Question

radioactive decay practice. one of each type of question will appear on the test. add the atomic number in the line beneath the atomic symbol.
alpha decay
$_{86}^{222}\text{rn} \
ightarrow \\, _2^4\text{he} + \underline{\quad\quad}$
$_{90}^{238}\text{th} \
ightarrow \underline{\quad\quad} + \\, _{88}^{226}\text{ra}$
$_{92}^{235}\text{u} \
ightarrow \\, _2^4\text{he} + \underline{\quad\quad}$
$_{94}^{239}\text{pu} \
ightarrow \underline{\quad\quad} + \underline{\quad\quad}$
beta decay
$_{14}^{32}\text{si} \
ightarrow \\, _{-1}^0\text{e} + \underline{\quad\quad}$
$_{38}^{89}\text{sr} \
ightarrow \underline{\quad\quad} + \underline{\quad\quad}$
$_{79}^{198}\text{au} \
ightarrow \\, _{-1}^0\text{e} + \underline{\quad\quad}$
$_{53}^{131}\text{i} \
ightarrow \underline{\quad\quad} + \underline{\quad\quad}$
electron capture
$_{27}^{57}\text{co} + \\, _{-1}^0\text{e} \
ightarrow \underline{\quad\quad}$
$_{31}^{68}\text{ga} + \underline{\quad\quad} \
ightarrow \underline{\quad\quad}$
$_{77}^{192}\text{co} + \\, _{-1}^0\text{e} \
ightarrow \underline{\quad\quad}$
$_{56}^{133}\text{ba} + \underline{\quad\quad} \
ightarrow \underline{\quad\quad}$
working backwards
$\underline{\quad\quad} \
ightarrow \\, _{56}^{140}\text{ba} + \\, _{-1}^0\text{e}$
$\underline{\quad\quad} \
ightarrow \\, _{90}^{230}\text{th} + \\, _2^4\text{he}$
$_{10}^{23}\text{ne} \
ightarrow \\, _{11}^{23}\text{na} + \underline{\quad\quad}$
$\underline{\quad\quad} + \\, _{-1}^0\text{e} \
ightarrow \\, _{79}^{200}\text{au}$

Explanation:

Step1: Recall Alpha Decay Rules

In alpha decay, the parent nucleus emits an alpha particle ($\ce{_2^4He}$). The mass number (top number) decreases by 4, and the atomic number (bottom number) decreases by 2. For $\ce{_{92}^{235}U \to _2^4He + ?}$:

• Mass number: $235 - 4 = 231$
• Atomic number: $92 - 2 = 90$

So the daughter nucleus is $\ce{_{90}^{231}Th}$.

Step2: Recall Beta Decay Rules

In beta decay (β⁻ decay), a neutron turns into a proton, emitting an electron ($\ce{_{-1}^0e}$). Mass number stays the same, atomic number increases by 1. For $\ce{_{14}^{32}Si \to _{-1}^0e + ?}$:

• Mass number: $32$ (unchanged)
• Atomic number: $14 + 1 = 15$

So the daughter nucleus is $\ce{_{15}^{32}P}$.

Step3: Recall Electron Capture Rules

In electron capture, a proton captures an electron ($\ce{_{-1}^0e}$), turning into a neutron. Mass number stays the same, atomic number decreases by 1. For $\ce{_{27}^{57}Co + _{-1}^0e \to ?}$:

• Mass number: $57$ (unchanged)
• Atomic number: $27 - 1 = 26$

So the daughter nucleus is $\ce{_{26}^{57}Fe}$.

Step4: Working Backwards (Beta Decay Example)

For $\ce{? \to _{56}^{140}Ba + _{-1}^0e}$ (beta decay, so parent has mass number 140, atomic number $56 - 1 = 55$). So the parent is $\ce{_{55}^{140}Cs}$.

Answer:

(Examples of filled answers:
Alpha decay: $\ce{_{92}^{235}U \to _2^4He + \ce{_{90}^{231}Th}}$;
Beta decay: $\ce{_{14}^{32}Si \to _{-1}^0e + \ce{_{15}^{32}P}}$;
Electron capture: $\ce{_{27}^{57}Co + _{-1}^0e \to \ce{_{26}^{57}Fe}}$;
Working backwards: $\ce{_{55}^{140}Cs \to _{56}^{140}Ba + _{-1}^0e}$)

(Note: Each blank follows the respective decay rules for mass/atomic number conservation.)