The equation for the alpha decay of 213At:
85213At --> 83209Bi + 24He
where the alpha particle is represented as a helium nucleus.
213Bi --> 209Tl + 4He, 213Po + e- 209Tl --> 209Pb + e- 213Po --> 209Pb + 4He 209Pb --> 209Bi + e- 209Bi --> 205Tl + 4He 205Tl, stable
When radium-226 undergoes alpha decay, it becomes radon-222. We write the equation like this: 88226Ra => 24He + 86222Rn Here we see the alpha particle written as a helium-4 nucleus, which is, in point of fact, what it is. Notice that the numbers that are subscripted are equal on both sides of the equation, and the superscripted numbers are as well. They must balance for your equation to be correct.
The alpha decay of 21385 At (Astatine-213) results in the production of 20981 Tl (Thallium-209) as the daughter element. During this process, an alpha particle, which consists of 2 protons and 2 neutrons, is emitted from the original atom, decreasing its atomic number by 2 and its mass number by 4.
The balanced nuclear reaction associated with Bi-213 decaying into Tl-209 involves the emission of an alpha particle (helium-4 nucleus) and results in the daughter nucleus Th-209. The balanced nuclear reaction is: Bi-213 -> Tl-209 + He-4.
Let x represent one of the numbers and x+1, the next consecutive number. Set up the equation and solve: x + x + 1 = 213 (next simplify) 2x + 1 = 213 (next subtract 1 from each side of the equation) 2x = 212 (next divide both sides by 2 to solve for x) x = 106 (this is the first number, the next consecutive number is 107 {x+1}) 106 + 107 = 213 (solution proved)
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85% of 213 = 85% * 213 = 0.85 * 213 = 181.05
312% of 213= 312% * 213= 3.12 * 213= 664.56
21% of 213= 21% * 213= 0.21 * 213= 44.73
43 ÷ 213 = 43/213 ≈ 0.202
213 - 0.4*213 = 213 - 85.2 = 127.8
The only composite factor of 213 is 213.