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Calculate the amount of energy released during the `alpha`-decay of `._(92)^(238)Urarr_(90)^(234)Th+._(2)^(4)He` Given: atomic mass of `._(92)^(238)U=238.05079 u`, atomic mass of `._(90)^(234)Th=234.04363 u`, atomic mass `._(2)^(4)He=4.00260u , 1u=931.5 MeV//c^(2)`. Is this decay spontaneous?Give reason. |
Answer» The energy released in the `alpha`-decay is `Q= [m(._(92)^(238)U)-m(._(90)^(234)Th)-m(._(2)^(4)He)]c^(2)` `=[238.0579-234.04363-4.00260]xx931.5 MeV` `=0.00456xx931.5=4.25 MeV` As the `Q`-value is positive, the decay process is spontaneous. |
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