Problem

Find the energy released in the given fission reaction:

$${}_{0}^{1}\mathrm{n} + {}_{92}^{235}\mathrm{U} \rightarrow {}_{36}^{92}\mathrm{Kr} + {}_{56}^{141}\mathrm{Ba} + 3 {}_{0}^{1}\mathrm{n} + Q$$

Data

• Mass of reactants:
  • Neutron: $m_{{}_{0}^{1}\mathrm{n}} = 1.0087 \, \mathrm{u}$
  • Uranium: $m_{{}_{92}^{235}\mathrm{U}} = 235.0439 \, \mathrm{u}$
  • Total mass of reactants:

$$m_{\text{reactants}} = 236.0526 \, \mathrm{u}$$

• Mass of products:
  • Krypton: $m_{{}_{36}^{92}\mathrm{Kr}} = 91.8973 \, \mathrm{u}$
  • Barium: $m_{{}_{56}^{141}\mathrm{Ba}} = 140.9139 \, \mathrm{u}$
  • Neutrons: $3 \times m_{{}_{0}^{1}\mathrm{n}} = 3.0261 \, \mathrm{u}$
  • Total mass of products:

$$m_{\text{products}} = 235.8373 \, \mathrm{u}$$

• Mass-energy equivalence: $1 \, \mathrm{u} = 931.5 \, \mathrm{MeV}$

Prerequisite Concepts

1. Mass Defect ($\Delta m$):
   The difference between the total mass of reactants and the total mass of products.

$$\Delta m = m_{\text{reactants}} - m_{\text{products}}$$

2. Energy Released ($Q$):
   The energy released during the reaction is given by the mass-energy equivalence relation:

$$Q = \Delta m \cdot 931.5 \, \mathrm{MeV}$$

Solution

Step 1: Calculate the Mass Defect ($\Delta m$)

$$\Delta m = m_{\text{reactants}} - m_{\text{products}}$$

Substitute the given values:

$$\Delta m = 236.0526 \, \mathrm{u} - 235.8373 \, \mathrm{u}$$ $$\Delta m = 0.2153 \, \mathrm{u}$$

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Step 2: Calculate the Energy Released ($Q$)

Using the formula:

$$Q = \Delta m \cdot 931.5 \, \mathrm{MeV}$$

Substitute $\Delta m = 0.2153 \, \mathrm{u}$:

$$Q = 0.2153 \cdot 931.5$$ $$Q = 200.55 \, \mathrm{MeV}$$

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Answer

The energy released in the given fission reaction is:

$$Q = 200.55 \, \mathrm{MeV}$$

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