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Chapter 21: Problem 20

Each of the following nuclei undergoes either beta decay or positron emission. Predict the type of emission for each: \((\mathbf{a})\) tritium, \(_{1}^{3} \mathrm{H},(\mathbf{b})_{38}^{89} \mathrm{Sr},(\mathbf{c})\) iodine-120, \((\mathbf{d})\) (d) silver-102.

Short Answer

Expert verified
All four nuclei will undergo beta decay: Tritium (_{1}^{3}H), Strontium (_{38}^{89}Sr), Iodine-120 (_{53}^{120}I), and Silver-102 (_{47}^{102}Ag). This is because their neutron-to-proton ratios indicate that they are neutron-rich.

Step by step solution

01

Tritium (_{1}^{3}H)

To find the neutron-to-proton ratio, we calculate: Neutrons = mass number - atomic number Neutrons = 3 - 1 = 2 Ratio = Neutrons/Protons Ratio = 2/1 = 2 Tritium (_{1}^{3}H) is neutron-rich, as the ratio is higher than 1, so it will undergo beta decay.
02

Strontium (_{38}^{89}Sr)

Calculate the neutron-to-proton ratio: Neutrons = mass number - atomic number Neutrons = 89 - 38 = 51 Ratio = Neutrons/Protons Ratio = 51/38 ≈ 1.342 Strontium (_{38}^{89}Sr) is also neutron-rich, so it will undergo beta decay.
03

Iodine-120 (_{53}^{120}I)

Calculate the neutron-to-proton ratio: Neutrons = mass number - atomic number Neutrons = 120 - 53 = 67 Ratio = Neutrons/Protons Ratio = 67/53 ≈ 1.264 Iodine-120 (_{53}^{120}I) is neutron-rich, so it will undergo beta decay.
04

Silver-102 (_{47}^{102}Ag)

Calculate the neutron-to-proton ratio: Neutrons = mass number - atomic number Neutrons = 102 - 47 = 55 Ratio = Neutrons/Protons Ratio = 55/47 ≈ 1.170 Silver-102 (_{47}^{102}Ag) has a neutron-to-proton ratio close to the optimal ratio for stable nuclei (found mostly in lighter elements), but it is still considered neutron-rich. Thus, silver-102 will undergo beta decay. In conclusion, all four nuclei will undergo beta decay, as their neutron-to-proton ratios indicate that they are neutron-rich.

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Most popular questions from this chapter

In 2010, a team of scientists from Russia and the United States reported creation of the first atom of element 117, which is named tennessine, and whose symbol is Ts. The synthesis involved the collision of a target of \(_{97}^{249} \mathrm{Bk}\) with accelerated ions of an isotope which we will denote Q. The product atom, which we will call Z, immediately releases neutrons and forms \(_{97}^{249} \mathrm{Bk} :\) $$_{97}^{249} \mathrm{Bk}+\mathrm{Q} \longrightarrow \mathrm{Z} \longrightarrow_{117 \mathrm{Ts}}^{294 \mathrm{Ts}}+3_{0}^{1} \mathrm{n}$$ (a) What are the identities of isotopes Q and Z? (b) Isotope Q is unusual in that it is very long-lived (its half-life is on the order of 1019 yr) in spite of having an unfavorable neutron-to-proton ratio (Figure 21.1). Can you propose a reason for its unusual stability? (c) Collision of ions of isotope Q with a target was also used to produce the first atoms of livermorium, Lv. The initial product of this collision was \(_{116}^{296} \mathrm{Zn}\). What was the target isotope with which Q collided in this experiment?

It takes 4 \(\mathrm{h} 39\) min for a 2.00 - mg sample of radium-230 to decay to 0.25 \(\mathrm{mg.}\) What is the half-life of radium-230?

The cloth shroud from around a mummy is found to have \(\mathrm{a}^{14} \mathrm{C}\) activity of 9.7 disintegrations per minute per gram of carbon as compared with living organisms that undergo 16.3 disintegrations per minute per gram of carbon. From the half-life for \(^{14} \mathrm{C}\) decay, 5715 yr, calculate the age of the shroud.

One nuclide in each of these pairs is radioactive. Predict which is radioactive and which is stable: \((\mathbf{a})_{20}^{40} \mathrm{Ca}\) and \(_{20}^{45} \mathrm{Ca}\) , \((\mathbf{b})^{12} \mathrm{C}\) and \(^{14} \mathrm{C},\) \((\mathbf{c})\) lead-206 and thorium-230. Explain your choice in each case.

Give the symbol for \((\mathbf{a})\) a neutron, \((\mathbf{b})\) an alpha particle, \((\mathbf{c})\) gamma radiation.
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