Identify the missingf product in the given reaction `""_(92)""^(235)Uto""_(0)""^(1)nto?+""_(36)""^(92)Kr+3""_(0)""^(1)n` a)`B""_(56)""^(143)` b)`""_(56)""^(139)Ba` c)`""_(54)""^(139)Ba` d)`""_(54)""^(141)Ba`

Options

A. `._(56)^(141)Ba`
B. `._(56)^(130)Ba`
C. `_(54)^(139)Ba`
D. `._(54)^(141)Ba`

Answer

To solve the problem, we need to identify the missing product in the nuclear reaction given:

\[ _{92}^{235}\text{U} + _{0}^{1}\text{n} \rightarrow ? + _{36}^{92}\text{Kr} + 3 \, _{0}^{1}\text{n} \]

### Step-by-Step Solution:

1. **Identify the Total Mass Number on the Left Side:**
- The mass number of Uranium (U) is 235.
- The mass number of the neutron (n) is 1.
- Therefore, the total mass number on the left side is:
\[ 235 + 1 = 236 \]

2. **Identify the Total Mass Number on the Right Side:**
- The mass number of Krypton (Kr) is 92.
- Each neutron has a mass number of 1, and there are 3 neutrons.
- Therefore, the total mass number on the right side is:
\[ 92 + 3 \times 1 = 92 + 3 = 95 \]

3. **Set Up the Equation for Mass Numbers:**
- Let the mass number of the missing product be \( A \).
- According to the conservation of mass number:
\[ A + 95 = 236 \]
- Solving for \( A \):
\[ A = 236 - 95 = 141 \]

4. **Identify the Total Atomic Number on the Left Side:**
- The atomic number of Uranium (U) is 92.
- The atomic number of the neutron (n) is 0.
- Therefore, the total atomic number on the left side is:
\[ 92 + 0 = 92 \]

5. **Identify the Total Atomic Number on the Right Side:**
- The atomic number of Krypton (Kr) is 36.
- Each neutron has an atomic number of 0, and there are 3 neutrons.
- Therefore, the total atomic number on the right side is:
\[ 36 + 3 \times 0 = 36 \]

6. **Set Up the Equation for Atomic Numbers:**
- Let the atomic number of the missing product be \( Z \).
- According to the conservation of atomic number:
\[ Z + 36 = 92 \]
- Solving for \( Z \):
\[ Z = 92 - 36 = 56 \]

7. **Identify the Missing Product:**
- We have determined that the missing product has a mass number of 141 and an atomic number of 56.
- This corresponds to the element Barium (Ba), which has the symbol \( _{56}^{141}\text{Ba} \).

### Conclusion:
The missing product in the reaction is:
\[ \boxed{_{56}^{141}\text{Ba}} \]