Certainly! Let's approach this step-by-step:
When a californium atom (Cl-252) undergoes nuclear fission, it typically splits into lighter elements along with a release of neutrons and energy. Given the options and the context, you need to identify the correct atomic numbers and mass numbers that balance the nuclear equation.
The initial and resulting particles in the nuclear reaction are:
1. A californium-252 atom ([tex]\( \ ^{252}_{88} \text{Cl} \)[/tex]) reacts with a neutron ([tex]\( \ ^{1}_{0} \text{n} \)[/tex]).
2. Products:
- A fragment with atomic number 43 and atomic mass 114.
- A fragment of tellurium-135 ([tex]\( \ ^{135}_{52} \text{Te} \)[/tex]).
- Three neutrons ([tex]\( \ ^{1}_{0} \text{n} \)[/tex]).
To balance the nuclear equation, the sum of the atomic numbers (protons) and mass numbers (neutrons and protons combined) on both the left and right sides must be equal.
Given information:
[tex]\[ _{98}^{252} \text{Cf} + \ _{0}^{1} \text{n} \rightarrow \ _{43}^{114} \text{Mo} + \ _{52}^{135} \text{Te} + 3 \ _{0}^{1} \text{n} \ \][/tex]
This gives us:
1. Total mass number on the left side: [tex]\(252 + 1 = 253 \)[/tex]
2. Total atomic number on the left side: [tex]\(98 + 0 = 98 \)[/tex]
Balancing the products:
- Mass numbers: [tex]\(114 (Mo) + 135 (Te) + 3 \cdot 1 (neutrons) = 114 + 135 + 3 = 252 \)[/tex]
- Atomic numbers: [tex]\(43 (Mo) + 52 (Te) + 0 = 95 \)[/tex]
If there were a missing product with mass number = 1 (1 neutron atom), it will balance both sides for fission nuclear mass conservation.
So, the balanced nuclear equation is:
[tex]\[ \ _{98}^{252} \text{Cf} + \ _{0}^{1} \text{n} \rightarrow \ _{53}^{114} \text{Mo} + \ _{52}^{135} \text{Te} + 3 \ _{0}^{1} \text{n} + \ _{1}^{n} \text{H} \ \][/tex]
