We are given the following atomic masses: 

$$\begin{gathered} {}_{92}{}^{238}\mathrm{U} = 238.05079 \mathrm{u} {}_2{}^4\mathrm{He} = 4.00260 \mathrm{u} \\ {}_{90}{}^{234}\mathrm{Th} = 234.04363 \mathrm{u} {}_1{}^1\mathrm{H} = 1.00783 \mathrm{u} \\ {}_{91}{}^{237}\mathrm{Pa} = 237.05121 \mathrm{u} \end{gathered}$$ 

Here the symbol Pa is for the element protactinium (Z = 91). 

(a) Calculate the energy released during the alpha decay of ${}_{92}{}^{238}\mathrm{U}.$ 

(b) Calculate the kinetic energy of the emitted α-particles. 

(c) Show that ${}_{92}{}^{238}\mathrm{U}$ cannot spontaneously emit a proton.

(a) The alpha decay of ${}_{92}{}^{238}\mathrm{U}$ is given by, 

                 ${}_{92}{}^{238}\mathrm{U} \stackrel{\alpha }{\to } {}_{90}{}^{234}\mathrm{Th} + {}_2{}^4\mathrm{He}$ 

The energy released in this process is given by 

              $\mathrm{Q} = ({\mathrm{M}}_{\mathrm{u}} - {\mathrm{M}}_{\mathrm{Th}} - {\mathrm{M}}_{\mathrm{He}}) {\mathrm{c}}^2$    ... (1)

Substituting the atomic masses as given in the data in equation (1), we find 
                          $\mathrm{Q} = (238.05079 - 234.04363 - 4.00260) \mathrm{u} \mathrm{x} {\mathrm{c}}^2$ 
   $$\begin{gathered} = (0.00456 \mathrm{u}) {\mathrm{c}}^2 \\ = (0.00456 \mathrm{u}) (931.5 \mathrm{MeV}/\mathrm{u}) \\ = 4.25 \mathrm{MeV} \end{gathered}$$ 

(b) The kinetic energy of the $\mathrm{\alpha }-$ particle is, 

 $$\begin{gathered} \boxed{{\mathrm{E}}_{\mathrm{\alpha }} \approx \left(\frac{\mathrm{A}-4}{\mathrm{A}}\right)\mathrm{Q}} \\ = \frac{234}{238}\times 4.25 \\ = 4.18 \mathrm{MeV} \end{gathered}$$ 

(c) If ${}_{92}{}^{238}\mathrm{U}$  spontaneously emits a proton, the decay process would be, 

             ${}_{92}{}^{238}\mathrm{U} \to {}_{91}{}^{237}\mathrm{Pa} + {}_1{}^1\mathrm{H}$ 

The Q for this process to happen is, 

$$\begin{gathered} = ({\mathrm{M}}_{\mathrm{U}} - {\mathrm{M}}_{\mathrm{Pa}} - {\mathrm{M}}_{\mathrm{H}}) {\mathrm{c}}^2 \\ = (238.05079 - 237.05121 - 1.00783) \mathrm{u} \times {\mathrm{c}}^2 \\ = (-0.00825 \mathrm{u}) {\mathrm{c}}^2 \\ = -(0.00825 \mathrm{u}) (931.5 \mathrm{MeV}/\mathrm{u}) \\ = -7.68 \mathrm{MeV} \end{gathered}$$ 

Since the Q-value for this process is negative, it cannot proceed spontaneously and will require 7.68 MeV energy.