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Question
- a 208 g sample of sodium - 24 decays to 13 g of sodium - 24 within 60 hours. what is the half - life of this radioactive isotope?
Step1: Calculate the number of half - lives
The formula for radioactive decay is \(N = N_0\times(\frac{1}{2})^n\), where \(N\) is the final amount, \(N_0\) is the initial amount, and \(n\) is the number of half - lives.
Given \(N_0 = 208\space g\), \(N = 13\space g\).
Substitute into the formula: \(13=208\times(\frac{1}{2})^n\).
Then \(\frac{13}{208}=(\frac{1}{2})^n\), and \(\frac{1}{16}=(\frac{1}{2})^n\).
Since \(\frac{1}{16}=(\frac{1}{2})^4\), so \(n = 4\).
Step2: Calculate the half - life
We know that the time \(t = 60\) hours and \(n = 4\).
The formula for the relationship between time \(t\), half - life \(T_{1/2}\) and number of half - lives \(n\) is \(t=n\times T_{1/2}\).
So \(T_{1/2}=\frac{t}{n}\).
Substitute \(t = 60\) hours and \(n = 4\) into the formula: \(T_{1/2}=\frac{60}{4}=15\) hours.
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The half - life of this radioactive isotope is \(15\) hours.