Chapter 16: Problem 104

Rubidium- 87, a beta emitter, is a product of positron emission. (a) Identify the parent nucleus of \({ }^{87} \mathrm{Rb}\). (b) When the parent nucleus named in part (a) decays, does the \(\mathrm{n} / \mathrm{p}\) ratio increase or decrease?

Short Answer

Expert verified

(a) The parent nucleus of \({ }^{87} \mathrm{Rb}\) is \({ }^{87} \mathrm{Sr}\) (Strontium-87). (b) When the parent nucleus decays via positron emission, the neutron-to-proton ratio (n/p) increases.

Step by step solution

Understand positron emission

Positron emission, or beta-plus decay, is a type of radioactive decay in which a proton in the nucleus of an atom is converted into a neutron, simultaneously releasing a positron and a neutrino. In terms of nuclear equations, this process can be represented as: \[p \rightarrow n + e^{+} + \nu_{e}\] In positron emission, the atomic number (Z) of the nucleus decreases by 1 while the mass number (A) remains unchanged.

Identify the parent nucleus of Rubidium-87

Given that Rubidium-87 (\({ }^{87} \mathrm{Rb}\)) is a product of positron emission, we can now determine the atomic and mass numbers of its parent nucleus. Before the decay, the parent nucleus had: 1. One more proton (since a proton is converted into a neutron) 2. The same mass number (since mass number is unchanged) Rubidium's atomic number is 37 (Z = 37), and the mass number is 87 (A = 87). So, the parent nucleus had Z = 37 + 1 = 38, and A = 87. Thus, the parent nucleus is \({ }^{87} \mathrm{Sr}\) (Strontium-87), which has 38 protons and 49 neutrons.

Analyze the change in \(\mathrm{n} / \mathrm{p}\) ratio upon decay

Now let's compare the neutron-to-proton (n/p) ratio before and after the positron emission: 1. Before decay (Strontium-87): n/p = 49/38 2. After decay (Rubidium-87): n/p = 50/37 Since 50/37 > 49/38, the neutron-to-proton ratio increases when the parent nucleus (\({ }^{87}\mathrm{Sr}\)) decays via positron emission.

Final answers

(a) The parent nucleus of \({ }^{87} \mathrm{Rb}\) is \({ }^{87} \mathrm{Sr}\) (Strontium-87). (b) When the parent nucleus decays via positron emission, the neutron-to-proton ratio (n/p) increases.

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Rubidium- 87, a beta emitter, is a product of positron emission. (a) Identify the parent nucleus of \({ }^{87} \mathrm{Rb}\). (b) When the parent nucleus named in part (a) decays, does the \(\mathrm{n} / \mathrm{p}\) ratio increase or decrease?

Short Answer

Step by step solution

Understand positron emission

Identify the parent nucleus of Rubidium-87

Analyze the change in \(\mathrm{n} / \mathrm{p}\) ratio upon decay

Final answers

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