Problem 67
Question
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.
Step-by-Step Solution
Verified Answer
The products of each step in the Bi-214 decay series are: \(Bi-214 \rightarrow Po-214 \rightarrow Pb-210 \rightarrow Bi-210 \rightarrow Po-210\).
1Step 1: 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.
2Step 2: 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.
3Step 3: 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.
4Step 4: 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
Key Concepts
Beta DecayAlpha DecayRadioactive Decay Process
Beta Decay
Beta decay is an interesting process that involves the transformation of a neutron into a proton, or vice versa, within an atomic nucleus. This process results in the emission of a beta particle, which can be an electron ( \( \beta^- \) decay) or a positron ( \( \beta^+ \) decay). In the context of the Bi-214 decay series, we focus on \( \beta^- \) decay, where an electron is emitted.
Here's what happens during \( \beta^- \) decay:
Understanding beta decay is vital for grasping how elements transform into their daughter products in radioactive series.
Here's what happens during \( \beta^- \) decay:
- One neutron in the nucleus converts into a proton.
- An electron, known as a beta particle, is emitted from the nucleus.
- This results in an increase in the atomic number since there is one more proton.
Understanding beta decay is vital for grasping how elements transform into their daughter products in radioactive series.
Alpha Decay
Alpha decay is a type of radioactive decay where an unstable nucleus releases an alpha particle. An alpha particle consists of two protons and two neutrons, equivalent to the nucleus of a helium atom. This process reduces the atomic number by 2 and the mass number by 4, turning the parent atom into a different element.
During the alpha decay of Po-214 in the bismuth-214 decay series:
During the alpha decay of Po-214 in the bismuth-214 decay series:
- The number of protons decreases by two.
- The number of neutrons decreases by two, too.
- The daughter nuclide is lead-210 (Pb-210), as the atomic number shifts from 84 to 82.
Radioactive Decay Process
Radioactive decay is the process by which an unstable atomic nucleus loses energy by radiation. Understanding this process means diving into a sequence of transformations that leads to stability over time.
In the Bi-214 decay series:
The radioactive decay process elucidates the foundational principles of nuclear chemistry and physics, showcasing how substances evolve at an atomic level.
In the Bi-214 decay series:
- It begins with beta decay, shifting from Bi-214 to Po-214.
- This is followed by alpha decay, transforming Po-214 to Pb-210.
- Subsequent beta decays return Pb-210 to Bi-210 and finally back to Po-210.
The radioactive decay process elucidates the foundational principles of nuclear chemistry and physics, showcasing how substances evolve at an atomic level.
Other exercises in this chapter
Problem 65
Predict whether each of the following nuclides is stable or unstable (radioactive). If the nuclide is unstable, predict the type of radioactivity you would expe
View solution Problem 66
Each of the following isotopes has been used medically for the purpose indicated. Suggest reasons why the particular element might have been chosen for this pur
View solution Problem 68
The curie (Ci) is a commonly used unit for measuring nuclear radioactivity: 1 curie of radiation is equal to \(3.7 \times 10^{10}\) decay events per second (the
View solution Problem 71
Define third-life in a similar way to half-life, and determine the "third- life" for a nuclide that has a half-life of 31.4 years.
View solution