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[From: ] [author: ] [Date: 11-05-18] [Hit: ]
In decay,where N(t) is what is left,1 - (1/2)^(3/(30*365.25*1440))as there are 365.=1.6.......
Cesium-137 undergoes beta decay and has a half life of 30 years. How many beta particles are emitted by a 14.0 gram sample of cesium-137 in three minutes?

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14g of Cesium-137 contains 14/137 moles of Cesium = 14/137 * 6.02 * 10^23 atoms = 6.15 * 10^22 atoms

In decay,

N(t) = N0 (1/2)^(t/half-life)

where N(t) is what is left, undecayed
N0 is starting quantity
t is time

The fraction that decays is then

1 - (1/2)^(3/(30*365.25*1440)) as there are 365.25*1440 minutes in a year
=1.32*10^-7

and the amount of particles that decay is

6.15 * 10^22 * 1.32 * 10^-7
=8.1 * 10 ^15
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