To determine the particle that completes the given nuclear equation:
[tex]\[
{}_{94}^{239} Pu + {}_0^1 n \rightarrow {}_{40}^{100} Zr + ? + 2 \cdot {}_0^1 n
\][/tex]
we need to follow these steps:
1. Calculate the total mass number and atomic number before and after the reaction:
- Before the reaction:
- Plutonium ([tex]\( {}_{94}^{239} Pu \)[/tex]): Mass number is 239, atomic number is 94.
- Neutron ([tex]\( {}_0^1 n \)[/tex]): Mass number is 1, atomic number is 0.
- Total mass number before reaction = 239 (Pu) + 1 (neutron) = 240.
- Total atomic number before reaction = 94 (Pu) + 0 (neutron) = 94.
- After the reaction:
- Zirconium ([tex]\( {}_{40}^{100} Zr \)[/tex]): Mass number is 100, atomic number is 40.
- Two neutrons ([tex]\( 2 \cdot {}_0^1 n \)[/tex]): Mass number is [tex]\( 2 \cdot 1 = 2 \)[/tex], atomic number is [tex]\( 2 \cdot 0 = 0 \)[/tex].
- Let's denote the unknown particle with its mass number [tex]\( A \)[/tex] and atomic number [tex]\( Z \)[/tex].
- Total mass number after the reaction = 100 (Zr) + 2 (from 2 neutrons) + [tex]\( A \)[/tex].
- Total atomic number after the reaction = 40 (Zr) + 0 (from 2 neutrons) + [tex]\( Z \)[/tex].
2. Establish equation for conservation of mass and charge:
- Using conservation of mass number:
[tex]\[
240 = 100 + 2 + A \implies A = 240 - 102 \implies A = 138
\][/tex]
- Using conservation of atomic number:
[tex]\[
94 = 40 + 0 + Z \implies Z = 94 - 40 \implies Z = 54
\][/tex]
So, we deduced that the mass number [tex]\( A \)[/tex] of the unknown particle is 138 and the atomic number [tex]\( Z \)[/tex] is 54.
3. Determine the identity of the particle:
- A particle with atomic number [tex]\( Z = 54 \)[/tex] and mass number [tex]\( A = 138 \)[/tex] is Xenon (Xe).
4. Compare with the given options to find the correct one:
- [tex]\({ }_{54}^{138} Xe\)[/tex]
- [tex]\({ }_{54}^{139} Xe\)[/tex]
- [tex]\({ }_{54}^{119} Xe\)[/tex]
- [tex]\({ }_4^{138} Te\)[/tex]
Therefore, the correct option is [tex]\({ }_{54}^{138} Xe\)[/tex].
