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Chapter 25: Problem 46

Both \(\beta^{-}\) and \(\beta^{+}\) emissions are observed for artificially produced radioisotopes of low atomic numbers, but only \(\beta^{-}\) emission is observed with naturally occurring radioisotopes of high atomic number. Why do you suppose this is so?

Short Answer

Expert verified
\(\beta^{+}\) emission is less observed in nature because it requires a lower neutron-to-proton ratio and is less energetically favourable. High atomic number elements prefer \(\beta^{-}\) decay to increase stability by decreasing their proton count.

Step by step solution

01

Understanding the Difference between \(\beta^{-}\) and \(\beta^{+}\) decay

In \(\beta^{-}\) decay, a neutron inside the nucleus transforms into a proton and emits an electron (and an electron antineutrino). This increases the atomic number by 1. \(\beta^{+}\) decay is the opposite, where a proton inside the nucleus changes into a neutron and emits a positron (and an electron neutrino), thereby decreasing the atomic number by 1.
02

Connection to Atomic number

For elements with high atomic numbers, they have a high proton count which can cause instability due to repulsion between protons. To reduce this repulsion and increase the stability of the nucleus, it prefers the transformation process that allows it to reduce its proton count i.e., \(\beta^{-}\) decay. Conversely, lighter or artificially produced elements with low atomic numbers might have an excess of neutrons, leading to \(\beta^{+}\) decay to increase their proton count.
03

Understanding why \(\beta^{+}\) decay rarely occurs in nature

In nature, isotopes that undergo \(\beta^{+}\) decay are rare because it requires the atom to have a lower neutron-to-proton ratio. Most naturally occurring isotopes have a higher neutron-to-proton ratio which favours \(\beta^{-}\) decay. Additionally, \(\beta^{+}\) decay is less energetically favourable because it requires the nucleus to possess sufficient energy to convert a proton to a neutron, which is a heavier particle.

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Most popular questions from this chapter

The following reactions are carried out with HCl(aq) containing some tritium ( \(_{1}^{3} \mathrm{H}\) ) as a tracer. Would you expect any of the tritium radioactivity to appear in the \(\mathrm{NH}_{3}(\mathrm{g}) ?\) In the \(\mathrm{H}_{2} \mathrm{O} ?\) Explain. $$\begin{array}{c} \mathrm{NH}_{3}(\mathrm{aq})+\mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(\mathrm{aq}) \\ \mathrm{NH}_{4} \mathrm{Cl}(\mathrm{aq})+\mathrm{NaOH}(\mathrm{aq}) \longrightarrow\\\\\mathrm{NaCl}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(1)+\mathrm{NH}_{3}(\mathrm{g})\end{array}$$

A lunar rock was analyzed for argon by mass spectrometry and for potassium by atomic absorption. The results of these analyses showed that the sample contained \(3.02 \times 10^{-5} \mathrm{mL} \mathrm{g}^{-1}\) of argon and \(0.083 \%\) of potassium. The half-life of potassium- 40 is \(1.248 \times\) \(10^{9} \mathrm{y} \cdot\) Calculate the age of the lunar rock.

The disintegration rate for a sample containing \(_{27}^{60} \mathrm{Co}\) as the only radioactive nuclide is 6740 dis \(\mathrm{h}^{-1}\). The half-life of 20 Co is 5.2 years. Estimate the number of atoms of \(_{27}^{60}\) Co in the sample.

\(^{222} \mathrm{Rn}\) is an \(\alpha\) -particle emitter with a half-life of 3.82 days. Is it hazardous to be near a flask containing this isotope? Under what conditions might \(^{222} \mathrm{Rn}\) be hazardous?

For medical uses, radon-222 formed in the radioactive decay of radium-226 is allowed to collect over the radium metal. Then, the gas is withdrawn and sealed into a glass vial. Following this, the radium is allowed to disintegrate for another period, when a new sample of radon- 222 can be withdrawn. The procedure can be continued indefinitely. The process is somewhat complicated by the fact that radon-222 itself undergoes radioactive decay to polonium- 218 , and so on. The half-lives of radium-226 and radon-222 are \(1.60 \times 10^{3}\) years and 3.82 days, respectively.(a) Beginning with pure radium- \(226,\) the number of radon-222 atoms present starts at zero, increases for a time, and then falls off again. Explain this behavior. That is, because the half-life of radon-222 is so much shorter than that of radium- \(226,\) why doesn't the radon-222 simply decay as fast as it is produced, without ever building up to a maximum concentration?(b) Write an expression for the rate of change \((d \mathrm{D} / d t)\) in the number of atoms (D) of the radon- 222 daughter in terms of the number of radium- 226 atoms present initially ( \(\mathrm{P}_{0}\) ) and the decay constants of the parent \(\left(\lambda_{\mathrm{p}}\right)\) and daughter \(\left(\lambda_{\mathrm{d}}\right)\) (c) Integration of the expression obtained in part (b) yields the following expression for the number of atoms of the radon-222 daughter (D) present at a time \(t\).$$\mathrm{D}=\frac{\mathrm{P}_{0} \lambda_{\mathrm{p}}\left(\mathrm{e}^{-\lambda_{\mathrm{p}} \times t}-\mathrm{e}^{-\lambda_{\mathrm{d}} \times t}\right)}{\lambda_{\mathrm{d}}-\lambda_{\mathrm{p}}}$$,Starting with \(1.00 \mathrm{g}\) of pure radium- \(226,\) approximately how long will it take for the amount of radon222 to reach its maximum value: one day, one week, one year, one century, or one millennium?
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