{K{E_i}}}\\K{E_{in}} &= \left( {\dfrac{1}{2}{m_{He}}{v_{He}}^2 + \dfrac{1}{2}{m_u}{v_{Ua}}^2} \right) - 0\\K{E_{in}} &= \dfrac{1}{2}\left( {{m_{He}}{v_{He}}^2 + {m_u}{{\left( {\dfrac{{ - {m_{He}}{v_{He}}}}{{{m_u}}}} \right)}^2}} \right)....from\,eq\,(1)\end{aligned}\)

\(\begin{aligned}2K{E_{in}} &= {m_U}{v_U}^2 + \dfrac{{{m_{}}^2{v_{1a}}^2}}{{{m_2}}}\\2K{E_{in}} &= {v_{He}}^2\left( {{m_{He}} + \dfrac{{{m_{He}}^2}}{{{m_u}}}} \right)\end{aligned}\)

\(\begin{aligned}{v_{He}} &= \sqrt {\dfrac{{2K{E_{in}}}}{{\left( {{m_{He}} + \dfrac{{{m_{He}}^2}}{{{m_U}}}} \right)}}} \\{v_{He}} &= \sqrt {\dfrac{{2\left( {8.4 \times 1{0^{ - 13}}} \right)}}{{\left( {6.68 \times 1{0^{ - 27}} + \dfrac{{{{\left( {6.68 \times 1{0^{ - 27}}} \right)}^2}}}{{3.92 \times 1{0^{ - 25}}}}} \right)}}} \\{v_{He}} &= \pm 1.57 \times 1{0^7}\,m/s\\{v_{He}} &= - 1.57 \times 1{0^7}\,m/s\end{aligned}\)

Hence the velocity of the helium atom after explosion is \( - 1.57 \times 1{0^7}\,m/s\)

Putting the value of the velocity in equation 1

\(\begin{aligned} - {M_{He}}{V_{He}} &= {M_U}{V_{Ua}}\\{V_{Ua}} &= \dfrac{{ - {M_{He}}{V_{He}}}}{{{M_U}}}\\{V_{Ua}} &= \dfrac{{ - \left( {6.68 \times 1{0^{ - 27}}} \right)\left( {1.57 \times 1{0^7}} \right)}}{{\left( {3.92 \times 1{0^{ - 25}}} \right)}}\\{V_{Ua}} &= 2.68\, \times 1{0^5}m/s\end{aligned}\)

Therefore, the velocity of the uranium atom is \(2.68\, \times 1{0^5}m/s\)

The kinetic energy does each nucleus carry away

\(\begin{aligned}K{E_{He}} &= \dfrac{1}{2}m{V^2}\\ &= \dfrac{1}{2}\left( {6.68x1{0^{ - 27}}} \right){\left( {1.57x1{0^7}} \right)^2}\\ &= 8.23 \times 1{0^{ - 13}}\end{aligned}\)

The kinetic energy of Helium atom is \(8.23 \times 1{0^{ - 13}}\)

Similarly for the uranium atom it will be

\(\begin{aligned}K{E_U} &= \dfrac{1}{2}m{V^2}\\ &= \dfrac{1}{2}\left( {3.92x1{0^{ - 25}}} \right){\left( {2.68x1{0^5}} \right)^2}\\ &= 1.41 \times 1{0^{ - 14}}J\end{aligned}\)

Therefore, the kinetic energy of uranium atom is \(1.41 \times 1{0^{ - 14}}J\)