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6.7: homework assignment
score: 9.5/10 answered: 9/10
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question 3
0.5/
the half - life of palladium - 100 is 4 days. after 24 days a sample of palladium - 100 has been reduced to a
mass of 6 mg.
what was the initial mass (in mg) of the sample?
what is the mass (in mg) 7 weeks after the start?
you may enter the exact value or round to 4 decimal places.
question help: video

Explanation:

Step1: Calculate the number of half - lives for 24 days

The formula for the number of half - lives \(n=\frac{t}{T}\), where \(t\) is the time elapsed and \(T\) is the half - life. Given \(t = 24\) days and \(T=4\) days, then \(n=\frac{24}{4}=6\).
The radioactive decay formula is \(A = A_0(\frac{1}{2})^n\), where \(A\) is the final amount, \(A_0\) is the initial amount, and \(n\) is the number of half - lives. We know \(A = 6\) mg and \(n = 6\). Substituting into the formula: \(6=A_0(\frac{1}{2})^6\).

Step2: Solve for the initial mass \(A_0\)

From \(6=A_0(\frac{1}{2})^6\), we can rewrite it as \(A_0=6\times2^6\).
Since \(2^6 = 64\), then \(A_0=6\times64 = 384\) mg.

Step3: Calculate the number of half - lives for 7 weeks

First, convert 7 weeks to days. Since 1 week = 7 days, 7 weeks \(t=7\times7 = 49\) days.
Using \(n=\frac{t}{T}\) with \(T = 4\) days, \(n=\frac{49}{4}=12.25\).
Using the formula \(A = A_0(\frac{1}{2})^n\), with \(A_0 = 384\) and \(n = 12.25\).
\(A=384\times(\frac{1}{2})^{12.25}\)
We know that \((\frac{1}{2})^{12.25}=(\frac{1}{2})^{12}\times(\frac{1}{2})^{0.25}\)
\((\frac{1}{2})^{12}=\frac{1}{4096}\), \((\frac{1}{2})^{0.25}=\frac{1}{\sqrt[4]{2}}\approx0.8409\)
\(A = 384\times\frac{1}{4096}\times0.8409\)
\(A=\frac{384\times0.8409}{4096}\)
\(A=\frac{322.9056}{4096}\approx0.0788\) mg

Answer:

The initial mass is \(384\) mg. The mass 7 weeks after the start is approximately \(0.0788\) mg.