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Chapter 40: Problem 55

The radon isotope \({ }^{222} \mathrm{Rn}\), which has a half-life of 3.825 days, is used for medical purposes such as radiotherapy. How long does it take until \({ }^{222} \mathrm{Rn}\) decays to \(10.00 \%\) of its initial quantity?

Short Answer

Expert verified
Answer: It takes approximately 16.195 days for the radon isotope ${ }^{222} \mathrm{Rn}$ to decay to 10.00% of its initial quantity.

Step by step solution

01

Understand the radioactive decay formula

The radioactive decay formula is: \(N(t) = N_0 e^{-\lambda t}\) Where: - \(N(t)\) is the remaining quantity of radioactive material at time \(t\) - \(N_0\) is the initial quantity of radioactive material - \(\lambda\) is the decay constant - \(t\) is the time elapsed
02

Calculate the decay constant with the given half-life

The decay constant can be calculated using the half-life formula: \(\lambda = \frac{\ln 2}{T_{1/2}}\) Where: - \(T_{1/2}\) is the half-life of the radioactive material In this case, \(T_{1/2} = 3.825\) days. Plugging this value into the formula, we get: \(\lambda = \frac{\ln 2}{3.825} \approx 0.1812 \, \mathrm{day}^{-1}\)
03

Set the equation N(t) = 10% of N0 and solve for t

We are given that the remaining quantity should be 10% of the initial quantity which translates to the equation: \(N(t) = 0.1 N_0\) To solve for \(t\), we will use both equations and substitute like this: \(0.1 N_0 = N_0 e^{-\lambda t}\) The initial quantity \(N_0\) will cancel out, and we are left with: \(0.1 = e^{-0.1812 t}\)
04

Solve for t using logarithms

To solve for \(t\), take the natural logarithm of both sides: \(\ln 0.1 = -0.1812 t\) Now, isolate \(t\) by dividing both sides by \(-0.1812\): \(t = \frac{\ln 0.1 }{-0.1812} \approx 16.195 \, \mathrm{days}\)
05

Final answer

So, it takes approximately \(16.195\) days for the radon isotope \({ }^{222} \mathrm{Rn}\) to decay to 10.00% of its initial quantity.

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Most popular questions from this chapter

What is the total energy released in the decay \(n \rightarrow p+e^{-}+\bar{\nu}_{e} ?\)

The most common isotope of uranium, \({ }_{92}^{238} \mathrm{U},\) produces radon \({ }_{86}^{222} \mathrm{Rn}\) through the following sequence of decays: $$\begin{array}{c}{ }^{238} \mathrm{U} \rightarrow{ }^{234} \mathrm{Th}+\alpha,{ }^{234} \mathrm{Th} \rightarrow{ }^{234} \mathrm{~Pa}+\beta^{-}+\bar{\nu}_{e}, \\\\{ }_{91}^{234} \mathrm{~Pa} \rightarrow{ }_{92}^{234} \mathrm{U}+\beta+\bar{\nu}_{e},{ }^{234} \mathrm{U} \rightarrow{ }^{230} \mathrm{Th}+\alpha ,\\\\{ }_{91}^{230} \mathrm{Th} \rightarrow{ }_{90}^{226} \mathrm{Ra}+\alpha,{ }_{88}^{226} \mathrm{Ra} \rightarrow{ }_{86}^{222} \mathrm{Rn}+\alpha,\end{array}$$. A sample of \({ }_{92}^{238} \mathrm{U}\) will build up equilibrium concentrations of its daughter nuclei down to \({ }_{88}^{226} \mathrm{Ra} ;\) the concentrations of each are such that each daughter is produced as fast as it decays. The \({ }_{88}^{226} \mathrm{Ra}\) decays to \({ }_{86}^{222} \mathrm{Rn},\) which escapes as a gas. (The \(\alpha\) particles also escape, as helium; this is a source of much of the helium found on Earth.) In high concentrations, the radon is a health hazard in buildings built on soil or foundations containing uranium ores, as it can be inhaled. a) Look up the necessary data, and calculate the rate at which \(1.00 \mathrm{~kg}\) of an equilibrium mixture of \({ }_{92}^{238} \mathrm{U}\) and its first five daughters produces \({ }_{86}^{222} \mathrm{Rn}\) (mass per unit time). b) What activity (in curies per unit time) of radon does this represent?

Pu decays with a half-life of 24,100 yr via a 5.25 \(\mathrm{MeV}\) alpha particle. If you have a \(1.00 \mathrm{~kg}\) spherical sample of \({ }^{239} \mathrm{Pu},\) find the initial activity in \(\mathrm{Bq} .\)

a) What is the energy released in the fusion reaction \({ }_{1}^{2} \mathrm{H}+{ }_{1}^{2} \mathrm{H} \rightarrow{ }_{2}^{4} \mathrm{He}+Q ?\) b) The oceans have a total mass of water of \(1.50 \cdot 10^{16} \mathrm{~kg}\), and \(0.0300 \%\) of this quantity is deuterium, \({ }_{1}^{2} \mathrm{H} .\) If all the deuterium in the oceans were fused by controlled fusion into \({ }_{2}^{4} \mathrm{He},\) how many joules of energy would be released? c) World power consumption is about \(1.00 \cdot 10^{13} \mathrm{~W}\). If consumption were to stay constant and all problems arising from ocean water consumption (including those of political, meteorological, and ecological nature) could be avoided, how many years would the energy calculated in part (b) last?

The mean lifetime for a radioactive nucleus is \(4300 \mathrm{~s}\) What is its half-life?
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