To solve this nuclear fission equation, we need to determine the missing element and its mass number and atomic number on the right-hand side of the equation. Let's break it down step-by-step:
1. Identify the given elements and particles:
- Left-hand side:
- Californium (Cf) with mass number 252 and atomic number 98: [tex]\({ }_{98}^{252} Cf\)[/tex]
- A neutron with mass number 1 and atomic number 0: [tex]\({ }_0^1 n\)[/tex]
- Right-hand side:
- Tellurium (Te) with mass number 136 and atomic number 52: [tex]\({ }_{52}^{136} Te\)[/tex]
- Three neutrons, each with mass number 1 and atomic number 0: [tex]\(3{ }_0^1 n\)[/tex]
2. Calculate the total mass number and atomic number on the left-hand side:
[tex]\[
\text{Total mass number (LHS)} = 252 + 1 = 253
\][/tex]
[tex]\[
\text{Total atomic number (LHS)} = 98 + 0 = 98
\][/tex]
3. Calculate the total mass number and atomic number contributed by known products on the right-hand side:
[tex]\[
\text{Mass number from Te and neutrons (RHS)} = 136 + 3 \times 1 = 136 + 3 = 139
\][/tex]
[tex]\[
\text{Atomic number from Te (RHS)} = 52
\][/tex]
4. Determine the mass number and atomic number of the missing fragment:
- Mass number of the missing fragment:
[tex]\[
\text{Mass number (missing)} = \text{Total mass number (LHS)} - \text{Mass number (RHS)} = 253 - 139 = 114
\][/tex]
- Atomic number of the missing fragment:
[tex]\[
\text{Atomic number (missing)} = \text{Total atomic number (LHS)} - \text{Atomic number (RHS)} = 98 - 52 = 46
\][/tex]
5. Identify the element with atomic number 46:
- The element with atomic number 46 is Palladium (Pd).
6. Complete the nuclear equation:
[tex]\[
{ }_{98}^{252} Cf +{ }_0^1 n \longrightarrow { }_{46}^{114} Pd + { }_{52}^{136} Te +3{ }_0^1 n
\][/tex]
So, the completed equation with the missing isotope filled in is:
[tex]\[
{ }_{98}^{252} Cf +{ }_0^1 n \longrightarrow { }_{46}^{114} Pd + { }_{52}^{136} Te +3{ }_0^1 n
\][/tex]
