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Chapter 43: Problem 13

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

Short Answer

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

Step by step solution

01

Understanding Alpha Decay

In an alpha (\(\alpha\)) decay, the nucleus emits an alpha particle, which consists of 2 protons and 2 neutrons (a helium-4 nucleus). This means the new atom will have an atomic number reduced by 2 and a mass number reduced by 4.
02

Calculating Alpha Decay of Pu-239

For the alpha decay of Pu-239, the atomic number decreases from 94 to 92 (since 2 protons are lost), and the mass number decreases from 239 to 235 (since 4 nucleons in total are lost). Thus, the resulting nuclide is \(^{235}_{92}\mathrm{U}\).
03

Understanding Beta-minus Decay

In a beta-minus (\(\beta^{-}\)) decay, a neutron is converted into a proton while emitting an electron (\(e^{-}\)) and an antineutrino. This results in an increase of the atomic number by 1, while the mass number remains unchanged.
04

Calculating Beta-minus Decay of Na-24

During the \(\beta^{-}\) decay of Na-24, its atomic number increases from 11 to 12 (turning it into magnesium) while the mass number stays the same. The resulting nuclide is \(^{24}_{12}\mathrm{Mg}\).
05

Understanding Beta-plus Decay

In a beta-plus (\(\beta^{+}\)) decay, a proton is converted into a neutron while emitting a positron (\(e^{+}\)) and a neutrino. This results in a decrease of the atomic number by 1, while the mass number remains unchanged.
06

Calculating Beta-plus Decay of O-15

For the \(\beta^{+}\) decay of O-15, the atomic number decreases from 8 to 7 (turning it into nitrogen) while the mass number remains 15. The resulting nuclide is \(^{15}_{7}\mathrm{N}\).

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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 type of nuclear decay where the atom's nucleus releases an alpha particle. An alpha particle contains 2 protons and 2 neutrons, similar to a helium-4 nucleus. When an atom undergoes alpha decay, its atomic number decreases by 2 since it loses 2 protons. Additionally, the mass number decreases by 4 because the atom loses 2 neutrons and 2 protons, totaling 4 nucleons.
  • Alpha particles are relatively large and can be stopped by a sheet of paper.
  • This form of decay is common among heavy elements such as uranium and radium.
For example, in the alpha decay of plutonium-239 ( Pu-239):
  • The atomic number reduces from 94 (Pu) to 92, resulting in uranium (U).
  • The mass number decreases from 239 to 235.
Thus, the resulting nuclide is uranium-235 ( U-235). This process happens because the nucleus seeks a more stable configuration by releasing excess protons and neutrons.
Beta-Minus Decay
Beta-minus decay is a process that involves the conversion of a neutron into a proton, with the emission of an electron (known as beta particle) and an antineutrino. The exciting part is that the atomic number of the atom increases by one because a new proton is formed in place of the neutron, while the mass number remains unchanged.
  • This decay can penetrate further than alpha decay and requires a thin layer of aluminum to stop it.
  • Beta-minus decay is commonly observed in nuclides with an excess of neutrons.
During the beta-minus decay of sodium-24 ( Na-24):
  • The atomic number increases from 11 to 12, thus converting sodium (Na) into magnesium (Mg).
  • The mass number remains the same at 24.
Consequently, the resulting nuclide is magnesium-24 ( Mg-24). This transformation helps balance the ratio of protons to neutrons, thereby stabilizing the atom.
Beta-Plus Decay
Beta-plus decay occurs when a proton is converted into a neutron, emitting a positron (the electron's antimatter counterpart) and a neutrino in the process. This transformation results in the decrease of the atomic number by one, while, similar to beta-minus decay, the mass number doesn't change.
  • Beta-plus decay is typical in isotopes with an excess of protons.
  • This type of decay is less penetrating than beta-minus decay and is effectively blocked by a plastic or glass container.
Take the example of oxygen-15 ( O-15) undergoing beta-plus decay:
  • The atomic number decreases from 8 to 7, transforming oxygen (O) into nitrogen (N).
  • The mass number stays constant at 15.
Thus, the end product is nitrogen-15 ( N-15). This decay pathway assists the nucleus in attaining a more stable energy state by adjusting its proton-to-neutron ratio.

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Most popular questions from this chapter

The atomic mass of \(_{12}^{25} \mathrm{Mg}\) is 24.985837 \(\mathrm{u},\) and the atomic mass of \(^{25}_{13}\) Al is 24.990429 u. (a) Which of these nuclei will decay into the other? (b) What type of decay will occur? Explain how you determined this. (c) How much energy (in MeV) is released in the decay?

\(\mathrm{A}^{60} \mathrm{Co}\) 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 RBE for these \(\gamma\) rays is 0.70\(?\) (d) What exposure time is required for an equivalent dose of 200 rem?

BIO 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 \(^{90}\) Sr decay? (b) What percentage of the original level of \(^{90}\) 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?

(a) Prove that when a particle with mass \(m\) and kinetic energy \(K\) collides with a stationary particle with mass \(M,\) the total kinetic energy \(K_{\mathrm{cm}}\) in the center-of-mass coordinate system (the energy available to cause reactions) is $$K_{\mathrm{cm}}=\frac{M}{M+m} K$$ Assume that the kinetic energies of the particles and nuclei are much lower than their rest energies. (b) If \(K_{\text { th }}\) is the minimum, or threshold, kinetic energy to cause an endoergic reaction to occur in the situation of part (a), show that $$K_{\mathrm{th}}=-\frac{M+m}{M} Q$$

Use conservation of mass-energy to show that the energy released in \(\beta^{+}\) decay is positive whenever the neutral atomic mass of the original atom is at least two electron masses greater than that of the final atom. (See the hint in Problem \(43.49 .\))
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