53

​Q53. Given m $\left({}_{92}{}^{238}\cup \right)$ = 238. 05079u
         m $\left({}_1{}^1\mathrm{H}\right) = 1,00783\mathrm{u}, \mathrm{m} \left({}_{90}{}^{234}\mathrm{Th}\right) =234.04363\mathrm{u}$
        m $\left({}_{91}{}^{237}\mathrm{Pa}\right) =237.05121\mathrm{u}, \mathrm{m} \left({}_2{}^4\mathrm{He}\right) =4.00260\mathrm{u}$
          From the data given above, show that 

       (i) ${}_{92}{}^{238}\cup$ cannot spontaneously emit a proton.

      (ii) Alpha decay of ${}_{92}{}^{238}\cup$ is energetically possible. Calculate the energy released during the process.