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The barium isotope 133ba has a half-life of 10.5 years. a sample begins with 1.1×1010 133ba atoms. how many are left after (a) 5 years, (b) 30 years, and (c) 180 years?
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Jan 27, 2019
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The barium isotope 133ba has a half-life of 10.5 years. a sample begins with 1.1×1010 133ba atoms. how many are left after (a) 5 years, (b) 30 years, and (c) 180 years?
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Matt Fichman
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Given
: Half-life of
133Ba = t1/2 = 10.5 years.
The radio-active materials obeys 1st order dissociation kinetics. Therefore we have:
k = 0.693 / t1/2 = 0.693 / 10.5 = 0.066 years-1.
Also,
where, Co = initial concentration =
1.1×10^10 atoms
Ct = conc. of Ba at time t.
......................................................................................................................
Answer 1
: For t =
5 years
Therefore,
= 0.1432
Therefore, Co/Ct = 1.3908
Therefore, Ct = 7.9086 X 10^9 atoms.
Number of 133Ba atoms left after 5 years = 7.9086 X 10^9.
....................................................................................................................
Answer 2
: For t = 30 years
Therefore,
= 0.8597
Therefore, Co/Ct = 7.2402
Therefore, Ct = 1.5193 X 10^9 atoms.
Number of 133Ba atoms left after 30 years = 1.5193 X 10^9.
........................................................................................................................
Answer 3
: t = 180 years
Therefore,
= 5.1585
Therefore, Co/Ct = 1.44 X 10^5
Therefore, Ct = 7.6367 X10^4 atoms.
Number of 133Ba atoms left after 180 years = 7.6367 X10^4.
Fscheidl
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Feb 2, 2019
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Fscheidl
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