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Nitrogen-13, with a half-life of \(10.0 \mathrm{~min}\), decays by beta emission. (a) Write down the decay equation to determine the daughter product and whether the beta particle is a positron or electron. (b) If a sample of pure \({ }^{13} \mathrm{~N}\) has a mass of \(1.50 \mathrm{~g}\) at a certain time, what is the activity 35.0 min later? (c) What percentage of the sample is \({ }^{13} \mathrm{~N}\) at this time? (d) What alternative process could have happened to the nitrogen-13? Write down its decay equation and determine the daughter product for this process.

Short Answer

Expert verified
(a) \( ^{13}N \rightarrow ^{13}C + e^+ + \nu_e \). (b) Use \( A = A_0 e^{-\lambda t} \). (c) Calculate \( N/N_0 \) and convert to %. (d) Electron capture: \( ^{13}N + e^- \rightarrow ^{13}C + \nu_e \).

Step by step solution

01

Write the Beta Decay Equation for Nitrogen-13

Nitrogen-13 decays by beta-plus decay, which means it transforms by emitting a positron. The beta decay equation for \( ^{13}N \) can be written as \( ^{13}N \rightarrow ^{13}C + e^+ + u_e \). Here, \( e^+ \) is a positron and \( u_e \) is a neutrino. The daughter product is Carbon-13.
02

Calculate Activity 35 Minutes Later

Activity (\(A\)) can be calculated using the formula \( A = A_0 e^{-\lambda t} \). The decay constant \( \lambda \) is calculated by \( \lambda = \frac{\ln(2)}{t_{1/2}} \), where \( t_{1/2} = 10 \) minutes. Calculate \( \lambda \), then substitute \( \lambda \) and \( t = 35 \) minutes into the activity formula to get the final activity.
03

Determine Remaining Percentage of Nitrogen-13

The remaining percentage of \(^{13}N\) can be found using the formula \( N = N_0 e^{-\lambda t} \). First calculate \( N/N_0 \), which represents the fraction of \(^{13}N\) left. Then convert it into a percentage.
04

Identify Alternative Decay Process for Nitrogen-13

An alternative decay process for \(^{13}N\) is electron capture. In this process, one proton in the nucleus captures an electron, transforming into a neutron. The reaction is \( ^{13}N + e^- \rightarrow ^{13}C + u_e \), with Carbon-13 as the daughter product.

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Key Concepts

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

Nitrogen-13
Nitrogen-13 is a radioactive isotope of nitrogen. It is denoted as \(^ {13}N\), where the number 13 represents its atomic mass, which is the sum of its protons and neutrons. This isotope contains 7 protons and 6 neutrons.
Nitrogen-13 is commonly used in positron emission tomography (PET) scans in the medical field due to its short half-life and the emission of positrons. This makes it ideal for imaging and diagnosing medical conditions.
It typically originates from the bombardment of natural nitrogen with high-energy particles such as protons. This process is often performed in a cyclotron, a type of particle accelerator.
Half-life
The half-life of Nitrogen-13 is 10 minutes. This means that every 10 minutes, half of the atoms in a sample of \(^ {13}N\) will decay into more stable products.
The concept of a half-life is crucial in understanding radioactive decay as it describes the rate at which instability reduces.
  • In 10 minutes, 50% of the original sample remains.
  • In 20 minutes, 25% of the original sample is left.
  • In 30 minutes, 12.5% remains, and so on.
This rapid decay makes Nitrogen-13 particularly useful in medical diagnostics, allowing healthy tissue to absorb radioactive isotopes quickly, without prolonged exposure.
Positron Emission
Positron emission is a type of beta decay where a proton inside the nucleus transforms into a neutron, emitting a positron and a neutrino. For Nitrogen-13, this process can be written as: \(^ {13}N \rightarrow ^ {13}C + e^+ + u_e \)
Here, \( e^+ \) represents the positron, which is essentially an anti-electron with the same mass as an electron but a positive charge.
During this process:
  • The atomic number decreases by 1, turning the nitrogen into carbon.
  • The mass number remains unchanged at 13.
  • The emitted positron (along with the neutrino) travels a short distance before it encounters an electron, leading to annihilation and the release of gamma rays, which can be detected externally.
Such outcomes are what make positron emission so valuable in PET scans for mapping functional processes in the body.
Radioactive Decay
Radioactive decay is a natural process by which unstable atomic nuclei lose energy by emitting radiation. For isotopes like Nitrogen-13, this decay occurs via positron emission, transforming them into more stable elements like Carbon-13.
This transformation involves the emission of particles and energy. Understanding radioactive decay helps predict the behavior of radioactive isotopes over time.
Besides positron emission, Nitrogen-13 can also decay through a process known as electron capture, described by the reaction: \(^ {13}N + e^- \rightarrow ^{13}C + u_e \).
  • In electron capture, a proton captures an electron and converts to a neutron.
  • This also results in the formation of Carbon-13 as the daughter product.
The study of such decays provides insights into nuclear reactions and the stability of elements, assisting in various scientific and medical fields.

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Most popular questions from this chapter

Prove that the number \(N\) of radioactive nuclei remaining in a sample after an integer number \((n)\) of half-lives has elapsed is \(N=\frac{N_{o}}{2^{n}}=\left(\frac{1}{2}\right)^{n} N_{0} .\) Here \(N_{0}\) stands for the initial number of nuclei.

A cancer treatment called the gamma knife (see Insight 29.1, Biological and Medical Application of Radiation) uses focused \({ }^{60}\) Co sources to treat tumors. Each \({ }^{60}\) Co nucleus emits two gamma rays, of energy \(1.33 \mathrm{MeV}\) and \(1.17 \mathrm{MeV}\), in quick succession. Assume that \(50.0 \%\) of the total gamma-ray energy is absorbed by a tumor. Further assume that the total activity of the \({ }^{60}\) Co sources is \(1.00 \mathrm{mCi}\), the tumor's mass is \(0.100 \mathrm{~kg}\), and the patient is exposed to the gamma radiation for an hour. Determine the effective radiation dose received by the tumor. (Since the \({ }^{60}\) Co half-life is 5.3 years, changes in its activity during treatment are negligible.)

Neutron activation analysis was performed on small pieces of hair that had been taken from the exiled Napoleon after he died on the island of St. Helena in 1821\. This procedure involves exposing the samples to a source of neutrons. Some (stable) arsenic nuclei, if present in the sample, will absorb a neutron. In Napoleon's case the samples did contain abnormally high levels of arsenic, which supported the theory that his death was not due to natural causes. (a) These results came from studying beta emissions of the resulting \({ }^{76} \mathrm{As},\) nucleus. Write the nuclear equation for the neutron absorption and use it to determine the arsenic isotope initially present in the hair. (b) Write the nuclear equation for the subsequent beta decay of \({ }^{76} \mathrm{As}\). Use it to determine the nucleus after this decay.

A particular radioactive sample undergoes \(2.50 \times 10^{6}\) decays/s. What is the activity of the sample in (a) curies and (b) becquerels?

Polonium-214 can decay by alpha decay. (a) The product of its decay has how many fewer protons than polonium- 214:(1) zero, (2) one, (3) two, or (4) four? (b) Write the nuclear equation for this decay and determine the daughter nucleus.

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