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The nucleus \(_{8}^{15} \mathrm{O}\) has a half-life of 2.0 min. \(_{8}^{19}\mathrm{O}\) has a half-life of about 0.5 min. (a) If, at some, time, a sample contains equal amounts of \(_{8}^{15} \mathrm{O}\) and \(_{8}^{19} \mathrm{O},\) what is the ratio of \(_{8}^{15} \mathrm{O}\) to \(_{8}^{19} \mathrm{O}\) after 2.0 min? (b) After 10.0 min?

Short Answer

Expert verified
(a) 8:1 after 2 min; (b) 32768:1 after 10 min.

Step by step solution

01

Understand Half-life

The half-life of a radioactive substance is the time it takes for half of the material to decay. In this problem, 1. For _{8}^{15} \( \mathrm{O} \), the half-life is 2.0 minutes,2. For \( _{8}^{19} \mathrm{O} \), the half-life is 0.5 minutes.We need to find how much of each substance remains after given time intervals.
02

Formulate General Decay Formula

The decay of a radioactive element can be described by the formula:\( N(t) = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}}\)where \(N(t)\) represents the amount remaining at time \(t\), \(N_0\) represents the initial amount, and \(T_{1/2}\) is the half-life.
03

Calculate Remaining \( _{8}^{15} O \) after 2 Min

Since the half-life of \( _{8}^{15} \mathrm{O} \) is 2.0 minutes, after 2 minutes, half of it will remain:\( N_{\,_{8}^{15}O}(2) = \frac{N_0}{2}\)Therefore, quantity remaining = \(\frac{N_0}{2}\).
04

Calculate Remaining \( _{8}^{19} O \) after 2 Min

Since the half-life of \( _{8}^{19} \mathrm{O} \) is 0.5 minutes, after 2 minutes, we calculate the decay through two half-lives:\( N_{\,_{8}^{19}O}(2) = N_0 \left( \frac{1}{2} \right)^{\frac{2}{0.5}} = N_0 \left( \frac{1}{2} \right)^4 = \frac{N_0}{16}\)
05

Calculate Ratio After 2 Min

The ratio of \( _{8}^{15} \mathrm{O} \) to \( _{8}^{19} \mathrm{O} \) after 2 minutes is:\( \frac{N_{\,_{8}^{15}O}(2)}{N_{\,_{8}^{19}O}(2)} = \frac{\frac{N_0}{2}}{\frac{N_0}{16}} = 8\)So, the ratio is 8:1.
06

Calculate Remaining \( _{8}^{15} O \) after 10 Min

After 10 minutes, we calculate using \(T_{1/2} = 2\) minutes:\( N_{\,_{8}^{15}O}(10) = N_0 \left( \frac{1}{2} \right)^{\frac{10}{2}} = \frac{N_0}{32}\)
07

Calculate Remaining \( _{8}^{19} O \) after 10 Min

After 10 minutes, we calculate using \(T_{1/2} = 0.5\) minutes:\( N_{\,_{8}^{19}O}(10) = N_0 \left( \frac{1}{2} \right)^{\frac{10}{0.5}} = N_0 \left( \frac{1}{2} \right)^{20} = \frac{N_0}{1048576}\)
08

Calculate Ratio After 10 Min

The ratio of \( _{8}^{15} \mathrm{O} \) to \( _{8}^{19} \mathrm{O} \) after 10 minutes is:\( \frac{N_{\,_{8}^{15}O}(10)}{N_{\,_{8}^{19}O}(10)} = \frac{\frac{N_0}{32}}{\frac{N_0}{1048576}} = \frac{1048576}{32} = 32768\)So, the ratio is 32768:1.

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Key Concepts

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

Half-life
In radioactive decay, understanding the concept of half-life is crucial. It refers to the period needed for half of a sample of a radioactive substance to decay into another substance. This measure provides insight into how long an isotope remains active before significant decay occurs.
For example, in our exercise, the half-life of the isotope \(_{8}^{15} \mathrm{O}\) is 2.0 minutes, meaning that in 2 minutes, half of its sample would have decayed into a different element. Similarly, \(_{8}^{19} \mathrm{O}\) has a half-life of 0.5 minutes, indicating that in only 30 seconds, half of it transforms into another substance.
Understanding half-life helps us calculate how long it will take for a radioactive substance to reduce to a particular amount. It's a pivotal concept in radiochemistry and nuclear physics because it lays the foundation for safely handling radioactive materials and determining timelines for their natural decay.
Decay Formula
The decay formula is essential for calculating how quickly a radioactive isotope decays over time. This mathematical equation is expressed as: \[ N(t) = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \]Here, \(N(t)\) stands for the amount of substance remaining at time \(t\), \(N_0\) is the initial quantity of the substance, and \(T_{1/2}\) is the half-life of the substance.
Using this formula, you can predict how much of an isotope will remain after a certain time has passed. For instance, with \(_{8}^{15} \mathrm{O}\), after 2 minutes, its amount reduces to half because its half-life is 2 minutes. For \(_{8}^{19} \mathrm{O}\), with a half-life of 0.5 minutes, the calculations after 2 minutes show significant decay via four half-lives.
The decay formula's elegance lies in its ability to model exponential decay, making it an essential tool in sciences that deal with radioactivity. It helps predict decay rates accurately, which is vital for various practical applications such as nuclear medicine and radioactive dating.
Isotope Calculation
Calculating isotopic decay and ratios requires a firm grip on the half-life and decay formula concepts. In the given exercise, equal initial amounts of \(_{8}^{15} \mathrm{O}\) and \(_{8}^{19} \mathrm{O}\) are used. Over time, as these isotopes decay, their remains can be calculated using the decay formula.
For \(_{8}^{15} \mathrm{O}\), after 2 minutes, it decays through one half-life, leaving half its initial amount. Meanwhile, \(_{8}^{19} \mathrm{O}\) undergoes four half-lives in the same period, leaving just \( \frac{1}{16} \) of its initial amount. The ratio between them at this time is 8:1.
When extended to 10 minutes, calculations show \(_{8}^{15} \mathrm{O}\) reduces through five half-lives to \( \frac{1}{32} \) of its initial amount. \(_{8}^{19} \mathrm{O}\) undergoes twenty half-lives, shrinking to \( \frac{1}{1048576} \). The resulting dramatic ratio is 32768:1.
Accurate isotope calculation helps us discern how different isotopes behave under identical timeframes, guiding important decisions in fields such as radiotherapy and archaeology.

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Most popular questions from this chapter

Which of the following reactions obey the conservation of baryon number? (a) \(\mathrm{p}+\mathrm{p} \rightarrow \mathrm{p}+\mathrm{e}^{+},(\mathrm{b}) \mathrm{p}+\mathrm{n} \rightarrow 2 \mathrm{e}^{+}+\mathrm{e}^{-}\)(c) \(\mathrm{p} \rightarrow \mathrm{n}+\mathrm{e}^{-}+\overline{\nu}_{\mathrm{c}},(\mathrm{d}) \mathrm{p}+\overline{\mathrm{p}} \rightarrow 2 \gamma\)

Comparison of energy released per gram of fuel. (a) When gasoline is burned, it releases \(1.3 \times 10^{8}\) J per gallon \((3.788\) L) of energy. Given that the density of gasoline is \(737 \mathrm{kg} / \mathrm{m}^{3},\) express the quantity of energy released in \(\mathrm{J} / \mathrm{g}\) of fuel. (b) During fission, when a neutron is absorbed by a \(^{235} \mathrm{U}\) nucleus, about 200 \(\mathrm{MeV}\) of energy is released for each nucleus that undergoes fission. Express this quantity in \(\mathrm{J} / \mathrm{g}\) of fuel. (c) In the proton-proton chain that takes place in stars like our sun, the overall fusion reaction can be summarized as six protons fusing to form one 4 He nucleus with two leftover protons and the liberation of 26.7 \(\mathrm{MeV}\) of energy. The fuel is the six protons. Express the energy produced here in units of J/g of fuel. Notice the huge difference between the two forms of nuclear energy, on the one hand, and the chemical energy from gasoline, on the other. (d) Our sun produces energy at a meas- ured rate of \(3.92 \times 10^{26} \mathrm{W}\) . If its mass of \(1.99 \times 10^{30} \mathrm{kg}\) were all gasoline, how long could it last before consuming all its fuel? (Historical note: Before the discovery of nuclear fusion and the vast amounts of energy it releases, scientists were confused. They knew that the earth was at least many millions of years old, but could not explain how the sun could survive that long if its energy came from chemical burning.)

Density of the nucleus. (a) Using the empirical formula for the radius of a nucleus, show that the volume of a nucleus is directly proportional to its nucleon number \(A .\) (b) Give a reasonable argument concluding that the mass \(m\) of a nucleus of nucleon number \(A\) is approximately \(m=m_{\mathrm{p}} A,\) where \(m_{\mathrm{p}}\) is the mass of a proton. (c) Use the results of parts (a) and (b) to show that all nuclei should have about the same density. Then calculate this density in \(\mathrm{kg} / \mathrm{m}^{3},\) and compare it with the density of lead (which is 11.4 \(\mathrm{g} / \mathrm{cm}^{3} )\) and a neutron star (about \(10^{17} \mathrm{kg} / \mathrm{m}^{3} ) .\)

(a) Calculate the total binding energy (in MeV) of the nuclei of \(^{56} \mathrm{Fe}\) (of atomic mass 55.934937 \(\mathrm{u} )\) and of \(^{207} \mathrm{Pb}\) (of atomic mass 206.975897 u). (b) Calculate the binding energy per nucleon for each of these atoms.(c) How much energy would be needed to totally take apart each of these nuclei? (d) For which of these atoms are the individual nucleons more tightly bound? Explain your reasoning.

Thorium series. The following decays make up the thorium decay series (the \(X\) 's are unknowns for you to identify): $$^{232} \mathrm{Th} \stackrel{\alpha}{\longrightarrow} X_{1}, \quad^{228} \mathrm{Ra} \stackrel{\beta^{-}}{\longrightarrow}^{228} \mathrm{Ac}, \quad X_{2} \stackrel{\beta^{-}}{\longrightarrow}^{228} \mathrm{Th}$$ $$^{228} \mathrm{Th} \stackrel{x_{3}}{\longrightarrow}^{224} \mathrm{Ra}, \quad^{224} \mathrm{Ra} \stackrel{\alpha}{\longrightarrow}^{220} \mathrm{Rn}, \quad^{220} \mathrm{Rn} \stackrel{\alpha}{\longrightarrow} X_{4}$$ \(X_{5} \stackrel{\alpha}{\longrightarrow}^{212} \mathrm{Pb},\) and \(^{212 \mathrm{Pb}} \stackrel{x_{6}}{\longrightarrow}^{212} \mathrm{Bi} .\) The \(^{212} \mathrm{Bi}\) then decays by an \(\alpha\) decay and a \(\beta^{-}\) decay, which can occur in either order \((\alpha\) followed by \(\beta\) or \(\beta\) followed by \(\alpha)\) . (a) Identify each of the six unknowns \((X_{1}, X_{2},\) etc. \()\) by nucleon number, atomic number, neutron number, and name. (b) Write out the decays of \(^{212} \mathrm{Bi}\) and indicate the end product of this series. (For some guidance, see the discussion under "Decay Series" in Section \(30.3 . )(\mathrm{c})\) Draw a Segre chart for the thorium series, similar to the one shown in Figure \(30.5 .\)

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