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Chapter 14: Problem 22

The half-life of the radioactive isotope phosphorus- 32 is 14.3 days. How long does it take for a sample of phosphorus-32 to lose \(99 \%\) of its radioactivity?

Short Answer

Expert verified
It will take approximately 67.1 days for a sample of phosphorus-32 to lose 99% of its radioactivity.

Step by step solution

01

Understand the Problem

The problem has provided the half-life of phosphorus-32 as 14.3 days. The task is to calculate how long it will take for a sample of this isotope to lose 99% of its radioactivity. This will involve understanding the concept of half-life and calculating the number of half-lives that would result in a 99% decrease.
02

Calculate the number of half-lives

The formula for exponential decay can be used to calculate the number of half-lives needed for a 99% decrease. It is given by \(N = N0(1/2)^n\), where \(N\) is the final amount, \(N0\) is the initial amount, and \(n\) is the number of half-lives. Since we are looking for a 99% decrease, \(N\) would be 1% of \(N0\), which can be written as \(N = 0.01 N0\). Now, substituting into the equation we get: \(0.01 = (1/2)^n\). Then, take the natural logarithm of both sides to solve for \(n\). By doing this you isolate \(n\) on one side of the equation.
03

Solve for the time

Now that we have the number of half-lives, multiplying this by the half-life of phosphorus-32 will give the total time required. We can substitute the known value for the half-life of phosphorus-32 as 14.3 days into our solution to find out the total time it takes for phosphorus-32 to lose 99% of its radioactivity.

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Most popular questions from this chapter

The decomposition of nitric oxide occurs through two parallel reactions: $$\mathrm{NO}(\mathrm{g}) \longrightarrow \frac{1}{2} \mathrm{N}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \quad k_{1}=25.7 \mathrm{s}^{-1}$$ $$\mathrm{NO}(\mathrm{g}) \longrightarrow \frac{1}{2} \mathrm{N}_{2} \mathrm{O}(\mathrm{g})+\frac{1}{4} \mathrm{O}_{2}(\mathrm{g}) \quad k_{2}=18.2 \mathrm{s}^{-1}$$ (a) What is the reaction order for these reactions? (b) Which reaction is the slow reaction? (c) If the initial concentration of \(\mathrm{NO}(\mathrm{g})\) is \(2.0 \mathrm{M},\) what is the concentration of \(\mathrm{N}_{2}(\mathrm{g})\) after 0.1 seconds? (d) If the initial concentration of \(\mathrm{NO}(\mathrm{g})\) is \(4.0 \mathrm{M},\) what is the concentration of \(\mathrm{N}_{2} \mathrm{O}(\mathrm{g})\) after 0.025 seconds?

What are the similarities and differences between the catalytic activity of platinum metal and of an enzyme?

The following data were obtained for the dimerization of 1,3 -butadiene, \(2 \mathrm{C}_{4} \mathrm{H}_{6}(\mathrm{g}) \longrightarrow \mathrm{C}_{8} \mathrm{H}_{12}(\mathrm{g}),\) at 600 K: \(t=0 \min ,\left[\mathrm{C}_{4} \mathrm{H}_{6}\right]=0.0169 \mathrm{M} ; 12.18 \mathrm{min}, 0.0144 \mathrm{M} ; 24.55 \mathrm{min}, 0.0124 \mathrm{M} ; 42.50 \mathrm{min}, 0.0103 \mathrm{M}, 68.05 \min , 0.00845 \mathrm{M}.\) (a) What is the order of this reaction? (b) What is the value of the rate constant, \(k ?\) (c) At what time would \(\left[\mathrm{C}_{4} \mathrm{H}_{6}\right]=0.00423 \mathrm{M} ?\) (d) At what time would \(\left[\mathrm{C}_{4} \mathrm{H}_{6}\right]=0.0050 \mathrm{M} ?\)

For the reaction \(A+2 B \longrightarrow 2 C\), the rate of reaction is \(1.76 \times 10^{-5} \mathrm{M} \mathrm{s}^{-1}\) at the time when \([\mathrm{A}]=0.3580 \mathrm{M}.\) (a) What is the rate of formation of \(\mathrm{C}\) ? (b) What will \([\mathrm{A}]\) be 1.00 min later? (c) Assume the rate remains at \(1.76 \times 10^{-5} \mathrm{M} \mathrm{s}^{-1}\) How long would it take for \([\mathrm{A}]\) to change from 0.3580 to \(0.3500 \mathrm{M} ?\)

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