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Chapter 21: Problem 65

The radioactive isotope \({ }^{60} \mathrm{Co}_{27}\) which is used in the treatment of cancer can be made by (n, p) reaction. For this reaction, the target nucleus is (a) \({ }_{28} \mathrm{Ni}^{59}\) (b) \({ }_{27} \mathrm{Co}^{59}\) (c) \({ }_{28} \mathrm{Ni}^{60}\) (d) \({ }_{27} \mathrm{Co}^{60}\)

Short Answer

Expert verified
The target nucleus is \\( { }_{28} \text{Ni}^{59} \\).

Step by step solution

01

Understanding the (n, p) Reaction

In a (n, p) reaction, a neutron is captured by a nucleus and a proton is emitted. This changes the original nucleus into a new one by increasing the number of protons by one and keeping the mass number the same.
02

Identifying Target Nucleus for Reaction

In the reaction producing \({ }^{60} \text{Co}_{27}\), the neutron capture increases the atomic number of the target nucleus by 1 to become Cobalt (Co) from Nickel (Ni). Thus, the original target nucleus before the reaction must have 59 nucleons to match the resulting mass after the reaction.
03

Matching Atomic Numbers and Mass Numbers

Given that in a (n, p) reaction the atomic number increases by 1, then the target nucleus must have an atomic number of Z - 1, where Z is 27 for Cobalt. Therefore, the target must be Nickel (Ni) with an atomic number of 28 - 1 = 27. Additionally, the mass number remains 60 in the reaction, implying the target mass number must be one less than the outcome.
04

Verifying Options

The correct target nucleus is \({ }_{28} \text{Ni}^{59}\), which captures a neutron to form \({ }^{60} \text{Co}_{27}\), aligning with both atomic and mass number requirements described in previous steps.

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Key Concepts

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

Nuclear Reactions
Nuclear reactions are processes that change the composition of an atomic nucleus. These reactions can involve the collision of a nucleus with another particle (like a neutron or proton), leading to a change in the nuclear structure. Unlike chemical reactions, nuclear reactions can result in a change in the identity of elements and the release or absorption of a significant amount of energy.

Types of nuclear reactions include:
  • Fusion: Combining two lighter nuclei to form a more massive nucleus, releasing energy.
  • Fission: Splitting a heavy nucleus into lighter nuclei, often releasing energy in the process.
  • (n, p) Reactions: A special type of nuclear reaction where a neutron is captured, and a proton is emitted.
Each type of reaction involves changes at the nuclear level and can have various applications, such as energy production and medical treatments.
Atomic Nucleus
The atomic nucleus is the core of an atom, comprising protons and neutrons. These particles are collectively known as nucleons. The nucleus is surrounded by a cloud of electrons, which are involved in chemical reactions, but it is the properties of the nucleus that are crucial for nuclear reactions.

Key characteristics of nuclei include:
  • Atomic Number (Z): The number of protons in a nucleus, determining the element's identity. For example, Carbon has an atomic number of 6.
  • Mass Number (A): The total number of protons and neutrons. It roughly equals the atomic mass.
  • Isotopes: Variants of a given element with different neutron numbers but the same atomic number.
These properties dictate how a nucleus behaves in reactions, including stability and the potential for radioactive decay.
Neutron Capture
Neutron capture occurs when a neutron is absorbed by the nucleus of an atom. This process increases the mass number by 1 without changing the element, unless accompanied by another nuclear event, like proton emission. In neutron capture reactions, the mass of the nucleus is altered, but not its charge until further transformation.

Important points about neutron capture include:
  • Typically happens in elements with a high atomic number and can occur naturally or be induced by neutron flux in nuclear reactors.
  • This process is a key part of the (n, p) reaction, changing the atomic structure by creating a short-lived isotope that may undergo decay.
  • Neutron capture plays a crucial role in the creation of heavier elements within stars, through slow (s-process) and rapid (r-process) nucleosynthesis.
Understanding neutron capture is essential for manipulating nuclear reactions and understanding astrophysical phenomena.
Proton Emission
Proton emission is a type of radioactive decay where a proton is ejected from the nucleus of an atom. This process decreases the atomic number by one and involves a change of the element itself, as a new nucleus with fewer protons is formed. Proton emission is relatively rare compared to other nuclear reactions like alpha and beta decay.

Key aspects of proton emission include:
  • Occurs in proton-rich nuclei, often with more protons than neutrons, leading to instability.
  • As a result of this reaction, the element is transformed into another element. For example, Carbon could become Boron after emitting a proton.
  • This process is used in specialized radiation therapy treatments and contributes to our understanding of nuclear stability and isotopic chains.
Proton emission provides insights into the forces that hold nuclei together and the limits of atomic stability, offering valuable information for both theoretical and applied nuclear physics.

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Most popular questions from this chapter

\({ }_{92} \mathrm{U}^{238}\) is radioactive and it emits \(\alpha\) and \(\beta\) particles to form \({ }_{82} \mathrm{~Pb}^{206} .\) The number of \(\alpha\) and \(\beta\) particles are respectively (a) 8,6 (b) 6,8 (c) 4,2 (d) 8,4

Lead is the final product formed by a series of changes in which the rate determining stage is the radioactive decay of uranium-238. This radioactive decay is a first order reaction with a half-life of \(4.5 \times 10^{9}\) years. What would be the age of a rock sample originally lead free, in which the molar proportion of uranium to lead is now \(1: 3\) ? (a) \(1.5 \times 10^{9}\) years (b) \(2.25 \times 10^{9}\) years (c) \(4.5 \times 10^{9}\) years (d) \(9.0 \times 10^{9}\) years

Which of the following is easily stopped by air? (a) uv rays (b) X-rays (c) \(\alpha\) rays (d) \(\gamma\) rays

The binding energy of an element is \(64 \mathrm{meV}\). If \(\mathrm{BE} /\) Nucleon is \(6.4\), the number of nucleons are (a) 10 (b) 64 (c) 16 (d) \(6.4\)

Assuming the age of the earth to be \(10^{10}\) years, what fraction of the original amount of \(_{92} \mathrm{U}^{238}\) is still in existence on earth \(\left(\mathrm{t}_{1 / 2}\right.\) of \(_{92} \mathrm{U}^{238}=4.51 \times 10^{9}\) years \() ?\) (a) \(10 \%\) (b) \(20 \%\) (c) \(30 \%\) (d) \(40 \%\)
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