Question
Write the nuclear equation for the following radioactive decay: a) Ra 223 emits an alpha b) Po 215 emits an alpha and Gamma c) the product from (b) emits a beta the product from (c) emits a beta and Gamma
Answers
GPT-4o mini
To write the nuclear equations for the given radioactive decays, we need to follow the conventions for alpha (α) and beta (β) emissions.
### a) Ra-223 emits an alpha particle
Radium-223 (^223Ra) emits an alpha particle (α), which is equivalent to a helium nucleus ( ^4He). The equation is:
\[
^{223}_{88}\text{Ra} \rightarrow \, ^{219}_{86}\text{Rn} + \, ^{4}_{2}\text{He}
\]
Here, Ra-223 decays into Radon-219 (^219Rn) and an alpha particle (^4He).
---
### b) Po-215 emits an alpha and Gamma
Polonium-215 (^215Po) emits an alpha particle (α) and a gamma ray (γ). The equation is:
\[
^{215}_{84}\text{Po} \rightarrow \, ^{211}_{82}\text{Pb} + \, ^{4}_{2}\text{He} + \gamma
\]
In this step, Polonium-215 decays to Lead-211 (^211Pb) and emits an alpha particle and gamma radiation.
---
### c) The product from (b) emits a beta
Now we take the product from (b), which is Lead-211 (^211Pb), and it emits a beta particle (β). The equation is:
\[
^{211}_{82}\text{Pb} \rightarrow \, ^{211}_{83}\text{Bi} + \, ^{0}_{-1}\text{e}
\]
Here, Lead-211 decays to Bismuth-211 (^211Bi) while emitting a beta particle.
---
### d) The product from (c) emits a beta and Gamma
The next step involves Bismuth-211 (^211Bi) emitting another beta particle and gamma ray. The equation is:
\[
^{211}_{83}\text{Bi} \rightarrow \, ^{211}_{84}\text{Po} + \, ^{0}_{-1}\text{e} + \gamma
\]
In this last decay, Bismuth-211 decays into Polonium-211 (^211Po), emitting a beta particle and gamma radiation.
---
To summarize, the complete sequence of nuclear decay equations is:
1. \(^{223}_{88}\text{Ra} \rightarrow \, ^{219}_{86}\text{Rn} + \, ^{4}_{2}\text{He}\)
2. \(^{215}_{84}\text{Po} \rightarrow \, ^{211}_{82}\text{Pb} + \, ^{4}_{2}\text{He} + \gamma\)
3. \(^{211}_{82}\text{Pb} \rightarrow \, ^{211}_{83}\text{Bi} + \, ^{0}_{-1}\text{e}\)
4. \(^{211}_{83}\text{Bi} \rightarrow \, ^{211}_{84}\text{Po} + \, ^{0}_{-1}\text{e} + \gamma\)
### a) Ra-223 emits an alpha particle
Radium-223 (^223Ra) emits an alpha particle (α), which is equivalent to a helium nucleus ( ^4He). The equation is:
\[
^{223}_{88}\text{Ra} \rightarrow \, ^{219}_{86}\text{Rn} + \, ^{4}_{2}\text{He}
\]
Here, Ra-223 decays into Radon-219 (^219Rn) and an alpha particle (^4He).
---
### b) Po-215 emits an alpha and Gamma
Polonium-215 (^215Po) emits an alpha particle (α) and a gamma ray (γ). The equation is:
\[
^{215}_{84}\text{Po} \rightarrow \, ^{211}_{82}\text{Pb} + \, ^{4}_{2}\text{He} + \gamma
\]
In this step, Polonium-215 decays to Lead-211 (^211Pb) and emits an alpha particle and gamma radiation.
---
### c) The product from (b) emits a beta
Now we take the product from (b), which is Lead-211 (^211Pb), and it emits a beta particle (β). The equation is:
\[
^{211}_{82}\text{Pb} \rightarrow \, ^{211}_{83}\text{Bi} + \, ^{0}_{-1}\text{e}
\]
Here, Lead-211 decays to Bismuth-211 (^211Bi) while emitting a beta particle.
---
### d) The product from (c) emits a beta and Gamma
The next step involves Bismuth-211 (^211Bi) emitting another beta particle and gamma ray. The equation is:
\[
^{211}_{83}\text{Bi} \rightarrow \, ^{211}_{84}\text{Po} + \, ^{0}_{-1}\text{e} + \gamma
\]
In this last decay, Bismuth-211 decays into Polonium-211 (^211Po), emitting a beta particle and gamma radiation.
---
To summarize, the complete sequence of nuclear decay equations is:
1. \(^{223}_{88}\text{Ra} \rightarrow \, ^{219}_{86}\text{Rn} + \, ^{4}_{2}\text{He}\)
2. \(^{215}_{84}\text{Po} \rightarrow \, ^{211}_{82}\text{Pb} + \, ^{4}_{2}\text{He} + \gamma\)
3. \(^{211}_{82}\text{Pb} \rightarrow \, ^{211}_{83}\text{Bi} + \, ^{0}_{-1}\text{e}\)
4. \(^{211}_{83}\text{Bi} \rightarrow \, ^{211}_{84}\text{Po} + \, ^{0}_{-1}\text{e} + \gamma\)