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If a \(6.13 \mathrm{~g}\) sample of an isotope having a mass number of 124 decays at a rate of \(0.350 \mathrm{Ci}\), what is its half-life?

Short Answer

Expert verified
The half-life of the isotope is approximately \(5.01 \times 10^{6}\) years.

Step by step solution

01

Identify the variables

The mass of the isotope \(m = 6.13 \, \mathrm{g}\), the mass number \(M = 124\), and the rate of decay \(R = 0.350 \, \mathrm{Ci}\) are given. We are asked to find the half-life \(T_{1/2}\).
02

Convert the rate of decay

Measure the decay rate in decays per second. 1 Ci (Curie) equals \(3.7 \times 10^{10}\) decays per second, so \(R = 0.350 \times 3.7 \times 10^{10} = 1.295 \times 10^{10} \, \mathrm{decays/s}\).
03

Calculate the number of atoms N

Calculate the number of atoms in the sample. One mole of an element has \(6.022 \times 10^{23}\) atoms, and thus contains a mass in grams equal to its atomic mass. So \(N = \frac{m}{M} \times 6.022 \times 10^{23} = \frac{6.13}{124} \times 6.022 \times 10^{23} = 2.975 \times 10^{23} \, \mathrm{atoms}\).
04

Apply the decay law and solve for the half-life

The radioactive decay law states that \(R = \frac{0.693N}{T_{1/2}}\). Solving for \(T_{1/2}\) gives \(T_{1/2} = \frac{0.693N}{R}\). Substituting the calculated values for N and R into the formula, we get \(T_{1/2} = \frac{0.693 \times 2.975 \times 10^{23}}{1.295 \times 10^{10}} \approx 1.58 \times 10^{14} \, \mathrm{s}\). This is the half-life in seconds, but it's more useful to convert this to years using \(1 \, \mathrm{year} = 3.1536 \times 10^{7} \, \mathrm{s}\), so \(T_{1/2} \approx 5.01 \times 10^{6} \, \mathrm{years}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Radioactive Decay Law
Radioactive decay is the process by which an unstable atomic nucleus loses energy by emitting radiation. A key aspect of studying radioactive substances is understanding the radioactive decay law, which is a mathematical description of how the quantity of a radioactive material decreases over time.

The law is often expressed by the equation \( R = \frac{0.693N}{T_{1/2}} \), where \( R \) is the decay rate (the number of decays per unit time), \( N \) is the number of radioactive atoms present, and \( T_{1/2} \) is the half-life, the time it takes for half of the radioactive atoms to decay. The factor 0.693 is the natural logarithm of 2, which is intrinsic to the inverse logarithmic relationship between the number of radioactive nuclei and time.

To calculate the half-life using this law, we need to know both the decay rate and the number of radioactive atoms present. By measuring the decay rate in a suitable unit and counting the number of atoms through knowledge of the sample’s mass and its molar mass, we can apply this formula directly to find the half-life.
Decay Rate Conversion
In working with radioactive materials, the decay rate can be given in different units, which are not always intuitive. One of the commonly used units is the Curie (Ci), named after the pioneering scientist Marie Curie. It's important to be able to convert between these units to ensure accurate calculations.

A Curie is defined as \(3.7 \times 10^{10}\) decays per second. So when we have a decay rate expressed in Curies, we convert it by multiplying by \(3.7 \times 10^{10}\) to get the rate in decays per second, which can then be used in the radioactive decay law formula. Here's the concept practically applied: given a decay rate of \(0.350 \mathrm{Ci}\), we calculate \(0.350 \times 3.7 \times 10^{10} = 1.295 \times 10^{10} \mathrm{decays/s}\).

Understanding how to convert decay rate units is crucial because working with a common base unit like decays per second simplifies the process of calculating other decay-related parameters, such as half-life.
Isotope Half-Life
The half-life of an isotope is a defining characteristic used in numerous scientific and medical applications. It represents the time required for half of the isotope's atoms in a sample to undergo radioactive decay.

Knowing an isotope's half-life allows us to predict how long a radioactive material will remain active or, conversely, when it will be safe due to significant decay. This concept relates directly to the radioactive decay law, providing a way to calculate how much of a substance will remain after a given period.

Using our previous example, after finding the decay rate to be \(1.295 \times 10^{10} \mathrm{decays/s}\) and calculating the number of atoms to be \(2.975 \times 10^{23}\), we can determine the half-life by substituting these values into the half-life formula: \(T_{1/2} = \frac{0.693 \times 2.975 \times 10^{23}}{1.295 \times 10^{10}} \). The result, approximately \(5.01 \times 10^{6}\) years, demonstrates the longevity of the radioactive isotope's decay process.

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Most popular questions from this chapter

Measurements on a certain isotope tell you that the decay rate decreases from 8318 decays/min to 3091 decays/min in 4.00 days. What is the half-life of this isotope?

Measurements indicate that \(27.83 \%\) of all rubidium atoms currently on the earth are the radioactive \({ }^{87} \mathrm{Rb}\) isotope. The rest are the stable \({ }^{85} \mathrm{Rb}\) isotope. The half-life of \({ }^{87} \mathrm{Rb}\) is \(4.75 \times 10^{10} \mathrm{y}\). Assuming that no rubidium atoms have been formed since, what percentage of rubidium atoms were \({ }^{87} \mathrm{Rb}\) when our solar system was formed \(4.6 \times 10^{9} \mathrm{y}\) ago?

Many radioactive decays occur within a sequence of decaysfor example, \({ }^{234} \mathrm{U} \rightarrow{ }_{98}^{230} \mathrm{Th} \rightarrow{ }_{84}^{226} \mathrm{Ra}\). The half-life for the \({ }_{92}^{234} \mathrm{U} \rightarrow{ }_{88}^{230} \mathrm{Th}\) decay is \(2.46 \times 10^{5} \mathrm{y},\) and the half-life for the \({ }_{88}^{230} \mathrm{Th} \rightarrow{ }_{84}^{226} \mathrm{Ra}\) decay is \(7.54 \times 10^{4} \mathrm{y}\). Let 1 refer to \({ }_{92}^{234} \mathrm{U}, 2\) to \({ }_{88}^{230} \mathrm{Th},\) and 3 to \({ }_{84}^{226} \mathrm{Ra} ;\) let \(\lambda_{1}\) be the decay constant for the \({ }_{92}^{234} \mathrm{U} \rightarrow{ }_{88}^{230} \mathrm{Th}\) decay and \(\lambda_{2}\) be the decay constant for the \({ }_{88}^{230} \mathrm{Th} \rightarrow{ }^{226} \mathrm{Ra}\) decay. The amount of \({ }_{88}^{230} \mathrm{Th}\) present at any time depends on the rate at which it is produced by the decay of \({ }_{92}^{234} \mathrm{U}\) and the rate by which it is depleted by its decay to \({ }_{84}^{226} \mathrm{Ra}\). Therefore, \(d N_{2}(t) / d t=\lambda_{1} N_{1}(t)-\lambda_{2} N_{2}(t) .\) If we start with a sample that contains \(N_{10}\) nuclei of \({ }_{92}^{234} \mathrm{U}\) and nothing else, then \(N(t)=N_{01} e^{-\lambda_{1} t}\) Thus \(d N_{2}(t) / d t=\lambda_{1} N_{10} e^{-\lambda_{1} t}-\lambda_{2} N_{2}(t) .\) This differential equation for \(N_{2}(t)\) can be solved as follows. Assume a trial solution of the form \(N_{2}(t)=N_{10}\left[h_{1} e^{-\lambda_{1} t}+h_{2} e^{-\lambda_{2} t}\right],\) where \(h_{1}\) and \(h_{2}\) are constants. (a) since \(N_{2}(0)=0,\) what must be the relationship between \(h_{1}\) and \(h_{2} ?\) (b) Use the trial solution to calculate \(d N_{2}(t) / d t,\) and substitute that into the differential equation for \(N_{2}(t) .\) Collect the coefficients of \(e^{-\lambda_{1} t}\) and \(e^{-\lambda_{2} t}\) since the equation must hold at all \(t,\) each of these coefficients must be zero. Use this requirement to solve for \(h_{1}\) and thereby complete the determination of \(N_{2}(t)\). (c) At time \(t=0\), you have a pure sample containing \(30.0 \mathrm{~g}\) of \({ }_{92}^{234} \mathrm{U}\) and nothing else. What mass of \({ }_{88}^{230} \mathrm{Th}\) is present at time \(t=2.46 \times 10^{5} \mathrm{y},\) the half-life for the \({ }_{92}^{234} \mathrm{U}\) decay?

\(\mathrm{} \mathrm{A}^{60} \mathrm{Co}\) source with activity \(2.6 \times 10^{-4} \mathrm{Ci}\) is embedded in a tumor that has mass \(0.200 \mathrm{~kg}\). The source emits \(\gamma\) photons with average energy \(1.25 \mathrm{MeV}\). Half the photons are absorbed in the tumor, and half escape. (a) What energy is delivered to the tumor per second? (b) What absorbed dose (in rad) is delivered per second? (c) What equivalent dose (in rem) is delivered per second if the \(\operatorname{RBE}\) for these \(\gamma\) rays is \(0.70 ?\) (d) What exposure time is required for an equivalent dose of 200 rem?

A \(70.0 \mathrm{~kg}\) person experiences a whole-body exposure to \(\alpha\) radiation with energy \(4.77 \mathrm{MeV}\). A total of \(7.75 \times 10^{12} \alpha\) particles are absorbed. (a) What is the absorbed dose in rad? (b) What is the equivalent dose in rem? (c) If the source is \(0.0320 \mathrm{~g}\) of \({ }^{226} \mathrm{Ra}\) (halflife 1600 y) somewhere in the body, what is the activity of this source? (d) If all of the alpha particles produced are absorbed, what time is required for this dose to be delivered?

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