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The thorium 232, radioactive decay series, beginning with thorium 232 with atomic number of 90, and ending with lead, 208.
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Of course, in the following sequence, alpha decay, followed by 2 beta decay, and then 4 alpha decay, 2 beta decay, and 1 alpha decay.
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So we need to write an equation for each step in this series.
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So, of course, we need to start first with the alpha decay of thorium.
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232.
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Take note that in alpha decay, an alpha particle is produced, which is represented by alpha symbol with atomic number of 2 and mass number of 4.
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So let me write that down first.
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Thorium 232 with a mass number of 232 and atomic number of 90.
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It's producing an alpha particle with atomic number of 2 and mass number of 4.
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So to balance this equation, we need to add an element, say, x.
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And then for the mass number, we simply subtract 4 from 2, 3, 2 .2, so we'll get 288.
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And then for the atomic number, let's subtract 2 from 90, then we'll get 88.
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Then to find out the identity of x, we use the periodic table.
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And based on the periodic table, the element with atomic number of 88 is radium represented as ra.
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Okay, let me center this one.
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So we have more space.
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And then for the beta decay, the radium 288 will undergo beta decay.
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So take note that in beta decay, beta particle represented as beta with atomic number of negative.
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1 and mass number of 0 is 3d.
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So let me write it down.
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So we have a radium 228 -28 -i -mean and atomic number of 88.
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So it produces alpha, i mean beta particle during beta -d -k...