the radioactive nuclide of plutonium Pu(mass no= 239 ,proton=94),
decays by alpha-particle emission with a half-life of 2.4X10^4 years.
The alpha-particle energy is 8.2X10^ -13 J
A small power souce to generate 2.5W is to be made from the plutonium isotope.
1) show that the rate of decay of the plutonium must be at least 3 X 10^12 Bq?
atomic physics!! 唔識呀
2008-03-22 5:25 am
回答 (1)
2008-03-22 8:11 am
✔ 最佳答案
Let A be the activity of plutonium in BqIn 1 s, the number of alpha particles emitted
= A
Energy emitted = 8.2x10^-13A
Thus, 8.2x10^-13A = 2.5
i.e. A = 2.5/8.2x10^-13 Bq = 3x10^12 Bq
收錄日期: 2021-04-29 17:26:56
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