Question

part 2 (1 point) oxygen - 15 undergoes the decay shown in diagram (a). write a balanced equation for this nuclear decay. see hint part 3 (1 point) uranium - 238 undergoes the decay shown in diagram (b). write a balanced equation for this nuclear decay. see hint

Explanation:

Part 2: Oxygen - 15 Decay

Step 1: Identify Decay Type (Positron Emission)

Oxygen - 15 typically undergoes positron emission. In positron emission, a proton in the nucleus is converted into a neutron, and a positron ($_{+1}^0e$ or $_{+1}^0\beta$) and a neutrino are emitted. The general form of positron emission is $_{Z}^{A}X
ightarrow_{Z - 1}^{A}Y+_{+1}^0e+
u_e$. For Oxygen - 15, the atomic number $Z = 8$, mass number $A=15$. After positron emission, the new element will have $Z=8 - 1=7$ (Nitrogen) and mass number $A = 15$.

Step 2: Write the Balanced Equation

So the balanced nuclear equation is $_{8}^{15}\text{O}
ightarrow_{7}^{15}\text{N}+_{+1}^0e$ (or $_{8}^{15}\text{O}
ightarrow_{7}^{15}\text{N}+_{+1}^0\beta$).

Part 3: Uranium - 238 Decay

Step 1: Identify Decay Type (Alpha Decay)

From the diagram (with $_{2}^{4}\text{He}$ (alpha particle) being emitted), Uranium - 238 undergoes alpha decay. In alpha decay, the nucleus emits an alpha particle ($_{2}^{4}\text{He}$). The general form of alpha decay is $_{Z}^{A}X
ightarrow_{Z - 2}^{A - 4}Y+_{2}^{4}\text{He}$. For Uranium - 238, $Z = 92$, $A = 238$. After alpha decay, the new element will have $Z=92 - 2 = 90$ (Thorium) and $A=238 - 4=234$.

Step 2: Write the Balanced Equation

So the balanced nuclear equation is $_{92}^{238}\text{U}
ightarrow_{90}^{234}\text{Th}+_{2}^{4}\text{He}$.

Answer:

s:

• Part 2: $\boldsymbol{_{8}^{15}\text{O}

ightarrow_{7}^{15}\text{N}+_{+1}^0e}$ (or $_{8}^{15}\text{O}
ightarrow_{7}^{15}\text{N}+_{+1}^0\beta$)

• Part 3: $\boldsymbol{_{92}^{238}\text{U}

ightarrow_{90}^{234}\text{Th}+_{2}^{4}\text{He}}$