A freshly made radioactive sample gives a count rate of $8000$ counts per minute. After twenty days, it gives a count-rate of $500$ counts per minute. What is the half-life of the material?

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Solution

$ln \frac{A_{o}}{A_{t}} = k t ln (\frac{8000}{500}) = k \times 20 ln 16 = k \times 20 k = \frac{ln 16}{20} = \frac{ln {(2)}^{4}}{20} k = \frac{4 ln 2}{20} k = \frac{ln 2}{5} t_{1 / 2} = \frac{ln 2}{k} = \frac{ln 2}{ln 2} \times 5 t_{1 / 2} = 5 d a y s$

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A freshly made radioactive sample gives a count rate of 8000 counts per minute. After twenty days, it gives a count-rate of 500 counts per minute. What is the half-life of the material?

lnAoAt=ktln(8000500)=k×20ln16=k×20k=ln1620=ln(2)420k=4ln220k=ln25t1/2=ln2k=ln2ln2×5t1/2=5days

A freshly made radioactive sample gives a count rate of $8000$ counts per minute. After twenty days, it gives a count-rate of $500$ counts per minute. What is the half-life of the material?

$ln \frac{A_{o}}{A_{t}} = k t ln (\frac{8000}{500}) = k \times 20 ln 16 = k \times 20 k = \frac{ln 16}{20} = \frac{ln {(2)}^{4}}{20} k = \frac{4 ln 2}{20} k = \frac{ln 2}{5} t_{1 / 2} = \frac{ln 2}{k} = \frac{ln 2}{ln 2} \times 5 t_{1 / 2} = 5 d a y s$