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Just as the uranium series is called the "4n \(+2^{\prime \prime}\) series, the thorium series can be called the "4n" series and the actinium series the "4n \(+3 "\) series. A \(4 n+1 "\) series has also been established, with \(^{241} \mathrm{Pu}\) as the parent nuclide. To which series does each of the following belong: (a) \(\frac{214}{83} \mathrm{Bi} ;\) (b) \(\frac{216}{84} \mathrm{Po} ;\) (c) \(\frac{215}{85} \mathrm{At}\) (d) \(\frac{235}{92} \mathrm{U} ?\)

Short Answer

Expert verified
(a) 214/83 Bi does not belong to any series. (b) 216/84 Po belongs to the thorium (4n) series. (c) 215/85 At does not belong to any series. (d) 235/92 U does not belong to any series.

Step by step solution

01

Calculate 'n' for each isotope

The value of 'n' for each isotope can be calculated from the mass number using the equation \( n = \frac{A}{4} \), where 'A' is the mass number of the isotope.
02

Determine the decay series for 214/83 Bi

For the isotope Bi-214 (214/83 Bi), the mass number is 214. So \( n = \frac{214}{4} = 53.5 \). Since 'n' is not an integer, it does not belong to any of the specified series.
03

Determine the decay series for 216/84 Po

For the isotope Po-216 (216/84 Po), the mass number is 216. So \( n = \frac{216}{4} = 54 \). Since 'n' is an integer and 216 is a multiple of 4 ('4n' series), the isotope belongs to the thorium series.
04

Determine the decay series for 215/85 At

For the isotope At-215 (215/85 At), the mass number is 215. So \( n = \frac{215}{4} = 53.75 \). Since 'n' is not an integer, the isotope does not belong to any of the specified series.
05

Determine the decay series for 235/92 U

For the isotope U-235 (235/92 U), the mass number is 235. So \( n = \frac{235}{4} = 58.75 \). This doesn't fall into any of the specified series, since 'n' is not an integer.

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Key Concepts

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

Thorium Series
The thorium series, also known as the "4n" series, is a sequence of radioactive decay processes where the mass numbers of the isotopes in the series are exact multiples of four. This means that any isotope that belongs to this series can have their mass number expressed as \( A = 4n \), where 'n' is an integer.

This series starts with thorium-232 and ends with lead-208, including several intermediate radioactive elements such as radium, radon, polonium, and thallium. Thorium-232 itself is the parent nuclide, which means it is the substance that undergoes radioactive decay to produce another isotope.

In answering the given exercise, if we encounter an isotope like polonium-216, with a mass number of 216, since 216 divided by four equals an integer (54), this curious finding places it directly within the thorium series. It highlights how interconnected decay pathways generate specific elements over time. Understanding the thorium series provides essential insights into nuclear reactions and radioactive decay chaining that occur naturally.
Uranium Series
The uranium series, also referred to as the "4n+2" series, is a set of isotopes with mass numbers that can be expressed as \( A = 4n + 2 \). This series is also sometimes called the uranium-radium series.

The sequence has uranium-238 as its initial parent nuclide. It progresses through a variety of radioactive isotopes, eventually reaching a stable end product, lead-206. Elements like radium, radon, and polonium are part of this intricate process of decay within this series.

Given an example from the exercise, isotopes like bismuth-214 were assessed numerically to determine if they fit into the series. However, if dividing the mass number doesn't yield a formula fit as "4n+2", then we realize it is not a member of the uranium series. Grasping the uranium series is important for understanding long-term radioactive decay and the formation of lead.
Actinium Series
The actinium series, or the "4n+3" series, consists of isotopes whose mass numbers follow the formula \( A = 4n + 3 \). This decay chain commences with uranium-235 and concludes with the stable nuclide lead-207.

This pathway includes several notable decays and isotopes, such as actinium itself, thorium, and others. The series is relatively present in nature and used to understand radiation and decay in geological and nuclear fields.

When evaluating isotopes such as astatine-215 from our exercise, checking whether it adheres to the "4n+3" criteria is key. If dividing the mass number 215 by four gives a result plus three that is an integer, it is a part of the actinium series. This exercise provides a practical application of the principles behind radioactive series, making it easier to classify and understand these decay chains.

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Most popular questions from this chapter

A sample containing \(_{88}^{224} \mathrm{Ra},\) which decays by \(\alpha\) -particle emission, disintegrates at the following rate, expressed as disintegrations per minute or counts per minute \((\mathrm{cpm}): t=0,1000 \mathrm{cpm} ; t=1 \mathrm{h}\) \(992 \mathrm{cpm} ; t=10 \mathrm{h}, 924 \mathrm{cpm} ; t=100 \mathrm{h}, 452 \mathrm{cpm}\) \(t=250 \mathrm{h}, 138 \mathrm{cpm} .\) What is the half-life of this nuclide?

Write nuclear equations to represent (a) the decay of \(^{214} \mathrm{Ra}\) by \(\alpha\) -particle emission (b) the decay of \(^{205}\) At by positron emission (c) the decay of \(^{212} \mathrm{Fr}\) by electron capture (d) the reaction of two deuterium nuclei (deuterons) to produce a nucleus of \(\frac{3}{2} \mathrm{He}\). (e) the production of \({243}_{97} \mathrm{Bk}\) get by the \(\alpha\) -particle bombardment of\({241}_{95} \mathrm{Am}\) (f) a nuclear reaction in which thorium-232 is bombarded with \(\alpha\) particles, producing a new nuclide and four neutrons.

A nuclide has a decay constant of \(4.28 \times 10^{-4} \mathrm{h}^{-1}\). If the activity of a sample is \(3.14 \times 10^{5} \mathrm{s}^{-1},\) how many atoms of the nuclide are present in the sample? (a) \(2.64 \times 10^{12} ;\) (b) \(7.34 \times 10^{8}\) (c) \(2.04 \times 10^{5}\) (d) \(4.40 \times 10^{10} ;\) (e) none of these.

Given a radioactive nuclide with \(t_{1 / 2}=1.00 \mathrm{h}\) and a current disintegration rate of 1000 atoms \(s^{-1}\), three hours from now the disintegration rate will be (a) 1000 atoms \(s^{-1} ;\) (b) 333 atoms \(s^{-1} ;\) (c) 250 atoms \(s^{-1}\); (d) 125 atoms \(s^{-1}\)

The immediate decay product of element 118 is thought to be element \(116 .\) Write a complete nuclear equation for this reaction.

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