after 42 days a 2.0 g sample of phosphorus - ...
after 42 days a 2.0 g sample of phosphorus - 32 contains only 0.25 g of the isotope. what is the half - life of phosphorus - 32? days
Answer
# Explanation:
## Step1: Determine the number of half - lives
Use the formula \(N = N_0\times\left(\frac{1}{2}\right)^n\), where \(N = 0.25\space g\), \(N_0=2.0\space g\).
Substitute the values into the formula: \(0.25 = 2.0\times\left(\frac{1}{2}\right)^n\).
Then \(\frac{0.25}{2.0}=\left(\frac{1}{2}\right)^n\), \(\frac{1}{8}=\left(\frac{1}{2}\right)^n\).
Since \(\frac{1}{8}=\left(\frac{1}{2}\right)^3\), so \(n = 3\).
## Step2: Calculate the half - life
We know that \(t=42\space days\) and \(n = 3\).
Use the formula \(t=n\times t_{1/2}\), then \(t_{1/2}=\frac{t}{n}\).
Substitute \(t = 42\space days\) and \(n = 3\) into the formula: \(t_{1/2}=\frac{42}{3}\).
# Answer:
\(14\)