Question

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?

Answer 1

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, $k = \frac{2.303}{t} log \frac{Co}{Ct}$
where, Co = initial concentration = 1.1×10^10 atoms
Ct = conc. of Ba at time t.
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Answer 1: For t = 5 years
$0.066 = \frac{2.303}{5} log \frac{Co}{Ct}$
Therefore, $log\frac{Co}{Ct}$ = 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.
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Answer 2: For t = 30 years
$0.066 = \frac{2.303}{30} log \frac{Co}{Ct}$
Therefore, $log\frac{Co}{Ct}$ = 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.
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Answer 3: t = 180 years
$0.066 = \frac{2.303}{180} log \frac{Co}{Ct}$
Therefore, $log\frac{Co}{Ct}$ = 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.