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Evaluate the amount of heat produced during a day by a `beta^(-)`- active `Na^(24)` preparation of mass `m= 1.0 mg`. The beta particles are assumed to possess an average kinetic energy equal to `1//3` of the highest possible energy of the given decay. The half-life of `Na^(24)` is `T= 15 hours`. |
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Answer» The masses are `Na^(24)= 24-0.00903 amu` and `Mg^(24)= 24-0.01496 amu` The reaction is `Na^(24) rarr Mg^(24)+e^(-)+bar(v)_(e )` The maximum `K.E` of electrons is `0.00593xx93MeV= 5.52MeV` Average `K.E.` according to the problem is then `(5.52)/(3)= 1.84MeV` The initial number of `Na^(24)` is `(10^(-3)xx6.023xx10^(23))/(24)=2.51xx10^(19)` The fraction decying in a day is `1-(2)^(-24//15)= 0.67` Hence the heat produced in a day is `0.67xx2.51xx10^(19)xx1.84xx1.602xx10^(-13)` Joul `= 4.95MJ` |
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