## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 29 - Problems - Page 1117: 24

#### Answer

We can write the reaction: $^{22}_{11}Na \rightarrow ^{22}_{10}Ne + ^{0}_{+1}e$ The daughter nuclide is $~~^{22}_{10}Ne$

#### Work Step by Step

During positron emission, an atom releases a positron and a proton is converted to a neutron. Therefore, in the daughter nuclide, the total number of nucleons stays the same, and the number of protons decreases by 1. The daughter nuclide has 22 nucleons and 10 protons. The element with 10 protons in the nucleus is neon. We can write the reaction: $^{22}_{11}Na \rightarrow ^{22}_{10}Ne + ^{0}_{+1}e$ The daughter nuclide is $~~^{22}_{10}Ne$

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