We are given the following atomic masses: 92238U=238.05079 u   24He=4.00260 u   90234Th=234.04363  u   11H=1.00783  u  91237Pa=237.05121  u. Take  1 u≡931.5  MeVHere the symbol Pa is for the element protactinium (Z = 91). Mark the CORRECT statement(s)

The energy released in  α-decay is given by  Q=(Mu−MTh−MHe)c2Substituting the atomic masses as given in the data, we findQ=(238.05079−234.04363−4.00260)u×c2=(0.00456 u)c2=(0.00456 u)c2 =(0.00456 u)(931.5  MeV/u)=4.25  MeV .If   92238U spontaneously emits a proton, the decay process would be   92238U→ 91237Pa+ 11HThe Q for this process to happen is =(MU−MPa−MH)c2=(238.05079−237.05121−1.00783)u×c2=(−0.00825 u)c2 =−(0.00825 u)(931.5  MeV/u)    =−7.68  MeVThus, the Q of the process is negative and therefore it cannot proceed spontaneously. We will have to supply energy of 7.68 MeV to a  92238U  nucleus to make it emit a proton.