Question
The count rate from a radioactive sample falls from $4.0 \times 10^6$ per second to $1.0 \times 10^6$ per second in 20 hours. What will be the count rate 100 hours after the beginning?
✓
Answer
$\text{A}_0=4\times10^5$ disintegration/sec$\text{A}'=1\times10^6$ dis/sec; t = 20 hours.
$\text{A}'=\frac{\text{A}_0}{2^{\frac{\text{t}}{\text{t}_{\frac{1}{2}}}}}\Rightarrow2^{\frac{\text{t}}{\text{t}_{\frac{1}{2}}}}=\frac{\text{A}_0}{\text{A}'}\Rightarrow2^{\frac{\text{t}}{\text{t}_{\frac{1}{2}}}}=4$
$\Rightarrow\frac{\text{t}}{\text{t}_{\frac{1}{2}}}=2\Rightarrow\text{t}^{\frac{1}{2}}=\frac{\text{t}}{2}=\frac{20\text{ hours}}{2}=10\text{ hours}.$
$\text{A}''=\frac{\text{A}_0}{2^{\frac{\text{t}}{\text{t}_{\frac{1}{2}}}}}\Rightarrow\text{A}''=\frac{4\times10^6}{2^{\frac{100}{10}}}$
$=0.00390625\times10^6=3.9\times10^3$ dintegrations/sec.
$\text{A}'=\frac{\text{A}_0}{2^{\frac{\text{t}}{\text{t}_{\frac{1}{2}}}}}\Rightarrow2^{\frac{\text{t}}{\text{t}_{\frac{1}{2}}}}=\frac{\text{A}_0}{\text{A}'}\Rightarrow2^{\frac{\text{t}}{\text{t}_{\frac{1}{2}}}}=4$
$\Rightarrow\frac{\text{t}}{\text{t}_{\frac{1}{2}}}=2\Rightarrow\text{t}^{\frac{1}{2}}=\frac{\text{t}}{2}=\frac{20\text{ hours}}{2}=10\text{ hours}.$
$\text{A}''=\frac{\text{A}_0}{2^{\frac{\text{t}}{\text{t}_{\frac{1}{2}}}}}\Rightarrow\text{A}''=\frac{4\times10^6}{2^{\frac{100}{10}}}$
$=0.00390625\times10^6=3.9\times10^3$ dintegrations/sec.
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