Question

name_________________________per____
the following atoms all undergo alpha particle emission. write the
complete nuclear equation.
$_{88}^{226}\text{ra}\
ightarrow\\ _{2}^{4}\text{he}+$ ______ $_{92}^{238}\text{u}\
ightarrow$ ____ + ____
the following atoms all undergo beta decay. write the complete nuclear
equation.
$_{14}^{32}\text{si}\
ightarrow\\ _{-1}^{0}\text{e}+$ ______ $_{55}^{137}\text{cs}\
ightarrow$ ____ + ____
the following undergo electron capture. write the complete nuclear
equation.
$_{48}^{109}\text{cd}+\\ _{-1}^{0}\text{e}\
ightarrow$ ______ $_{53}^{125}\text{i}+$ ______ $\
ightarrow$ ______
working backwards
$_{26}^{55}\text{fe}+$ ______ $\
ightarrow\\ _{25}^{55}\text{mn}$ ______ $\
ightarrow\\ _{91}^{233}\text{pa}+\\ _{2}^{4}\text{he}$

Explanation:

Step1: Recall Alpha Decay Rules

In alpha decay, the nucleus emits an alpha particle ($\ce{_2^4He}$). The mass number (top) decreases by 4, and the atomic number (bottom) decreases by 2.

Step2: Solve for $\ce{_{88}^{226}Ra}$ Decay

For $\ce{_{88}^{226}Ra}$: Mass number: $226 - 4 = 222$; Atomic number: $88 - 2 = 86$. The element with atomic number 86 is Radon (Rn), so the product is $\ce{_{86}^{222}Rn}$.

Step3: Solve for $\ce{_{92}^{238}U}$ Decay

For $\ce{_{92}^{238}U}$: Mass number: $238 - 4 = 234$; Atomic number: $92 - 2 = 90$. The element with atomic number 90 is Thorium (Th), so the equation is $\ce{_{92}^{238}U
ightarrow _{90}^{234}Th + _2^4He}$.

Step4: Recall Beta Decay Rules

In beta decay, a neutron turns into a proton, emitting a beta particle ($\ce{_{-1}^0e}$). Mass number stays the same, atomic number increases by 1.

Step5: Solve for $\ce{_{14}^{32}Si}$ Decay

For $\ce{_{14}^{32}Si}$: Mass number: 32 (unchanged); Atomic number: $14 + 1 = 15$. Element 15 is Phosphorus (P), so product is $\ce{_{15}^{32}P}$.

Step6: Solve for $\ce{_{55}^{137}Cs}$ Decay

For $\ce{_{55}^{137}Cs}$: Mass number: 137 (unchanged); Atomic number: $55 + 1 = 56$. Element 56 is Barium (Ba), so equation is $\ce{_{55}^{137}Cs
ightarrow _{56}^{137}Ba + _{-1}^0e}$.

Step7: Recall Electron Capture Rules

In electron capture, a proton captures an electron, turning into a neutron. Mass number stays the same, atomic number decreases by 1.

Step8: Solve for $\ce{_{48}^{109}Cd}$ Capture

For $\ce{_{48}^{109}Cd}$: Mass number: 109 (unchanged); Atomic number: $48 - 1 = 47$. Element 47 is Silver (Ag), so product is $\ce{_{47}^{109}Ag}$.

Step9: Solve for $\ce{_{53}^{125}I}$ Capture

For $\ce{_{53}^{125}I}$: It captures $\ce{_{-1}^0e}$. Mass number: 125 (unchanged); Atomic number: $53 - 1 = 52$. Element 52 is Tellurium (Te), so equation is $\ce{_{53}^{125}I + _{-1}^0e
ightarrow _{52}^{125}Te}$.

Step10: Solve "Working Backwards" Reactions

Reaction 1: $\ce{_{26}^{55}Fe + ?

ightarrow _{25}^{55}Mn}$
Mass number: 55 (Fe) +? = 55 (Mn) →? mass = 0. Atomic number: 26 +? = 25 →? atomic = -1. So the particle is $\ce{_{-1}^0e}$ (beta particle, or electron capture? Wait, Fe to Mn: atomic number decreases by 1, so Fe captures an electron? Wait, $\ce{_{26}^{55}Fe + _{-1}^0e
ightarrow _{25}^{55}Mn}$ (electron capture: proton + e⁻ → neutron, atomic number -1, mass number same).

Reaction 2: $\ce{?

ightarrow _{91}^{233}Pa + _2^4He}$
Mass number:? = 233 + 4 = 237. Atomic number:? = 91 + 2 = 93. Element 93 is Neptunium (Np), so $\ce{_{93}^{237}Np
ightarrow _{91}^{233}Pa + _2^4He}$ (alpha decay: Np emits alpha to become Pa).

Answer:

Alpha Decay:

1. $\ce{_{88}^{226}Ra

ightarrow _2^4He + \boxed{_{86}^{222}Rn}}$

1. $\ce{_{92}^{238}U

ightarrow \boxed{_{90}^{234}Th} + \boxed{_2^4He}}$

Beta Decay:

1. $\ce{_{14}^{32}Si

ightarrow _{-1}^0e + \boxed{_{15}^{32}P}}$

1. $\ce{_{55}^{137}Cs

ightarrow \boxed{_{56}^{137}Ba} + \boxed{_{-1}^0e}}$

Electron Capture:

1. $\ce{_{48}^{109}Cd + _{-1}^0e

ightarrow \boxed{_{47}^{109}Ag}}$

1. $\ce{_{53}^{125}I + \boxed{_{-1}^0e}

ightarrow \boxed{_{52}^{125}Te}}$

Working Backwards:

1. $\ce{_{26}^{55}Fe + \boxed{_{-1}^0e}

ightarrow _{25}^{55}Mn}$

1. $\ce{\boxed{_{93}^{237}Np}

ightarrow _{91}^{233}Pa + _2^4He}$