The parent nuclide of the thorium decay series is \(_{90}^{232}\) Th. The first four decays are as follows: alpha emission, beta emission, beta emission, and alpha emission. Write the nuclear equations for this series of emissions.

Short Answer

Expert verified
The nuclear equations representing the four stages of decay are: \(_{90}^{232}\) Th --> \(_{88}^{228}\) Ra + \(\alpha_1\), \(_{88}^{228}\) Ra --> \(_{89}^{228}\) Ac + \(\beta_1\), \(_{89}^{228}\) Ac --> \(_{90}^{228}\) Th + \(\beta_2\), and \(_{90}^{228}\) Th --> \(_{88}^{224}\) Ra + \(\alpha_2\).

Step by step solution

01

Alpha emission 1

The first decay is an alpha emission. This means that the atomic number will reduce by 2 and the mass number by 4. Therefore, the product of the first decay is \(_{88}^{228}\) Ra.
02

Beta emission 1

The second decay is a beta emission. This implies that the atomic number will increase by 1 with no change in mass number. Therefore, the product of the second decay is \(_{89}^{228}\) Ac.
03

Beta emission 2

The third decay is also a beta emission. Like in the previous step, the atomic number will increase by 1 with no change in mass number. Therefore, the product of the third decay is \(_{90}^{228}\) Th.
04

Alpha emission 2

The fourth decay is an alpha emission. This means the atomic number will decrease by 2 and the mass number by 4. Therefore, the product of the fourth decay is \(_{88}^{224}\) Ra.

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