Question

question 9
an unknown radioactive element decays into non - radioactive substances. in 960 days the radioactivity of a sample decreases by 77 percent.
(a) find the decay constant ( k ). (round your answer to 5 decimal places.)
( k approx )
(b) what is the half - life of the element? (round your answer to two decimal places)
half - life: (days)
(c) how long will it take for a sample of 100 mg to decay to 77 mg? (round your answer to two decimal places)
time needed: (days)
question help: video post to forum
submit question

Explanation:

Step1: Use the radioactive decay formula

The radioactive decay formula is \(A(t)=A_0e^{-kt}\). If the radioactivity decreases by \(77\%\), then \(A(t)=(1 - 0.77)A_0=0.23A_0\) and \(t = 960\) days. Substitute into the formula: \(0.23A_0=A_0e^{-960k}\). Divide both sides by \(A_0\) (since \(A_0
eq0\)) to get \(0.23=e^{-960k}\).

Step2: Solve for \(k\)

Take the natural logarithm of both sides: \(\ln(0.23)=\ln(e^{-960k})\). Using the property \(\ln(e^x)=x\), we have \(\ln(0.23)=- 960k\). Then \(k=-\frac{\ln(0.23)}{960}\). Calculate \(\ln(0.23)\approx - 1.46963\), so \(k=\frac{1.46963}{960}\approx0.00153\).

Step3: Find the half - life formula

The half - life formula is \(T=\frac{\ln(2)}{k}\). Substitute \(k = 0.00153\) into the formula: \(T=\frac{\ln(2)}{0.00153}\). Since \(\ln(2)\approx0.693147\), then \(T=\frac{0.693147}{0.00153}\approx453.04\) days.

Step4: Solve for \(t\) when \(A(t) = 77\), \(A_0=100\)

Use the formula \(A(t)=A_0e^{-kt}\), so \(77 = 100e^{-0.00153t}\). Divide both sides by \(100\) to get \(0.77=e^{-0.00153t}\). Take the natural logarithm of both sides: \(\ln(0.77)=\ln(e^{-0.00153t})\). Using \(\ln(e^x)=x\), we have \(\ln(0.77)=-0.00153t\). Then \(t=-\frac{\ln(0.77)}{0.00153}\). Calculate \(\ln(0.77)\approx - 0.2600\), so \(t=\frac{0.2600}{0.00153}\approx169.93\) days.

Answer:

(a) \(k\approx0.00153\)
(b) half - life: \(453.04\) days
(c) time needed: \(169.93\) days