Of course! Let's complete the given nuclear fusion equation step-by-step.
The original equation is:
[tex]\[ {}_{88}^{226} Ra \rightarrow {}_{B}^{Rn} + {}_{2}^{4} He \][/tex]
We know that in nuclear reactions, both the atomic number (subscript) and the mass number (superscript) must be conserved. Let's use this principle to find the missing atomic and mass numbers.
### Step-by-Step Solution
1. Identify the Initial Atomic and Mass Numbers:
- The original Radium (Ra) has an atomic number of 88 and a mass number of 226.
[tex]\[ {}_{88}^{226} Ra \][/tex]
2. Identify the Known Product (Helium):
- The Helium (He) nucleus produced has an atomic number of 2 and a mass number of 4.
[tex]\[ {}_{2}^{4} He \][/tex]
3. Use the Conservation of Atomic Number:
- The total atomic number before the reaction is 88 (from Radium).
- After the reaction, the atomic numbers of the products must sum up to 88.
[tex]\[ 88 = B + 2 \][/tex]
- Solving for [tex]\( B \)[/tex] (atomic number of Rn):
[tex]\[ B = 88 - 2 \][/tex]
[tex]\[ B = 86 \][/tex]
4. Use the Conservation of Mass Number:
- The total mass number before the reaction is 226 (from Radium).
- After the reaction, the mass numbers of the products must sum up to 226.
[tex]\[ 226 = R + 4 \][/tex]
- Solving for [tex]\( R \)[/tex] (mass number of Rn):
[tex]\[ R = 226 - 4 \][/tex]
[tex]\[ R = 222 \][/tex]
### Conclusion
The completed nuclear fusion equation is:
[tex]\[ {}_{88}^{226} Ra \rightarrow {}_{86}^{222} Rn + {}_{2}^{4} He \][/tex]
So, the values are:
- [tex]\( B = 86 \)[/tex]
- [tex]\( R = 222 \)[/tex]
Thus, the correct answer is:
A: [tex]\( 86 \)[/tex]
B: [tex]\( 222 \)[/tex]
