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Chapter 31: Problem 26

Rado \(220 \mathrm{Rn}\) produces a daughter nucleus that is radioactive. The daughter, in turn, produces its own radioactive daughter, and so on. This process continues until lead \(83 \underline{208} \mathrm{~b}\) is reached. What are the total number \(N_{\alpha}\) of \(\alpha\) particles and the total number \(N_{\beta}\) of \(\beta\) particles that are generated in this series of radioactive decays?

Short Answer

Expert verified
3 \(\alpha\) particles and 2 \(\beta\) particles are generated.

Step by step solution

01

Identify the Parent and Final Nucleus

The sequence starts with radon-220 \(_{86}^{220} \mathrm{Rn}\) and ends with lead-208 \(_{82}^{208} \mathrm{Pb}\). We need to calculate the changes in atomic and mass numbers to determine the \( \alpha \) and \( \beta \) decays involved.
02

Determine the Change in Mass Number

The original radon-220 has a mass number of 220, while lead-208 has a mass number of 208. To find the total change, subtract the final mass number from the initial mass number: \( 220 - 208 = 12 \). This indicates how many \( \alpha \) particles, each removing 4 mass units, were emitted:\[ \frac{12}{4} = 3 \].Thus, 3 \( \alpha \) decays occur.
03

Calculate Total Atomic Number Change

The atomic number changes from 86 in radon-220 to 82 in lead-208. Each \( \alpha \) decay decreases the atomic number by 2. Therefore, with 3 \( \alpha \) decays, the atomic number should drop by \((3 \times 2 = 6) \), reaching 80.
04

Adjust Atomic Number for Beta Decays

Since the final atomic number is 82 and we need to account for each \( \beta^- \) decay that increases the atomic number by 1, calculate the number of \( \beta \) decays required:\( 82 - 80 = 2 \).Thus, 2 \( \beta \) decays are needed to adjust the atomic number from 80 to 82.
05

Summarize the Particle Counts

From the analysis:- \( N_{\alpha} = 3 \) - \( N_{\beta} = 2 \)So, the series of decays releases 3 \( \alpha \) particles and 2 \( \beta \) particles.

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Key Concepts

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

Alpha decay in nuclear decay series
Alpha decay is a type of radioactive decay where an unstable atom emits an alpha particle, consisting of 2 protons and 2 neutrons. This particle resembles a helium nucleus. When an alpha particle is emitted, the original atom loses 4 mass units and 2 atomic numbers.
In the case of radon-220 decaying to lead-208, alpha decay plays a critical role. Each alpha decay removes components that reduce the mass number and the atomic number of the original atom.
  • Mass number decreases by 4 with each emission.
  • Atomic number drops by 2 for each decay event.
Through three instances of alpha decay, radon-220's mass number reduces from 220 to 208. This shows how integral alpha decay is to transforming the parent atom into its stable daughter product.
Understanding beta decay
Beta decay is another frequent process in nuclear decay. Unlike alpha decay, beta decay involves the transformation of a neutron into a proton within the nucleus, resulting in the emission of a beta particle (electron or positron). This process alters the atomic number without changing the mass number.
During beta decay:
  • The atomic number increases by 1 if a beta-minus particle (electron) is released.
  • The mass number remains constant.
In the radon to lead decay series, beta decay compensates for the changes in atomic number due to alpha decay. It ensures the nucleus reaches the appropriate number of protons while finishing its decay chain.
The significance of mass number
The mass number of an atom is the total count of protons and neutrons within its nucleus. It is central to understanding nuclear reactions because it indicates how much matter is inside the nucleus.
In decay series problems, observing the mass number helps us understand how many particles like alpha are emitted. Since each alpha decay decreases the mass number by 4, one can trace how the mass number evolves from a parent to a stable daughter nucleus.
Role of atomic number
The atomic number is key to defining an element's identity. It specifies the number of protons in the nucleus, determining the element's name and position on the periodic table.
In nuclear decay, the atomic number is vital for tracking changes in an element's identity. Alpha decay causes the atomic number to fall by 2, while beta decay increases it by 1. By looking at these shifts, one can map the sequence of transformations that a nucleus undergoes through radioactive decay.
Understanding radioactive decay sequences
Radioactive decay sequences describe a chain of transformations where an unstable nucleus undergoes a series of emissions to achieve stability. These emissions can be alpha, beta, or other less common types of decay.
This decay sequence for radon-220 to lead-208 is a perfect example, which includes a series of 3 alpha decays and 2 beta decays. Each step in the sequence moves the nucleus closer to a stable configuration:
  • Alpha decays gradually decrease both the mass and atomic numbers.
  • Beta decays adjust the atomic number, fine-tuning the balance of protons and neutrons.
Following these sequences explains how complex transformations converge towards forming stable elements, providing insight into nuclear chemistry.

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Most popular questions from this chapter

Find the energy released when lead \(\underset{82}^{211} \mathrm{~Pb}\) (atomic mass \(=210.988735 \mathrm{u}\) ) undergoes \(\beta^{-}\) decay to become bismuth \(\underset{83}{211} \mathrm{Bi}\) (atomic mass \(\left.=210.987255 \mathrm{u}\right)\).

The ratio \(r_{X} / r_{T}\) of the radius of an unknown nucleus \(A_{X}\) to a tritium nucleus \({ }_{1}^{3} T\) is \(\frac{r_{\mathrm{X}}}{r_{\mathrm{T}}}=1.10 .\) Both nuclei contain the same number of neutrons. Identify the unknown nucleus in the form \(A_{X}\). Use the periodic table on the inside of the back cover as needed.

The isotope \(\frac{198}{79} \mathrm{Au}\) (atomic mass \(=197.968 \mathrm{u}\) ) of gold has a half-life of 2.69 days and is used in cancer therapy. What mass (in grams) of this isotope is required to produce an activity of \(315 \mathrm{Ci} ?\)

Iodine \(\frac{131}{53} \mathrm{I}\) 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 \(\frac{131}{53}\) I remains after 30.0 days?

Concept Questions (a) Two radioactive nuclei \(\mathrm{A}\) and \(\mathrm{B}\) have half-lives of \(T_{1 / 2, \mathrm{~A}}\) and \(T_{1 / 2, \mathrm{~B}}\), where \(T_{1 / 2, \mathrm{~A}}\) is greater than \(T_{1 / 2, \mathrm{~B}}\). During the same time period, is the fraction of nuclei A that decay greater than, smaller than, or the same as the fraction of nuclei B that decay? (b) The numbers of these nuclei present initially are \(N_{0, \mathrm{~A}}\) and \(N_{0, \mathrm{~B}}\), the ratio of the two being \(N_{0, \mathrm{~A}} / N_{0, \mathrm{~B}}\). Is the ratio \(N_{\mathrm{A}} / N_{\mathrm{B}}\) of the number of nuclei present at a later time greater than, smaller than, or the same as \(N_{0, A} / N_{0, B}\) ? Justify your answers. Problem Two waste products from nuclear reactors are strontium \(9{ }_{38}^{90} \mathrm{Sr}\) \(\left(T_{1 / 2}=29.1 \mathrm{yr}\right)\) and cesium \({ }_{55}^{134} \mathrm{C}_{s}\left(T_{1 / 2}=2.06 \mathrm{yr}\right) .\) These two species are present initially in a ratio of \(N_{0, \mathrm{Sr}} / N_{0, \mathrm{Cs}}=7.80 \times 10^{-3}\). What is the ratio \(N_{\mathrm{Sr}} / N_{\mathrm{Cs}}\) fifteen years later? Verify that your answer is consistent with your answers to the Concept Questions.
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