Question

a radioactive isotope has a half life of 5 days. the change in the mass of 40 grams of this isotope was observed over a period of 15 days.
which table shows that the isotope decays by a constant percent rate?

Explanation:

Step1: Calculate the number of half - lives

The half - life \(T = 5\) days, and the total time \(t=15\) days. The number of half - lives \(n=\frac{t}{T}=\frac{15}{5} = 3\).

Step2: Use the radioactive decay formula \(m = m_0(\frac{1}{2})^n\)

Given \(m_0 = 40\) grams.

• After \(n = 1\) (at \(t = 5\) days): \(m=40\times(\frac{1}{2})^1=20\) grams.
• After \(n = 2\) (at \(t = 10\) days): \(m = 40\times(\frac{1}{2})^2=40\times\frac{1}{4}=10\) grams.
• After \(n = 3\) (at \(t = 15\) days): \(m=40\times(\frac{1}{2})^3=40\times\frac{1}{8} = 5\) grams.

Answer:

The second table (where at \(t = 0\) mass \(m = 40\), at \(t = 5\) mass \(m = 20\), at \(t = 10\) mass \(m = 10\), at \(t = 15\) mass \(m = 5\)) shows the correct decay.