Question

The disintegration energy Q for the nuclear fission of ²³⁵U → ¹⁴⁰Ce + ⁹⁴Zr + n is _____ MeV.

Given atomic masses:
²³⁵U = 235.0439 u 
¹⁴⁰Ce = 139.9054 u 
⁹⁴Zr = 93.9063 u 
n = 1.0086 u 

Value of c² = 931 MeV/u.

Answer 1

Correct Answer: 208

1. Total Reactant Mass (\(m_r\)):
\[ m_r = 235.0439 \, \text{u}. \]

2. Total Product Mass (\(m_p\)):
\[ m_p = 139.9054 + 93.9063 + 1.0086 = 234.8203 \, \text{u}. \]

3. Disintegration Energy (\(Q\)):
Calculated using the formula:
\[ Q = (m_r - m_p)c^2. \] 

Substitution:
\[ Q = (235.0439 - 234.8203) \times 931. \] 

Simplification:
\[ Q = 0.2236 \times 931 = 208.1716 \, \text{MeV}. \] 

Result: \(208 \, \text{MeV}\)