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Lead \(_{82}^{207} \mathrm{Pb}\) is a stable daughter nucleus that can result from either an \(\alpha\) decay or a \(\beta^{-}\) decay. Write the decay processes, including the chemical symbols and values for Z and A of the parent nuclei, for (a) the \(\alpha\) decay and \(\quad\) (b) the \(\beta^{-}\) decay.

Short Answer

Expert verified
(a) \(^{211}_{84}\mathrm{Po} \rightarrow {}^{207}_{82}\mathrm{Pb} + \alpha\); (b) \(^{207}_{81}\mathrm{Tl} \rightarrow {}^{207}_{82}\mathrm{Pb} + \beta^{-} + \overline{\nu}\).

Step by step solution

01

Understanding Alpha Decay

In alpha decay, a parent nucleus loses two protons and two neutrons by emitting an \(\alpha\) particle (\(^{4}_{2}\mathrm{He}\)). This causes the mass number \(A\) to decrease by 4 and the atomic number \(Z\) to decrease by 2.
02

Determine Parent Nucleus for Alpha Decay

Given the daughter nucleus is \(^{207}_{82}\mathrm{Pb}\) in \(\alpha\) decay, we calculate the parent isotope:\[A = 207 + 4 = 211\]\[Z = 82 + 2 = 84\]The parent nucleus is \(^{211}_{84}\mathrm{Po}\) (Polonium). The decay process:\[^{211}_{84}\mathrm{Po} \rightarrow {}^{207}_{82}\mathrm{Pb} + \alpha (^{4}_{2}\mathrm{He})\]
03

Understanding Beta Minus Decay

In \(\beta^{-}\) decay, a neutron in the parent nucleus is transformed into a proton, emitting a beta particle (\(\beta^{-}\)) and an antineutrino. This increases the atomic number \(Z\) by 1, while the mass number \(A\) stays the same.
04

Determine Parent Nucleus for Beta Minus Decay

For the \(\beta^{-}\) decay resulting in \(^{207}_{82}\mathrm{Pb}\), the parent nucleus:\[A = 207\]\[Z = 82 - 1 = 81\]The parent nucleus is \(^{207}_{81}\mathrm{Tl}\) (Thallium). The decay process:\[^{207}_{81}\mathrm{Tl} \rightarrow {}^{207}_{82}\mathrm{Pb} + \beta^{-} + \overline{u}\]
05

Summary of Decay Processes

In part (a), the \(\alpha\) decay starts with \(^{211}_{84}\mathrm{Po}\) and results in \(^{207}_{82}\mathrm{Pb}\); while in part (b), the \(\beta^{-}\) decay starts with \(^{207}_{81}\mathrm{Tl}\) and results in \(^{207}_{82}\mathrm{Pb}\).

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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 a nucleus emits an alpha particle. Essentially, an alpha particle is the same as a helium nucleus, represented by \( ^{4}_{2}\mathrm{He} \). These particles consist of two protons and two neutrons. As a result of this emission, both the atomic number \( Z \) and mass number \( A \) of the original atom change. Specifically, \( A \) decreases by 4, and \( Z \) drops by 2. This leads to the formation of a new element. For example, consider the decay of polonium-211 into lead-207. Here’s how it happens:
  • Start with polonium-211: \( ^{211}_{84} \mathrm{Po} \)
  • It emits an alpha particle \( ^{4}_{2}\mathrm{He} \)
  • We end up with lead-207: \( ^{207}_{82} \mathrm{Pb} \)
Understanding this process helps in observing that during alpha decay, the parent nucleus changes into a different element, not just a lighter isotope of itself.
Beta Decay
Beta decay is another kind of nuclear decay and comes in two main types: beta minus (\( \beta^{-} \)) and beta plus (\( \beta^{+} \)). Let's focus on beta minus decay, where a neutron in the nucleus turns into a proton. This change involves the emission of a beta particle, which is a high-speed electron (\( \beta^{-} \)), and an antineutrino symbolized by \( \overline{u} \). Here's what happens in beta minus decay:
  • The mass number \( A \) of the nucleus remains constant.
  • The atomic number \( Z \) increases by 1 because of the new proton.
To illustrate, look at the decay of thallium-207 to lead-207:
  • We start with thallium-207: \( ^{207}_{81} \mathrm{Tl} \)
  • It emits a beta particle and an antineutrino: \( \beta^{-} + \overline{u} \)
  • We end up with lead-207: \( ^{207}_{82} \mathrm{Pb} \)
In beta decay, the nucleus doesn't change its mass, but it transforms into an isotope of the next element with a higher atomic number.
Daughter Nucleus
During nuclear decay processes, the nucleus formed after the decay is known as the daughter nucleus. This new nucleus is the "offspring" of the original nucleus or the parent nucleus. Whether through alpha or beta decay, the parent nucleus transitions to a daughter nucleus while emitting alpha or beta particles:
  • In alpha decay, the daughter nucleus has a reduced atomic and mass number, leading to an entirely different element.
  • In beta decay, the daughter nucleus typically becomes a more stable isotope of an adjacent element on the periodic table, due to the alteration in atomic number.
Using our examples: - For alpha decay of polonium-211, the daughter nucleus is lead-207. - For beta minus decay of thallium-207, the daughter nucleus is also lead-207. The daughter nucleus is always more stable than the parent, contributing to the natural progression towards stability in nuclear reactions.

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Most popular questions from this chapter

Iodine \(\stackrel{131}{53} \mathrm{I}\) is used in diagnostic and therapeutic techniques in the treatment of thyroid disorders. This isotope has a half-life of 8.04 days. What percentage of an initial sample of \(\stackrel{131}{53} \mathrm{I}\) remains after 30.0 days?

Two radioactive nuclei A and B are present in equal numbers to begin with. Three days later, there are three times as many A nuclei as there are B nuclei. The half-life of species B is 1.50 days. Find the half-life of species A

Outside the nucleus, the neutron itself is radioactive and decays into a proton, an electron, and an antineutrino. The half-life of a neutron (mass \(=1.675 \times 10^{-27} \mathrm{kg} )\) outside the nucleus is 10.4 \(\mathrm{min}\) . On average, over what distance (in meters) would a beam of \(5.00-\mathrm{eV}\) neutrons travel before the number of neutrons decreased to 75.0\(\%\) of its initial value?

(a) Energy is required to separate a nucleus into its constituent nucleons, as Figure 31.3 indicates; this energy is the total binding energy of the nucleus. In a similar way one can speak of the energy that binds a single nucleon to the remainder of the nucleus. For example, separating nitrogen \(\stackrel{14}{7} \mathrm{N}\) into nitrogen \(\frac{13}{7} \mathrm{N}\) and a neutron takes energy equal to the binding energy of the neutron, as shown below: \(\stackrel{14}{7} \mathrm{N}+\) Energy \(\longrightarrow^{13}_{7} \mathrm{N}+_{0}^{1} \mathrm{n}\) Find the energy (in Mev) that binds the neutron to the \(\frac{14}{7}\) N nucleus by considering the mass of \(\frac{13}{7} \mathrm{N}\) (atomic mass \(=13.005738 \mathrm{u}\) and the mass of \(_{0}^{1} \mathrm{n}\) (atomic mass \(=\) 1.008665 \(\mathrm {u} )\) as compared to the mass of \(\frac{14}{7} \mathrm{N}\) (atomic mass \(=14.003 \quad 074\) u). (b) Similarly, one can speak of the energy that binds a single proton to the \(\frac{14}{7}\) N nucleus: $$ ^{14} \mathrm{N}+\text { Energy } \longrightarrow_{6}^{13} \mathrm{C}+\mathrm{IH}$$ Following the procedure outlined in part (a), determine the energy (in MeV) that binds the proton (atomic mass \(=1.007825\) u) to the \(\frac{14}{7} N\) nucleus. The atomic mass of carbon \(\frac{13}{6} \mathrm{C}\) is 13.003355 \(\mathrm{u}\) . (c) Which nucleon is more tightly bound, the neutron or the proton?

Multiple-Concept Example 11 reviews most of the concepts that are needed to solve this problem. Material found with a mummy in the arid highlands of southern Peru has a \(^{14}_{6} \mathrm{C}\) activity per gram of carbon that is 78.5\(\%\) of the activity present initially. How long ago (in years) did this individual die?

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