Question

question 4
if the half - life of phosphorus - 32 is 14.3 days, how much of a 200. mg sample of phosphorus - 32 remains after 71.5 days?
○ 100. mg
○ 50. mg
○ 12.5 mg
○ 6.25 mg

Explanation:

Step1: Calculate number of half - lives

The formula to find the number of half - lives \(n\) is \(n=\frac{t}{t_{1/2}}\), where \(t\) is the total time and \(t_{1/2}\) is the half - life.
Given \(t = 71.5\) days and \(t_{1/2}=14.3\) days.
\(n=\frac{71.5}{14.3}=5\)

Step2: Use the radioactive decay formula

The formula for radioactive decay is \(N = N_0\times(\frac{1}{2})^n\), where \(N_0\) is the initial amount, \(N\) is the remaining amount, and \(n\) is the number of half - lives.
Given \(N_0 = 200\) mg and \(n = 5\).
\(N=200\times(\frac{1}{2})^5\)
\((\frac{1}{2})^5=\frac{1}{32}\)
\(N = 200\times\frac{1}{32}=\frac{200}{32}=6.25\) mg

Answer:

6.25 mg