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Rudik [331]
3 years ago
13

The half-life of cesium-137 is 30 years. Suppose we have a

Mathematics
1 answer:
VMariaS [17]3 years ago
3 0
(a) If y(t) is the mass (in mg) remaining after t years, then y(t) = y(0) e^{kt}=100e^{kt}.

y(30) = 100 e^{30k} = \frac{1}{2}(100) \implies e^{30k} = \frac{1}{2} \implies k = -(\ln 2) /30 \implies \\ \\
y(t) = 100e^{-(\ln 2)t/30} = 100 \cdot 2^{-t/30}

(b) y(100) = 100 \cdot 2^{-100/30} \approx \text{9.92 mg}

(c)
100 e^{- (\ln 2)t/30} = 1\ \implies\ -(\ln 2) t / 30 = \ln \frac{1}{100}\ \implies\ \\ \\
t = -30 \frac{\ln 0.01}{\ln 2} \approx \text{199.3 years}
 
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