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Chapter 19: Problem 19

Write an equation describing the radioactive decay of each of the following nuclides. (The particle produced is shown in parentheses, except for electron capture, where an electron is a reactant.) a. 68 Ga (electron capture) b. 62 Cu (positron) c. 212 \(\mathrm{Fr}(\alpha)\) d. 129 \(\mathrm{Sb}(\beta)\)

Short Answer

Expert verified
The radioactive decay equations for the given nuclides are: a. Electron Capture: \[^{68}_{31}\textrm{Ga} + ^0_{-1}e \rightarrow ^{68}_{30}\textrm{Zn}\] b. Positron Decay: \[^{62}_{29}\textrm{Cu} \rightarrow ^{62}_{30}\textrm{Zn} + ^0_{+1}e\] c. Alpha Decay: \[^{212}_{87}\textrm{Fr} \rightarrow ^{208}_{85}\textrm{At} + ^4_2\alpha\] d. Beta Decay: \[^{129}_{51}\textrm{Sb} \rightarrow ^{129}_{52}\textrm{Te} + ^0_{-1}\beta\]

Step by step solution

01

a. 68 Ga (electron capture)

In electron capture, a proton within the nucleus captures an electron (usually from the innermost orbital) and is transformed into a neutron. The atomic number Z decreases by 1, while the mass number A remains unchanged. For this example, we have: Parent nuclide: \(^{68}_{31}\textrm{Ga}\) Electron capture: \(^{68}_{31}\textrm{Ga} + ^0_{-1}e \rightarrow ^{68}_{30}\textrm{Zn}\)
02

b. 62 Cu (positron)

In positron decay (also known as beta-plus decay), a neutron in the nucleus is transformed into a proton, releasing a positron in the process. The atomic number Z increases by 1, while the mass number A remains unchanged. For this example, we have: Parent nuclide: \(^{62}_{29}\textrm{Cu}\) Positron decay: \(^{62}_{29}\textrm{Cu} \rightarrow ^{62}_{30}\textrm{Zn} + ^0_{+1}e\)
03

c. 212 𝚽Fr(α)

In alpha decay, the parent nucleus releases an alpha particle (which consists of 2 protons and 2 neutrons). The atomic number Z decreases by 2, and the mass number A decreases by 4. For this example, we have: Parent nuclide: \(^{212}_{87}\textrm{Fr}\) Alpha decay: \(^{212}_{87}\textrm{Fr} \rightarrow ^{208}_{85}\textrm{At} + ^4_2\alpha\)
04

d. 129 𝚽Sb(β)

In beta decay (also known as beta-minus decay), a neutron in the nucleus is transformed into a proton, releasing an electron (beta particle) in the process. The atomic number Z increases by 1, while the mass number A remains unchanged. For this example, we have: Parent nuclide: \(^{129}_{51}\textrm{Sb}\) Beta decay: \(^{129}_{51}\textrm{Sb} \rightarrow ^{129}_{52}\textrm{Te} + ^0_{-1}\beta\)

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Most popular questions from this chapter

Iodine- 131 is used in the diagnosis and treatment of thyroid disease and has a half-life of 8.0 days. If a patient with thyroid disease consumes a sample of \(\mathrm{Na}^{131} \mathrm{I}\) containing $10 . \mu \mathrm{g}^{131 \mathrm{I}}\( how long will it take for the amount of \)^{131} \mathrm{I}$ to decrease to 1\(/ 100\) of the original amount?

There is a trend in the United States toward using coal-fired power plants to generate electricity rather than building new nuclear fission power plants. Is the use of coal-fired power plants without risk? Make a list of the risks to society from the use of each type of power plant.

A certain radioactive nuclide has a half-life of 3.00 hours. a. Calculate the rate constant in \(\mathrm{s}^{-1}\) for this nuclide. b. Calculate the decay rate in decays/s for 1.000 mole of this nuclide.

The mass ratios of 40 \(\mathrm{Ar}\) to 40 \(\mathrm{K}\) also can be used to date geologic materials. Potassium-40 decays by two processes: $$_{19}^{40} \mathrm{K}+_{-1}^{0} \mathrm{e} \longrightarrow_{\mathrm{i} 8}^{40} \mathrm{Ar}(10.7 \%)$$ $$_{19}^{40} \mathrm{K} \longrightarrow_{20}^{40} \mathrm{Ca}+_{-1}^{0} \mathrm{e}(89.3 \%)$$ $$t_{1 / 2}=1.27 \times 10^{9}$$ a. Why are \(^{40}\mathrm{Ar} /^{40} \mathrm{K}\) ratios used to date materials rather than \(^{40}\mathrm{Ca} / 40 \mathrm{K}\) ratios? b. What assumptions must be made using this technique? c. A sedimentary rock has an Ar \(^{40} \mathrm{K}\) ratio of \(0.95 .\) Calculate the age of the rock. d. How will the measured age of a rock compare to the actual age if some \(^{40}\) Ar escaped from the sample?

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 \(^{24} \mathrm{Am}\) is 433 years, and it decays by emitting \(\alpha\) particles. How many \(\alpha\) particles are emitted each second by a 5.00 -g sample of \(^{241} \mathrm{Am}\) ?
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