# Chapter 29 - Problems - Page 1117: 22

We can write the reaction: $^{232}_{90}Th \rightarrow ^{228}_{88}Ra+^{4}_{2}He$ The daughter nuclide is $~~^{228}_{88}Ra$

#### Work Step by Step

During $\alpha$ decay, an alpha particle (two protons and two neutrons) are released from the nucleus. Therefore, in the daughter nuclide, the total number of nucleons decreases by 4, and the number of protons decreases by 2. The daughter nuclide has 228 nucleons and 88 protons. The element with 88 protons in the nucleus is radium. We can write the reaction: $^{232}_{90}Th \rightarrow ^{228}_{88}Ra+^{4}_{2}He$ The daughter nuclide is $~~^{228}_{88}Ra$.

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