Complete the equation for the following fission process `._92 U^235 ._0 n^1 rarr ._38 Sr^90 +....`.

Options

A. `._54 Xe^143 + 3 ._0 n^1`
B. `._54 Xe^145`
C. `._57 Xe^142`
D. `._54 Xe^142 ._0 n^1`

Explanation

To complete the fission process equation given by:

\[ \ _{92}^{235}U + \ _{0}^{1}n \rightarrow \ _{38}^{90}Sr + \ ? \]

we need to determine the missing product of the fission reaction.

### Step 1: Identify the initial values
- Uranium-235 has an atomic number (Z) of 92 and a mass number (A) of 235.
- A neutron (n) with an atomic number of 0 and mass number of 1 is added.

### Step 2: Calculate the total atomic number and mass number
- The total atomic number before the reaction is:
\[ Z_{\text{initial}} = 92 + 0 = 92 \]
- The total mass number before the reaction is:
\[ A_{\text{initial}} = 235 + 1 = 236 \]

### Step 3: Write down the known products
- One of the products is Strontium-90, which has:
\[ Z_{Sr} = 38 \quad \text{and} \quad A_{Sr} = 90 \]

### Step 4: Set up the equation for conservation of atomic numbers
- Let the unknown product be represented as \(_{Z}^{A}X\).
- The conservation of atomic numbers gives us:
\[ Z_{\text{initial}} = Z_{Sr} + Z_{X} \]
\[ 92 = 38 + Z_{X} \]
Solving for \(Z_{X}\):
\[ Z_{X} = 92 - 38 = 54 \]

### Step 5: Set up the equation for conservation of mass numbers
- The conservation of mass numbers gives us:
\[ A_{\text{initial}} = A_{Sr} + A_{X} \]
\[ 236 = 90 + A_{X} \]
Solving for \(A_{X}\):
\[ A_{X} = 236 - 90 = 146 \]

### Step 6: Identify the element with \(Z = 54\)
- The element with atomic number 54 is Xenon (Xe).

### Step 7: Write the complete fission equation
- The complete fission reaction is:
\[ \ _{92}^{235}U + \ _{0}^{1}n \rightarrow \ _{38}^{90}Sr + \ _{54}^{146}Xe + 3 \ _{0}^{1}n \]
Here, we also include the three additional neutrons that are typically released during the fission process.

### Final Answer:
\[ \ _{92}^{235}U + \ _{0}^{1}n \rightarrow \ _{38}^{90}Sr + \ _{54}^{146}Xe + 3 \ _{0}^{1}n \]

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