question_answer

(a) If the \[\alpha \]-decay of \[^{238}U\]is energetically allowed (i.e., the decay products have a total mass less than the mass of \[^{238}U\]), what prevents \[^{238}U\] from decaying all at once ? Why is its half life so large? (b) The \[\alpha \]-particle faces a Coulomb barrier. A neutron being uncharged faces no such barrier. Why does the nucleus \[_{92}^{238}U\] not decay spontaneously by emitting a neutron?                                               

Answer:

                (a) The \[\alpha \]-decay is caused by the quantum mechanical tunnelling of the nucleus by the \[\alpha \]-particles. The rate of tunnelling is determined by the height of the nuclear potential barrier, its width etc. (b) The possible decay would be \[_{92}^{238}U\to _{92}^{237}U+_{0}^{1}n\] \[m(_{92}^{237}U)=237.04874\text{amu}\] Total mass \[{{m}_{n}}=\frac{1.00867\text{amu}}{=238.05741\text{amu}}\] \[m(_{92}^{238}U)=238.05081\text{amu}\] i.e., \[m(_{92}^{237}U)+{{m}_{2}}>m(_{92}^{238}U)\] Thus the spontaneous decay of \[_{92}^{238}U\] is not energetically possible, i.e., it will not emit a neutron spontaneously. Rather, energy would be needed to separate a neutron from \[_{92}^{238}U\].