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Chapter 19: Problem 22

The only stable isotope of fluorine is fluorine-19. Predict possible modes of decay for fluorine-21, fluorine-18, and fluorine- \(17 .\)

Short Answer

Expert verified
Fluorine-21 is likely to undergo beta-minus decay due to having more neutrons than protons. Fluorine-18 can achieve stability through beta-plus decay, aligning with the stable isotope Fluorine-19. Fluorine-17 can undergo either beta-plus decay to become Fluorine-18 or electron capture to become the stable Oxygen-17 isotope.

Step by step solution

01

1. Understand types of decay processes

There are mainly three types of decay processes: a. Alpha decay: In this process, a nucleus emits an alpha particle (consisting of 2 protons and 2 neutrons). b. Beta decay: This can be of two types, beta-minus decay and beta-plus decay. In beta-minus decay, a neutron is converted into a proton, while in beta-plus decay, a proton is converted into a neutron. c. Other decay processes: These include electron capture, gamma emission, etc.
02

2. Atomic composition of fluorine isotopes

First, let's find the atomic composition (number of protons and neutrons) of each fluorine isotope: - Fluorine-21: It has 9 protons and (21 - 9) = 12 neutrons. - Fluorine-18: It has 9 protons and (18 - 9) = 9 neutrons. - Fluorine-17: It has 9 protons and (17 - 9) = 8 neutrons.
03

3. Decay process for fluorine-21

Fluorine-21 has 12 neutrons and 9 protons. Having more neutrons than protons, this isotope can undergo beta-minus decay, where a neutron changes into a proton, resulting in neutron-neutron interactions to increase stability.
04

4. Decay process for fluorine-18

Fluorine-18 has 9 neutrons and 9 protons. Fluorine-19, the stable isotope, has 10 neutrons and 9 protons. To achieve stability, this isotope can undergo beta-plus decay, where a proton changes into a neutron, to increase neutron-proton interactions.
05

5. Decay process for fluorine-17

Fluorine-17 has 8 neutrons and 9 protons. This isotope can undergo beta-plus decay (proton changes to a neutron) to become the stable isotope fluorine-18 with 9 neutrons and 9 protons. Alternatively, it can undergo electron capture, where an electron is captured by the nucleus, changing a proton into a neutron and turning it into a stable oxygen-17 isotope with 8 protons and 9 neutrons. In conclusion, the likely decay processes for the given fluorine isotopes are: - Fluorine-21: Beta-minus decay - Fluorine-18: Beta-plus decay - Fluorine-17: Beta-plus decay or electron capture

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Most popular questions from this chapter

The radioactive isotope \({ }^{247} \mathrm{Bk}\) decays by a series of \(\alpha\) -particle and \(\beta\) -particle productions, taking \({ }^{247} \mathrm{Bk}\) through many transformations to end up as \({ }^{207} \mathrm{~Pb}\). In the complete decay series, how many \(\alpha\) particles and \(\beta\) particles are produced?

Consider the following information: i. The layer of dead skin on our bodies is sufficient to protect us from most \(\alpha\) -particle radiation. ii. Plutonium is an \(\alpha\) -particle producer. iii. The chemistry of \(\mathrm{Pu}^{4+}\) is similar to that of \(\mathrm{Fe}^{3+}\). iv. Pu oxidizes readily to \(\mathrm{Pu}^{4+}\). Why is plutonium one of the most toxic substances known?

Americium-241 is widely used in smoke detectors. The radiation released by this element ionizes particles that are then detected by a charged-particle collector. The half-life of \({ }^{241} \mathrm{Am}\) is 433 years, and it decays by emitting \(\alpha\) particles. How many \(\alpha\) particles are emitted each second by a \(5.00-\mathrm{g}\) sample of \({ }^{241} \mathrm{Am} ?\)

In addition to the process described in the text, a second process called the carbon-nitrogen cycle occurs in the sun: $$ \begin{aligned} { }_{1}^{1} \mathrm{H}+{ }_{6}^{12} \mathrm{C} \longrightarrow{ }_{7}^{13} \mathrm{~N}+{ }_{0}^{0} \gamma \\ { }_{7}^{13} \mathrm{~N} & \longrightarrow{ }_{6}^{13} \mathrm{C}+{ }_{+1}^{0} \mathrm{e} \\ { }_{1}^{1} \mathrm{H}+{ }_{6}^{13} \mathrm{C} &{ }_{7}^{14} \mathrm{~N}+{ }_{0}^{0} \gamma \\ { }_{1}^{1} \mathrm{H}+{ }_{7}^{14} \mathrm{~N} \longrightarrow &{ }_{8}^{15} \mathrm{O}+{ }_{0}^{0} \gamma \\ { }_{8}^{15} \mathrm{O} \longrightarrow{ }_{7}^{15} \mathrm{~N}+{ }_{+1}^{0} \mathrm{e} \\ { }_{1}^{1} \mathrm{H}+{ }_{7}^{15} \mathrm{~N} \longrightarrow{ }_{6}^{12} \mathrm{C}+{ }_{2}^{4} \mathrm{He}+{ }_{0}^{0} \gamma \\ \hline \end{aligned} $$ reaction: \(\quad 4{ }_{1}^{1} \mathrm{H} \longrightarrow{ }_{2}^{4} \mathrm{He}+2{ }_{+1}^{0} \mathrm{e}\) a. What is the catalyst in this process? b. What nucleons are intermediates? c. How much energy is released per mole of hydrogen nuclei in the overall reaction? (The atomic masses of \({ }_{1} \mathrm{H}\) and \({ }_{2}^{4} \mathrm{He}\) are \(1.00782 \mathrm{u}\) and \(4.00260 \mathrm{u}\), respectively. \()\)

The most significant source of natural radiation is radon- 222 . \({ }^{222} \mathrm{Rn}\), a decay product of \({ }^{238} \mathrm{U}\), is continuously generated in the earth's crust, allowing gaseous \(\mathrm{Rn}\) to seep into the basements of buildings. Because \({ }^{222} \mathrm{Rn}\) is an \(\alpha\) -particle producer with a relatively short half-life of \(3.82\) days, it can cause biological damage when inhaled. a. How many \(\alpha\) particles and \(\beta\) particles are produced when \({ }^{238} \mathrm{U}\) decays to \({ }^{222} \mathrm{Rn}\) ? What nuclei are produced when 222 \(\mathrm{Rn}\) decays? b. Radon is a noble gas so one would expect it to pass through the body quickly. Why is there a concern over inhaling \({ }^{222} \mathrm{Rn}\) ? c. Another problem associated with \({ }^{222} \mathrm{Rn}\) is that the decay of \({ }^{222} \mathrm{Rn}\) produces a more potent \(\alpha\) -particle producer \(\left(t_{1 / 2}=\right.\) \(3.11 \mathrm{~min}\) ) that is a solid. What is the identity of the solid? Give the balanced equation of this species decaying by \(\alpha\) -particle production. Why is the solid a more potent \(\alpha\) -particle producer? d. The U.S. Environmental Protection Agency (EPA) recommends that \({ }^{222} \mathrm{Rn}\) levels not exceed 4 pCi per liter of air \(\left(1 \mathrm{Ci}=1\right.\) curie \(=3.7 \times 10^{10}\) decay events per second; \(\left.1 \mathrm{pCi}=1 \times 10^{-12} \mathrm{Ci}\right) .\) Convert \(4.0 \mathrm{pCi}\) per liter of air into concentrations units of \({ }^{222} \mathrm{Rn}\) atoms per liter of air and moles of \(222 \mathrm{Rn}\) per liter of air.
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