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What nuclide is produced in the following radioactive decays? (a) \(\alpha\) decay of \({ }_{94}^{239} \mathrm{Pu} ;\) (b) \(\beta^{-}\)decay of \(_{11}^{24} \mathrm{Na} ;\) (c) \(\beta^{+}\)decay of \({ }_{8}^{15} \mathrm{O}\).

Short Answer

Expert verified
(a) \\({}_{92}^{235}\\mathrm{U}\\); (b) \\({}_{12}^{24}\\mathrm{Mg}\\); (c) \\({}_{7}^{15}\\mathrm{N}\\).

Step by step solution

01

Understand Alpha Decay

In alpha decay, the original nucleus emits an alpha particle, which consists of 2 protons and 2 neutrons. This results in a new element with atomic number reduced by 2 and mass number reduced by 4.
02

Apply Alpha Decay to \\({}_{94}^{239}\\mathrm{Pu}\\)

For \(^{239} \mathrm{Pu}\), applying alpha decay results in a mass number of \(239 - 4 = 235\) and an atomic number of \(94 - 2 = 92\), which is uranium, \(^{235}_{92} \mathrm{U}\).
03

Understand Beta Minus Decay

In beta minus (\(\beta^-\)) decay, a neutron in the nucleus is converted into a proton, emitting an electron (beta particle) and an antineutrino. The atomic number increases by 1, while the mass number remains unchanged.
04

Apply Beta Minus Decay to \\(_{11}^{24}\\mathrm{Na}\\)

The \(^{24} \mathrm{Na}\) undergoing \(\beta^-\) decay will have its atomic number increase from 11 to 12, resulting in \(^{24}_{12} \mathrm{Mg}\). The mass number remains 24.
05

Understand Beta Plus Decay

In beta plus (\(\beta^+\)) decay, a proton is converted into a neutron, emitting a positron and a neutrino. The atomic number decreases by 1, while the mass number remains unchanged.
06

Apply Beta Plus Decay to \\({}_{8}^{15}\\mathrm{O}\\)

For \(^{15} \mathrm{O}\), applying \(\beta^+\) decay results in the atomic number decreasing from 8 to 7, resulting in \(^{15}_{7} \mathrm{N}\). The mass number remains 15.

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Key Concepts

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

Alpha Decay
Alpha decay is a fascinating type of radioactive decay. During this process, an atomic nucleus ejects an alpha particle. What's an alpha particle, you ask? It's a tiny chunk consisting of 2 protons and 2 neutrons.
This means when a nucleus undergoes alpha decay,
  • its atomic number decreases by 2 (because it loses 2 protons),
  • and its mass number decreases by 4 (because it loses 2 protons and 2 neutrons, totaling 4).
After alpha decay, the original element transforms into a completely new one. An example is the decay of plutonium-239 (\({ }_{94}^{239} \mathrm{Pu}\)). By losing an alpha particle, it becomes uranium-235 (\(^{235}_{92} \mathrm{U}\)). Now the atom is no longer plutonium, but has turned into uranium!
Alpha decay is a common process for heavy elements, like plutonium, and it’s a crucial part of understanding nuclear stability.
Beta Minus Decay
Beta minus decay (\(\beta^-\)) is another intriguing type of radioactive decay. Here, a neutron within the nucleus is converted into a proton. This conversion involves the emission of an electron, which we commonly refer to as a beta particle, and an antineutrino.
As a result of beta minus decay,
  • the mass number of the element stays the same,
  • but the atomic number goes up by 1.
This increase in atomic number means the original element becomes a new one. A perfect example is sodium-24 (\({ }_{11}^{24} \mathrm{Na}\)), which, after emitting a beta particle, transforms into magnesium-24 (\({ }_{12}^{24} \mathrm{Mg}\)). The increase in the atomic number shows that a proton has been added!
Understanding beta minus decay is crucial, especially for those studying nuclear physics or chemistry. It demonstrates how elements can naturally transform, providing insight into the processes happening in stars and radioactive substances.
Beta Plus Decay
Beta plus decay (\(\beta^+\)) is when a proton in a nucleus turns into a neutron. During this process, a positron and a neutrino are emitted. You might be wondering, what is a positron? Simply put, it's the positively charged counterpart of an electron.
In beta plus decay,
  • the mass number stays unchanged,
  • while the atomic number decreases by 1.
This decay results in the original element turning into a different one. For instance, oxygen-15 (\({ }_{8}^{15} \mathrm{O}\)) becomes nitrogen-15 (\({ }_{7}^{15} \mathrm{N}\)) after a beta plus decay. Notice how the atomic number goes from 8 to 7? That's the proton turning into a neutron!
Beta plus decay is important in the study of particle physics and is used in medical applications like PET scans. It shows us the fascinating ways particles can transform, highlighting the dynamic nature of atomic structures.

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Most popular questions from this chapter

A thin uniform radioactive coating of an \(a\)-emitter is applied on a metallic sphere of radius ' \(a\) '. The sphere is connected to the ground by means of a thin wire of resistance ' \(R\) ' as shown in figure. Assume that, the sphere is uncharged at \(t=0\), the initial activity of \(a\) - emitter is \(A_{0}\) and the disintegration constant of radioactive material is \(\lambda=3 \sec ^{-1}\). Also assume that all the a-particles emitted go out of the sphere. Find the time (in sec) at which charge on the sphere is maximum. Use \(\ln 2=0.69\) and \(\frac{1}{4 \pi \varepsilon_{0} a}=2 R \lambda\).

Gold, \({ }_{79}^{198} \mathrm{Au}\), undergoes \(\beta^{-}\)decay to an excited state of \({ }_{80}^{198} \mathrm{Hg} .\) If the excited state decays by emission of a \(\gamma\) photon with energy \(0.412 \mathrm{MeV}\), what is the maximum kinetic energy of the electron emitted in the decay? This maximum occurs when the antineutrino has negligible energy. (The recoil energy of the \({ }_{80}^{198} \mathrm{Hg}\) nucleus can be ignored. The masses of the neutral atoms in their ground states are \(197.968225\) u for \({ }_{79}^{198} \mathrm{Au}\) and \(197.966752 \mathrm{u}\) for \(\left.{ }_{80}^{198} \mathrm{Hg} .\right)\)

Half lives of two isotopes \(X\) and \(Y\) of a material are known to be \(2 \times 10^{9}\) years and \(4 \times 10^{9}\) years respectively. If a planet was formed with equal number of these isotopes, estimate the current age of the planet, given that currently the material has \(20 \%\) of \(X\) and \(80 \%\) of \(Y\) by number.

A radioactive source in the form of a metal sphere of diameter \(3.2 \times 10^{-3} m\) emits \(\beta\)-particle at a constant rate of \(6.25 \times 10^{10}\) particle/sec. The source is electrically insulated and all the \(\beta\)-particle are emitted from the surface. The potential of the sphere will rise to \(1 V\) in time. Find the time in which potential of the sphere becomes \(1 V\).

Radioactive Fallout. One of the problems of in-air testing of nuclear weapons (or, even worse, the use of such weapons!) is the danger of radioactive fallout. One of the most problematic nuclides in such fallout is strontium- \(90\left({ }^{90} \mathrm{Sr}\right)\), which breaks down by \(\beta^{-}\)decay with a half-life of 28 years. It is chemically similar to calcium and therefore can be incorporated into bones and teeth, where, due to its rather long half-life, it remains for years as an internal source of radiation. (a) What is the daughter nucleus of the \({ }^{90} \mathrm{Sr}\) decay? (b) What percentage of the original level of \({ }^{90} \mathrm{Sr}\) is left after 56 years? (c) How long would you have to wait for the original level to be reduced to \(6.25 \%\) of its original value?

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