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(a)If alpha-decays of ._(92)U^(238) is energetically allowed (i.e., the decay products have a total mass less than tha mass of ._(92)U^(238)), what prevents ._(92)U^(238) form decaying all at once? Why is its half life so large? (b) The alpha-particle faces a Coulomb barrier. A neutron being unchanged faces no such barrier. Why does the nucleus ._(92)U^(238) not decay spontaneously, by emitting a neutron? |
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Answer» Solution :(a) As explained in theory, `alpha` -decay is caused by the quantum mechanical tunnelling of an alpha particle through a repulsive Coulomb barrier. The rate of tunneling would depend upon the height and width of the barrier. The decay cannot be all at once. And that is the reason why half life of `._(92)U^(238)` against `alpha` -decay is large, (b) The possible nuclear reaction is `._(92)U^(238)to._(92)U^(237)+._(0)n^(1)` The DATA shows that `m(._(92)U^(237))+m(n)` is GREATER than `m(._(92)U^(238))`. Therefore, the decay is not allowed energetically. Rather, some EXTERNAL energy has to be supplied to separated a neutron form `._(92)U^(238)`. |
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