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Ch.21 - Nuclear Chemistry
Chapter 21, Problem 15b

Decay of which nucleus will lead to the following products: (b) chromium-50 by positron emission?

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Identify the process of positron emission, which involves the conversion of a proton into a neutron, resulting in the emission of a positron (β⁺) and a neutrino.
Recognize that in positron emission, the atomic number decreases by 1 while the mass number remains unchanged.
Given the product is chromium-50 (Cr-50), which has an atomic number of 24, determine the original nucleus by increasing the atomic number by 1.
Calculate the atomic number of the original nucleus: 24 (Cr) + 1 = 25.
Identify the element with atomic number 25, which is manganese (Mn), and conclude that the original nucleus is manganese-50 (Mn-50).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Positron Emission

Positron emission is a type of radioactive decay in which a proton in the nucleus is transformed into a neutron, resulting in the emission of a positron (the antimatter counterpart of an electron) and a neutrino. This process decreases the atomic number of the nucleus by one while keeping the mass number unchanged, leading to the formation of a new element.
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Nuclear Decay and Stability

Nuclear decay refers to the process by which an unstable atomic nucleus loses energy by emitting radiation. The stability of a nucleus is influenced by the ratio of protons to neutrons; when this ratio is not optimal, the nucleus may undergo decay to achieve a more stable configuration. Understanding the stability of isotopes is crucial for predicting the products of decay processes.
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Chromium Isotopes

Chromium has several isotopes, with chromium-50 being one of them. Isotopes are variants of a chemical element that have the same number of protons but different numbers of neutrons. In the context of positron emission, identifying the parent isotope that decays to chromium-50 is essential for understanding the decay chain and the nuclear reactions involved.
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