Question

an unknown radioactive element decays into non - radioactive substances. in 240 days the radioactivity of a sample decreases by 40 percent.
(a) find the decay constant k. (round your answer to 5 decimal places.)
k≈
(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 78 mg? (round your answer to two decimal places)
time needed: (days)
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Explanation:

Step1: Find the decay constant \(k\)

The radioactive decay formula is \(A(t)=A_0e^{-kt}\). If the radioactivity decreases by \(40\%\), then \(A(t)=(1 - 0.4)A_0=0.6A_0\) and \(t = 240\) days.
Substitute into the formula: \(0.6A_0=A_0e^{-240k}\).
Divide both sides by \(A_0\): \(0.6=e^{-240k}\).
Take the natural logarithm of both sides: \(\ln(0.6)=- 240k\).
Solve for \(k\): \(k=-\frac{\ln(0.6)}{240}\).
Using a calculator, \(\ln(0.6)\approx - 0.51083\), so \(k=\frac{0.51083}{240}\approx0.00213\).

Step2: Find the half - life \(T\)

The formula for half - life is \(T=\frac{\ln(2)}{k}\).
Since \(k\approx0.00213\), then \(T=\frac{\ln(2)}{0.00213}\).
Using \(\ln(2)\approx0.69315\), \(T=\frac{0.69315}{0.00213}\approx325.42\) days.

Step3: Find the time \(t\) for \(A_0 = 100\) mg and \(A(t)=78\) mg

Substitute into \(A(t)=A_0e^{-kt}\): \(78 = 100e^{-0.00213t}\).
Divide both sides by \(100\): \(0.78=e^{-0.00213t}\).
Take the natural logarithm of both sides: \(\ln(0.78)=-0.00213t\).
Solve for \(t\): \(t=-\frac{\ln(0.78)}{0.00213}\).
Using \(\ln(0.78)\approx - 0.24847\), \(t=\frac{0.24847}{0.00213}\approx116.66\) days.

Answer:

(a) \(k\approx0.00213\)
(b) half - life: \(325.42\) days
(c) time needed: \(116.66\) days