An excited ${ } _ { 92 } ^ { 236 } \mathrm { U } ^ { * }$ nucleus undergoes fission into two fragments as follows:
$${ } _ { 92 } ^ { 236 } \mathrm { U } * \rightarrow { } _ { 56 } ^ { 144 } \mathrm { Ba } + { } _ { 36 } ^ { 92 } \mathrm { Kr }$$
The following atomic masses are known:
${ } _ { 36 } ^ { 92 } \mathrm { Kr } : \quad 91.926270 \mathrm { u }$
${ } _ { 56 } ^ { 144 } \mathrm { Ba } : \quad 143.922845 \mathrm { u }$
${ } _ { 92 } ^ { 236 } \mathrm { U } ^ { * } : \quad 236.045563 \mathrm { u }$
What is the reaction energy released during this fission reaction? $\left( 1 \mathrm { u } = 931.5 \mathrm { MeV } / \mathrm { c } ^ { 2 } \right)$

Question

Quizplus · Accepted Answer

The reaction energy released during the fission reaction can be calculated using the mass defect, which is the difference between the mass of the original nucleus and the sum of the masses of the fission fragments. The mass defect ($\Delta m$) is given by:$$\Delta m = m_{\text{U}^*} - (m_{\text{Ba}} + m_{\text{Kr}})$$Substituting the given masses:$$\Delta m = 236.045563 \, \text{u} - (143.922845 \, \text{u} + 91.926270 \, \text{u})$$$$\Delta m = 236.045563 \, \text{u} - 235.849115 \, \text{u}$$$$\Delta m = 0.196448 \, \text{u}$$The energy released ($E$) can be calculated using the equation $E = \Delta m \times c^2$, where $c$ is the speed of light. Since $1 \, \text{u} = 931.5 \, \text{MeV}/c^2$, the energy released is:$$E = 0.196448 \, \text{u} \times 931.5 \, \text{MeV}/\text{u} = 182.96 \, \text{MeV}$$Rounding to the nearest 10 MeV, the energy released is approximately 180 MeV.

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An Excited ${ } _ { 92 } ^ { 236 } \mathrm { U } ^ { * }$