Question 42.48: How much energy is released when a ^238U nucleus decays by e......
How much energy is released when a ^{238}U nucleus decays by emitting (a) an alpha particle and (b) a sequence of neutron, proton, neutron, proton? (c) Convince yourself both by reasoned argument and by direct calculation that the difference between these two numbers is just the total binding energy of the alpha particle. (d) Find that binding energy. Some needed atomic and particle masses are
\begin{array}{cccr} { }^{238} U & 238.05079 \,u & { }^{234} Th & 234.04363 \,u \\ { }^{237} U & 237.04873\, u & { }^4 He & 4.00260\, u \\ { }^{236} Pa & 236.04891 \,u & { }^1 H & 1.00783\, u \\ { }^{235} Pa & 235.04544 \,u & n & 1.00866 \,u \end{array}
(a) The nuclear reaction is written as { }^{238} U \rightarrow{ }^{234} Th +{ }^4 He . The energy released is
\begin{aligned}\Delta E_1 & =\left(m_{\mathrm{U}}-m_{\mathrm{He}}-m_{\mathrm{Th}}\right) c^2 \\& =(238.05079 \mathrm{u}-4.00260 \mathrm{u}-234.04363 \mathrm{u})(931.5 \mathrm{MeV} / \mathrm{u}) \\& =4.25 \mathrm{MeV}\end{aligned}
(b) The reaction series consists of { }^{238} U \rightarrow{ }^{237} U + n , followed by
\begin{aligned} { }^{237} U \rightarrow{ }^{236} Pa + p \\ { }^{236} Pa \rightarrow{ }^{235} Pa + n \\ { }^{235} Pa \rightarrow{ }^{234} Th + p \end{aligned}
The net energy released is then
\begin{aligned}\Delta E_2=\left(m_{238}-\right. & \left.m_{237}-m_n\right) c^2+\left(m_{237 \mathrm{U}}-m_{236 _\mathrm{~Pa}}-m_p\right) c^2 \\& +\left(m_{236 _\mathrm{~Pa}}-m_{235 _\mathrm{~Pa}}-m_n\right) c^2+\left(m_{235 _\mathrm{~Pa}}-m_{234 _\mathrm{Th}}-m_p\right) c^2\end{aligned}
=\left(m_{238_ \mathrm{U}}-2 m_n-2 m_p-m_{234 _\mathrm{Th}}\right) c^2
\begin{aligned}& =[238.05079 \mathrm{u}-2(1.00867 \mathrm{u})-2(1.00783 \mathrm{u})-234.04363 \mathrm{u}](931.5 \mathrm{MeV} / \mathrm{u}) \\& =-24.1 \mathrm{MeV}\end{aligned}(c) This leads us to conclude that the binding energy of the α particle is
\left|\left(2 m_n+2 m_p-m_{\mathrm{He}}\right) c^2\right|=|-24.1 \mathrm{MeV}-4.25 \mathrm{MeV}|=28.3 \mathrm{MeV}.