Question

problem 2 (____/9 pts)
a sample of radium has a weight of 1.5 mg and
its half - life is 6 years.

a. how much of the sample will remain after:
use units and box final answer.

i. 6 years?

ii. 3 years?

iii. 1 year?

Explanation:

Part i: After 6 years

Step1: Recall half - life concept

The half - life of a substance is the time it takes for half of the substance to decay. So after one half - life (6 years in this case), the amount remaining is half of the original amount.
The original amount \(N_0 = 1.5\space mg\), and the time \(t = 6\) years (which is equal to the half - life \(T_{1/2}=6\) years). The formula for the amount remaining \(N\) after \(n\) half - lives is \(N = N_0\times(\frac{1}{2})^n\), where \(n=\frac{t}{T_{1/2}}\). Here, \(n = \frac{6}{6}=1\).

Step2: Calculate the remaining amount

\(N=1.5\times(\frac{1}{2})^1\)
\(N = 1.5\times\frac{1}{2}=0.75\space mg\)

Step1: Determine the number of half - lives

The half - life \(T_{1/2} = 6\) years, and the time \(t = 3\) years. The number of half - lives \(n=\frac{t}{T_{1/2}}=\frac{3}{6}=0.5\).
The formula for the amount remaining is \(N = N_0\times(\frac{1}{2})^n\), with \(N_0 = 1.5\space mg\).

Step2: Calculate the remaining amount

\(N=1.5\times(\frac{1}{2})^{0.5}\)
Since \((\frac{1}{2})^{0.5}=\frac{1}{\sqrt{2}}\approx0.7071\)
\(N = 1.5\times\frac{1}{\sqrt{2}}\approx1.5\times0.7071\approx1.0607\space mg\) (or we can use the formula for exponential decay \(N = N_0e^{-\lambda t}\), where \(\lambda=\frac{\ln2}{T_{1/2}}\). First, \(\lambda=\frac{\ln2}{6}\), then \(N = 1.5e^{-\frac{\ln2}{6}\times3}=1.5e^{-\frac{\ln2}{2}}=1.5\times\frac{1}{\sqrt{2}}\approx1.06\space mg\))

Step1: Determine the number of half - lives

The half - life \(T_{1/2}=6\) years, time \(t = 1\) year. The number of half - lives \(n=\frac{t}{T_{1/2}}=\frac{1}{6}\).
The formula for the amount remaining is \(N = N_0\times(\frac{1}{2})^n\), with \(N_0 = 1.5\space mg\).

Step2: Calculate the remaining amount

\(N = 1.5\times(\frac{1}{2})^{\frac{1}{6}}\)
We know that \(a^{\frac{1}{n}}=\sqrt[n]{a}\), so \((\frac{1}{2})^{\frac{1}{6}}=\sqrt[6]{\frac{1}{2}} = 2^{-\frac{1}{6}}\approx0.8909\)
\(N=1.5\times0.8909\approx1.336\space mg\) (using the exponential decay formula \(N = N_0e^{-\lambda t}\), \(\lambda=\frac{\ln2}{6}\), \(t = 1\), so \(N = 1.5e^{-\frac{\ln2}{6}\times1}=1.5\times2^{-\frac{1}{6}}\approx1.34\space mg\))

Answer:

\(\boxed{0.75\space mg}\)

Part ii: After 3 years