A generic fission event is
`""^(235)U+ntoX+Y+2n`
Which of the following pairs cannot represent X and Y: (a) `""^(141)Xe` and `""^(93)Sr` (b) `""^(139)Cs` and `""^(94)Rb` (c) `""^(156)Nd` and `""^(79)Ge` d. `""^(121)"In"` and `""^(113")Ru`?

Answer

To solve the problem, we need to analyze the fission event represented by the equation:

\[ ^{235}\text{U} + n \rightarrow X + Y + 2n \]

In this reaction, the total mass number on the left side must equal the total mass number on the right side.

1. **Calculate the total mass number on the left side:**
- The mass number of \(^{235}\text{U}\) 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. **Calculate the total mass number on the right side:**
- The right side consists of the mass numbers of \(X\), \(Y\), and 2 neutrons (2n).
- The mass number of the two neutrons is \(2 \times 1 = 2\).
- Thus, the total mass number on the right side is:
\[ \text{mass}(X) + \text{mass}(Y) + 2 \]

3. **Set up the equation:**
- For the reaction to balance, we set the left side equal to the right side:
\[ 236 = \text{mass}(X) + \text{mass}(Y) + 2 \]
- Simplifying this gives:
\[ \text{mass}(X) + \text{mass}(Y) = 236 - 2 = 234 \]

4. **Evaluate each option:**
- **Option (a): \(^{141}\text{Xe}\) and \(^{93}\text{Sr}\)**
\[ 141 + 93 = 234 \quad \text{(valid)} \]
- **Option (b): \(^{139}\text{Cs}\) and \(^{94}\text{Rb}\)**
\[ 139 + 94 = 233 \quad \text{(invalid)} \]
- **Option (c): \(^{156}\text{Nd}\) and \(^{79}\text{Ge}\)**
\[ 156 + 79 = 235 \quad \text{(invalid)} \]
- **Option (d): \(^{121}\text{In}\) and \(^{113}\text{Ru}\)**
\[ 121 + 113 = 234 \quad \text{(valid)} \]

5. **Conclusion:**
- The pairs that cannot represent \(X\) and \(Y\) are:
- (b) \(^{139}\text{Cs}\) and \(^{94}\text{Rb}\)
- (c) \(^{156}\text{Nd}\) and \(^{79}\text{Ge}\)

Thus, the final answer is that options (b) and (c) cannot represent \(X\) and \(Y\).

---