/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 Cobalt has a stable isotope, \({... [FREE SOLUTION] | 91影视

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Chapter 40: Problem 6

Cobalt has a stable isotope, \({ }^{59} \mathrm{Co},\) and 22 radioactive isotopes. The most stable radioactive isotope is \({ }^{60} \mathrm{Co} .\) What is the dominant decay mode of this \({ }^{60}\) Co isotope? a) \(\beta^{+}\) b) \(\beta^{-}\) c) \(\mathrm{P}\) d) \(n\)

Short Answer

Expert verified
Answer: (b) \(\beta^{-}\)

Step by step solution

01

Determine the neutron and proton numbers for the isotopes

To determine the neutron and proton numbers for both \({ }^{59} \mathrm{Co}\) and \({ }^{60} \mathrm{Co}\), we will use the general notation for isotopes, \({ }^A_Z \mathrm{X}\), where A is the atomic mass number (total protons and neutrons), Z is the atomic number (number of protons), and X is the chemical element symbol. For cobalt (Co), the atomic number (Z) is 27. For \({ }^{59} \mathrm{Co}\): - Atomic mass number (A) = 59 - Number of protons (Z) = 27 - Number of neutrons (N) = A - Z = 59 - 27 = 32 For \({ }^{60} \mathrm{Co}\): - Atomic mass number (A) = 60 - Number of protons (Z) = 27 - Number of neutrons (N) = A - Z = 60 - 27 = 33
02

Calculate the neutron-to-proton ratios

Calculate the neutron-to-proton ratio (n/p ratio) for both \({ }^{59} \mathrm{Co}\) and \({ }^{60} \mathrm{Co}\) using the neutron and proton numbers found in Step 1. For \({ }^{59} \mathrm{Co}\): - n/p ratio = 32/27 鈮� 1.185 For \({ }^{60} \mathrm{Co}\): - n/p ratio = 33/27 鈮� 1.222
03

Determine the stability of \({ }^{60} \mathrm{Co}\)

Since the n/p ratio of \({ }^{60} \mathrm{Co}\) (1.222) is greater than that of \({ }^{59} \mathrm{Co}\) (1.185), indicating the neutron number is relatively higher, \({ }^{60} \mathrm{Co}\) is likely to move towards a lower n/p ratio through beta decay to increase stability.
04

Identify the dominant decay mode

To decrease the number of neutrons and increase stability, a neutron in the nucleus of \({ }^{60} \mathrm{Co}\) will convert into a proton through beta decay. When a neutron converts into a proton, it releases an electron called a beta particle, which is denoted as \(\beta^-\). Hence, the dominant decay mode of \({ }^{60} \mathrm{Co}\) is the following: - \(\beta^-\) decay Answer: (b) \(\beta^{-}\)

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Key Concepts

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

Beta Decay
Beta decay is a type of radioactive decay where an unstable atomic nucleus transforms into a more stable one by emitting a beta particle and an antineutrino (for beta-minus decay) or a positron and a neutrino (for beta-plus decay). The beta-minus decay occurs when a neutron in the nucleus is converted into a proton, an electron (the beta particle), and an antineutrino. Conversely, beta-plus decay is the process where a proton becomes a neutron, a positron, and a neutrino.

A common example, as seen in the exercise with cobalt-60, ) beta-minus decay is the prominent mode. This is evident because cobalt-60 has more neutrons than the stable isotope cobalt-59, indicating the need to convert a neutron into a proton to achieve greater stability.
Neutron-to-Proton Ratio
The neutron-to-proton ratio, often abbreviated as n/p ratio, is a crucial factor in understanding the stability of an isotope. Atoms with a n/p ratio that is too high or too low compared to the stable range for their size are likely to undergo radioactive decay to reach a more stable state. For light elements, a n/p ratio close to 1 is typically stable, but as atoms get heavier, they require a higher ratio to maintain nuclear stability.

In the given case of cobalt isotopes, ), the exercise illustrates that cobalt-60 has a slightly higher n/p ratio than cobalt-59, which is stable. This elevated n/p ratio suggests that cobalt-60 is more likely to undergo beta decay to reduce the excess of neutrons and adjust the ratio in favor of stability.
Nuclear Stability
Nuclear stability is determined by several factors, with the n/p ratio being one of the most significant. A stable nucleus has a balanced force between the attractive nuclear force and the repulsive electromagnetic force amongst protons. If there are too many or too few neutrons relative to protons, the nucleus will seek stability through various modes of decay, like alpha, beta, or gamma decay. The band of stability, a graph plotting stable nuclei, helps scientists predict the type of decay an unstable isotope may undergo.

In our example, the high n/p ratio of cobalt-60 renders it unstable. It seeks greater stability by undergoing beta-minus decay, reducing its overall n/p ratio. This shift towards a more favorable balance between neutrons and protons is a natural pursuit of stability by the atomic nucleus.

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Most popular questions from this chapter

The binding energy of \({ }_{2}^{3}\) He is lower than that of \({ }_{1}^{3} \mathrm{H} .\) Provide a plausible explanation, considering the Coulomb interaction between two protons in \({ }_{2}^{3}\) He.

Write down equations to describe the \(\beta^{-}\) -decay of the following atoms: a) \({ }^{60} \mathrm{Co}\) b) \({ }^{3} \mathrm{H}\) c) \({ }^{14} \mathrm{C}\)

a) What is the energy released in the fusion reaction \({ }_{1}^{2} \mathrm{H}+{ }_{1}^{2} \mathrm{H} \rightarrow{ }_{2}^{4} \mathrm{He}+Q ?\) b) The oceans have a total mass of water of \(1.50 \cdot 10^{16} \mathrm{~kg}\), and \(0.0300 \%\) of this quantity is deuterium, \({ }_{1}^{2} \mathrm{H} .\) If all the deuterium in the oceans were fused by controlled fusion into \({ }_{2}^{4} \mathrm{He},\) how many joules of energy would be released? c) World power consumption is about \(1.00 \cdot 10^{13} \mathrm{~W}\). If consumption were to stay constant and all problems arising from ocean water consumption (including those of political, meteorological, and ecological nature) could be avoided, how many years would the energy calculated in part (b) last?

Consider the Bethe-Weizs盲cker formula for the case of odd \(A\) nuclei. Show that the formula can be written as a quadratic in \(Z\) -and thus, that for any given \(A\), the binding energies of the isotopes having that \(A\) take a quadratic form, \(B=a+b Z+c Z^{2} .\) Find the most deeply bound isotope (the most stable one) having \(A=117\) using your result.

A nuclear reaction of the kind \({ }_{2}^{3} \mathrm{He}+{ }_{6}^{12} \mathrm{C} \rightarrow \mathrm{X}+\alpha\) is called a pick-up nuclear reaction. a) Why is it called a pick-up reaction, that is, what is picked up, what picked it up, and where did it come from? b) What is the resulting nucleus X? c) What is the \(\mathrm{Q}\) -value of this reaction? d) Is this reaction endothermic or exothermic?
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