The disintegration energy \( Q \) for the nuclear fission of \( ^{235}\text{U} \to ^{140}\text{Ce} + ^{94}\text{Zr} + n \) is _____ MeV.
Given atomic masses:
\[^{235}\text{U} = 235.0439 \, u, \quad ^{140}\text{Ce} = 139.9054 \, u, \quad ^{94}\text{Zr} = 93.9063 \, u, \quad n = 1.0086 \, u\]Value of \( c^2 = 931 \, \text{MeV/u} \).
Given atomic masses:
\[^{235}\text{U} = 235.0439 \, u, \quad ^{140}\text{Ce} = 139.9054 \, u, \quad ^{94}\text{Zr} = 93.9063 \, u, \quad n = 1.0086 \, u\]Value of \( c^2 = 931 \, \text{MeV/u} \).
Correct Answer: 208
Solution and Explanation
1. Calculate Total Mass of Reactants (\(m_r\)):
\[ m_r = 235.0439 \, \text{u}. \]
2. Calculate Total Mass of Products (\(m_p\)):
\[ m_p = 139.9054 + 93.9063 + 1.0086 = 234.8203 \, \text{u}. \]
3. Calculate Disintegration Energy (\(Q\)):
The disintegration energy \(Q\) is given by:
\[ Q = (m_r - m_p)c^2. \]
Substitute the values:
\[ Q = (235.0439 - 234.8203) \times 931. \]
Simplify:
\[ Q = 0.2236 \times 931 = 208.1716 \, \text{MeV}. \]
Answer: \(208 \, \text{MeV}\)
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