Question

the half - life of palladium - 100 is 4 days. after 20 days a sample of palladium - 100 has been reduced to a mass of 1 mg. what was the initial mass (in mg) of the sample? what is the mass (in mg) 6 weeks after the start? you may enter the exact value or round to 4 decimal places. question help: video

Explanation:

Step1: Find the number of half - lives

The half - life \(T = 4\) days. The time \(t = 20\) days. The number of half - lives \(n=\frac{t}{T}=\frac{20}{4}=5\)

The formula for radioactive decay is \(m = m_0(\frac{1}{2})^n\), where \(m\) is the final mass, \(m_0\) is the initial mass, and \(n\) is the number of half - lives.

We know \(m = 1\) mg and \(n = 5\). Substituting into the formula \(1=m_0(\frac{1}{2})^5\)

Step2: Solve for \(m_0\)

From \(1=m_0(\frac{1}{2})^5\), we can rewrite it as \(m_0=1\times2^5\)

Since \(2^5=32\), so the initial mass \(m_0 = 32\) mg.

Now, for the second part:
6 weeks \(=6\times7 = 42\) days. The number of half - lives \(n=\frac{42}{4}=10.5\)

Using the formula \(m = m_0(\frac{1}{2})^n\), with \(m_0 = 32\)

\(m=32\times(\frac{1}{2})^{10.5}\)

\(m = 32\times\frac{1}{2^{10}}\times\frac{1}{\sqrt{2}}\)

Since \(2^{10}=1024\) and \(\sqrt{2}\approx1.4142\)

\(m=\frac{32}{1024\times1.4142}\)

\(m=\frac{32}{1448.1408}\approx0.0221\)

Answer:

The initial mass is \(32\) mg. The mass 6 weeks after the start is approximately \(0.0221\) mg.