To solve the nuclear reaction given:
\[ ^{238}{92}U \rightarrow ^{4}{2}He + ^{A}_{Z}X \]
We need to balance both the atomic numbers (Z) and mass numbers (A) on each side of the equation.
- The mass number on the left side is 238 (from U).
- The mass number on the right side is the mass of the helium nucleus (4) plus the mass of the unknown element (A). Therefore:
\[ 238 = 4 + A \]
Solving for A:
\[ A = 238 - 4 = 234 \]
- Next, we consider the atomic numbers:
- The atomic number of Uranium (U) is 92.
- The atomic number of Helium (He) is 2.
So, for the unknown element:
\[ 92 = 2 + Z \]
Solving for Z:
\[ Z = 92 - 2 = 90 \]
Now we have:
\[ A = 234 \] \[ Z = 90 \]
The element with atomic number 90 is Thorium (Th).
Thus, the missing element in the blank is:
Blank 1: Th