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Chapter 19: Problem 67

In the bismuth-214 natural decay series, Bi-214 initially undergoes \(\beta\) decay, the resulting daughter emits an \(\alpha\) particle, and the succeeding daughters emit a \(\beta\) and a \(\beta\) particle in that order. Determine the product of each step in the Bi-214 decay series.

Short Answer

Expert verified
The products of each step in the Bi-214 decay series are: \(Bi-214 \rightarrow Po-214 \rightarrow Pb-210 \rightarrow Bi-210 \rightarrow Po-210\).

Step by step solution

01

Bi-214 β- decay

In this step, Bi-214 undergoes β- decay, which means one neutron is converted into a proton and an electron is emitted. To conserve the number of protons and neutrons, the resulting daughter nuclide will have one more proton and one fewer neutron. Bi-214 has 83 protons, so since one neutron converts to a proton, the new nuclide will have 84 protons: Bi-214 (Z=83, A=214) -> Daughter nuclide (Z=84, A=214) Checking the periodic table, an element with 84 protons is Polonium (Po). Thus, the daughter nuclide is Po-214.
02

Po-214 α decay

In this step, Po-214 emits an α particle, which is comprised of 2 protons and 2 neutrons. To conserve the number of protons and neutrons, the daughter nuclide will have 2 fewer protons and 2 fewer neutrons: Po-214 (Z=84, A=214) -> Daughter nuclide (Z=82, A=210) Checking the periodic table, an element with 82 protons is Lead (Pb). Thus, the daughter nuclide is Pb-210.
03

Pb-210 β- decay

In this step, Pb-210 undergoes β- decay, converting one neutron to a proton and emitting an electron. The resulting daughter nuclide will have one more proton and one fewer neutron: Pb-210 (Z=82, A=210) -> Daughter nuclide (Z=83, A=210) Checking the periodic table, an element with 83 protons is Bismuth (Bi). Thus, the daughter nuclide is Bi-210.
04

Bi-210 β- decay

In this final step, Bi-210 undergoes another β- decay, again converting a neutron into a proton and emitting an electron. The resulting daughter nuclide will have one more proton and one fewer neutron: Bi-210 (Z=83, A=210) -> Daughter nuclide (Z=84, A=210) Checking the periodic table, an element with 84 protons is Polonium (Po). Thus, the final daughter nuclide is Po-210. In conclusion, the Bi-214 decay series consists of the following products: 1. Bi-214 -> Po-214 2. Po-214 -> Pb-210 3. Pb-210 -> Bi-210 4. Bi-210 -> Po-210

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