What is the nuclear configuration of the daughter nucleus associated with the alpha decay of \(\mathrm{Hf}(A=157,\) \(Z=72\) )?

Short Answer

Expert verified
Answer: The nuclear configuration of the daughter nucleus after the alpha decay of Hf(A=157, Z=72) is Yb(A=153, Z=70).

Step by step solution

01

Understand the alpha decay process

In alpha decay, an alpha particle (\(^4_2\mathrm{He}\)) is emitted from the parent nucleus. An alpha particle has 2 protons and 2 neutrons.
02

Calculate the change in the atomic mass number (A) and atomic number (Z) during alpha decay

Since an alpha particle has 4 nucleons (A=4), the change in the atomic mass number will be: ΔA = 4 Similarly, since an alpha particle has 2 protons (Z=2), the change in atomic number (Z) will be: ΔZ = 2
03

Find the nuclear configuration of the daughter nucleus

After the alpha decay, the atomic mass number (A) and atomic number (Z) of the daughter nucleus can be found by subtracting the change in A and Z from the parent nucleus: Daughter nucleus A = Parent nucleus A - ΔA Daughter nucleus A = 157 - 4 Daughter nucleus A = 153 Daughter nucleus Z = Parent nucleus Z - ΔZ Daughter nucleus Z = 72 - 2 Daughter nucleus Z = 70
04

Identify the daughter nucleus element

Since the atomic number (Z) of the daughter nucleus is 70, we can look up the element with Z=70 in the periodic table. The element with atomic number 70 is Ytterbium (Yb).
05

Write the nuclear configuration of the daughter nucleus

The daughter nucleus has an atomic mass number (A) of 153 and an atomic number (Z) of 70. Therefore, it can be represented as: \(\mathrm{Yb}(A=153, Z=70)\) Hence, the nuclear configuration of the daughter nucleus associated with the alpha decay of Hafnium is \(\mathrm{Yb}(A=153, Z=70)\).

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Key Concepts

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

Nuclear Configuration
Understanding the nuclear configuration is crucial when discussing alpha decay. The nucleus of an atom consists of protons and neutrons, collectively known as nucleons. During alpha decay, a nucleus emits an alpha particle, which is essentially a helium-4 nucleus, composed of 2 protons and 2 neutrons. This emission causes the original nucleus to lose a total of 4 nucleons, altering both its atomic mass and atomic number, and resulting in a different nuclear configuration for the daughter nucleus. Therefore, the nuclear configuration refers to the specific arrangement and number of protons and neutrons in an atomic nucleus.

Nuclear configuration plays a pivotal role in the stability of an atom. Atoms aim for a balanced state, and alpha decay is one way a nucleus can move towards stability if it has an excess of protons or a high mass. By examining alpha decay, we get a glimpse into the natural processes helping to shape the elements around us.
Atomic Mass Number
The atomic mass number, denoted as 'A,' is a fundamental concept in understanding alpha decay. It represents the total number of nucleons (protons plus neutrons) in an atom's nucleus. An alpha particle has an atomic mass number of 4, which means that when an atom undergoes alpha decay, it will lose 4 from its atomic mass number. This reduction in 'A' is due to the ejection of an alpha particle.

For instance, in the provided exercise, the element Hafnium, which starts with an atomic mass number of 157, loses an alpha particle leading to a new atomic mass number of 153 for the daughter nucleus. It's important to note that, while the total mass of nucleons changes, the underlying mechanism doesn't alter the individual masses of the remaining protons and neutrons in the nucleus.
Atomic Number
The atomic number, represented by 'Z,' indicates the number of protons in an atom's nucleus, which directly defines the chemical element. Unlike neutrons, protons are electrically charged, and their number determines the atom's identity within the periodic table. In alpha decay, the atomic number decreases by 2 because the alpha particle contains 2 protons that are emitted from the nucleus.

In our exercise, Hafnium with an original atomic number of 72 undergoes alpha decay resulting in a daughter nucleus with an atomic number of 70. This change is significant as it transforms the element into a completely different one as listed in the periodic table, from Hafnium to Ytterbium.
Nucleons
Nucleons are the building blocks of an atomic nucleus and include both protons and neutrons. The number of nucleons within a nucleus bears a direct effect on the atom's mass and stability. Alpha decay is a process that influences nucleons by reducing their count, specifically by releasing an alpha particle composed of 2 protons and 2 neutrons.

The concept of nucleons is central to the topic of alpha decay, as the loss of 4 nucleons during the decay alters the nuclear structure and stability of the original atom. After this emission, the resulting daughter nucleus has fewer nucleons, which often leads to a more stable nuclear configuration.
Periodic Table
The periodic table is an organized chart of elements arranged by increasing atomic number. It is a fundamental tool in chemistry, that helps predict the properties of elements and their compounds. Each element on the periodic table is represented by its atomic number, which corresponds to the number of protons in its nucleus.

When studying alpha decay, the periodic table allows us to identify the new element formed after the process. Following the alpha decay of Hafnium in our exercise, we refer to the periodic table to find that the daughter nucleus with an atomic number of 70 is Ytterbium (Yb). Without the periodic table, understanding and identifying the outcomes of nuclear reactions like alpha decay would be much more complex.

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Most popular questions from this chapter

Using the table of isotopes in Appendix B, calculate the binding energies of the following nuclei. a) \({ }^{7} \mathrm{Li}\) b) \({ }^{12} \mathrm{C}\) c) \({ }^{56} \mathrm{Fe}\) d) \({ }^{85} \mathrm{Rb}\)

You have developed a grand unified theory which predicts the following things about the decay of the proton: (1) protons never get any older, in the sense that their probability of decay per unit time never changes, and (2) half the protons in any given collection of protons will have decayed in \(1.80 \cdot 10^{29}\) yr. You are given experimental facilities to test your theory: A tank containing \(1.00 \cdot 10^{4}\) tons of water and sensors to record proton decays. You will be allowed access to this facility for two years. How many proton decays will occur in this period if your theory is correct?

The most common isotope of uranium, \({ }_{92}^{238} \mathrm{U},\) produces radon \({ }_{86}^{222} \mathrm{Rn}\) through the following sequence of decays: $$\begin{array}{c}{ }^{238} \mathrm{U} \rightarrow{ }^{234} \mathrm{Th}+\alpha,{ }^{234} \mathrm{Th} \rightarrow{ }^{234} \mathrm{~Pa}+\beta^{-}+\bar{\nu}_{e}, \\\\{ }_{91}^{234} \mathrm{~Pa} \rightarrow{ }_{92}^{234} \mathrm{U}+\beta+\bar{\nu}_{e},{ }^{234} \mathrm{U} \rightarrow{ }^{230} \mathrm{Th}+\alpha ,\\\\{ }_{91}^{230} \mathrm{Th} \rightarrow{ }_{90}^{226} \mathrm{Ra}+\alpha,{ }_{88}^{226} \mathrm{Ra} \rightarrow{ }_{86}^{222} \mathrm{Rn}+\alpha,\end{array}$$. A sample of \({ }_{92}^{238} \mathrm{U}\) will build up equilibrium concentrations of its daughter nuclei down to \({ }_{88}^{226} \mathrm{Ra} ;\) the concentrations of each are such that each daughter is produced as fast as it decays. The \({ }_{88}^{226} \mathrm{Ra}\) decays to \({ }_{86}^{222} \mathrm{Rn},\) which escapes as a gas. (The \(\alpha\) particles also escape, as helium; this is a source of much of the helium found on Earth.) In high concentrations, the radon is a health hazard in buildings built on soil or foundations containing uranium ores, as it can be inhaled. a) Look up the necessary data, and calculate the rate at which \(1.00 \mathrm{~kg}\) of an equilibrium mixture of \({ }_{92}^{238} \mathrm{U}\) and its first five daughters produces \({ }_{86}^{222} \mathrm{Rn}\) (mass per unit time). b) What activity (in curies per unit time) of radon does this represent?

Consider the Bethe-Weizsäcker 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) 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?

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