Question

1. when isotope bismuth-213 emits an alpha particle:

a. write out the nuclear equation:
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b. which is the parent element?
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c. which is the daughter element?
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d. what new element results if the isotope, instead, emits a beta particle?
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Explanation:

Step1: Recall Alpha Decay Rules

Alpha decay involves emitting a helium nucleus ($\alpha$ particle: $^4_2\text{He}$). The parent isotope is bismuth-213, with atomic number 83 (Bi: $Z = 83$, $A = 213$). In alpha decay, mass number ($A$) decreases by 4, atomic number ($Z$) by 2.

Step2: Write Nuclear Equation

Parent: $^{213}_{83}\text{Bi}$. After emitting $^4_2\text{He}$, daughter has $A = 213 - 4 = 209$, $Z = 83 - 2 = 81$. Element with $Z = 81$ is thallium (Tl). So equation: $^{213}_{83}\text{Bi}
ightarrow ^{209}_{81}\text{Tl} + ^4_2\text{He}$.

Step3: Identify Parent Element

Parent is the original isotope: bismuth (Bi), from $^{213}_{83}\text{Bi}$.

Step4: Identify Daughter Element

Daughter is the product: thallium (Tl), from $^{209}_{81}\text{Tl}$.

Step5: Beta Decay (Part d)

Beta decay emits an electron ($^0_{-1}\text{e}$). Mass number ($A$) stays same, atomic number ($Z$) increases by 1. For Bi-213 ($Z = 83$), new $Z = 83 + 1 = 84$. Element with $Z = 84$ is polonium (Po), so $^{213}_{83}\text{Bi}
ightarrow ^{213}_{84}\text{Po} + ^0_{-1}\text{e}$.

Answer:

s:
a. Nuclear Equation: $\boldsymbol{^{213}_{83}\text{Bi}
ightarrow ^{209}_{81}\text{Tl} + ^4_2\text{He}}$
b. Parent Element: Bismuth (Bi)
c. Daughter Element: Thallium (Tl)
d. New Element (Beta Decay): Polonium (Po) (Equation: $\boldsymbol{^{213}_{83}\text{Bi}
ightarrow ^{213}_{84}\text{Po} + ^0_{-1}\text{e}}$)