Question

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 $c^2 = 931 \, \text{MeV/u}$.

Answer 1

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: $208 \, \text{MeV}$