To solve for the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], [tex]\(c\)[/tex], and [tex]\(d\)[/tex] in the nuclear reaction:
[tex]\[ { }_{92}^{235} U + { }_0^1 n \rightarrow { }_{56}^{141} Ba + { }_{cd}^{ab} X + 3\left({ }_0^1 n\right) \][/tex]
we need to ensure the conservation of both atomic mass (A) and atomic number (Z).
First, let's conserve the atomic mass (A):
- The left side of the equation has [tex]\(235\)[/tex] (from uranium) + [tex]\(1\)[/tex] (from the neutron), giving a total atomic mass of [tex]\(236\)[/tex].
- The right side must also sum to this total mass: [tex]\(141\)[/tex] (from barium) + [tex]\(a\)[/tex] (from X) + [tex]\(3 \times 1\)[/tex] (from the three neutrons, each with atomic mass 1).
Thus, the equation for the conservation of atomic mass becomes:
[tex]\[
235 + 1 = 141 + a + 3 \\
236 = 141 + a + 3 \\
236 = 144 + a \\
a = 236 - 144 \\
a = 92
\][/tex]
Now, let's conserve the atomic number (Z):
- The left side of the equation has [tex]\(92\)[/tex] (from uranium) + [tex]\(0\)[/tex] (from the neutron), giving a total atomic number of [tex]\(92\)[/tex].
- The right side must also sum to this total charge: [tex]\(56\)[/tex] (from barium) + [tex]\(c\)[/tex] (from X) + [tex]\(3 \times 0\)[/tex] (from the three neutrons, each with atomic number 0).
Thus, the equation for the conservation of atomic number becomes:
[tex]\[
92 + 0 = 56 + c + 0 \\
92 = 56 + c \\
c = 92 - 56 \\
c = 36
\][/tex]
Next, we need to identify [tex]\(b\)[/tex] and [tex]\(d\)[/tex] for element [tex]\( { }_{cd}^{ab} X \)[/tex]:
- Given [tex]\(a = 92\)[/tex] and [tex]\(c = 36\)[/tex], the element with atomic number 36 is Krypton (Kr).
- Thus, the mass number [tex]\(b\)[/tex] and the atomic number [tex]\(d\)[/tex] for Krypton are the same as those corresponding to the element's atomic number.
Therefore, the values are:
[tex]\[
a = 92 \\
b = 92 \\
c = 36 \\
d = 36
\][/tex]
So, the four-digit answer is:
[tex]\[ \boxed{9236} \][/tex]
