An excited $^{236} _{92}$ U^∗ nucleus undergoes fission into two fragments as follows: $^{236} _{92}$ U^∗ → $\begin{array} { c } 144 \\56\end{array}$ Ba + ${}^{92}_{36}$ Kr The following atomic masses are known: ${}^{92}_{36}$ Kr: 91.926270 u $\begin{array} { c } 144 \\56\end{array}$ Ba: 143.922845 u $^{236} _{92}$ U^∗: 236.045563 u
What is the reaction energy released during this fission reaction? (1 u = 931.5 MeV/c²)

Question

An excited

^{236} _{92}

U^∗ nucleus undergoes fission into two fragments as follows:

^{236} _{92}

U^∗ →

\begin{array} { c } 144 \\56\end{array}

Ba +

{}^{92}_{36}

Kr The following atomic masses are known:

{}^{92}_{36}

Kr: 91.926270 u

\begin{array} { c } 144 \\56\end{array}

Ba: 143.922845 u

^{236} _{92}

U^∗: 236.045563 u
What is the reaction energy released during this fission reaction? (1 u = 931.5 MeV/c²)

Quizplus · Accepted Answer

First, we need to calculate the total mass before and after the fission reaction:

Before: 236.045563 u
After: 91.926270 u + 143.922845 u = 235.849115 u

The mass difference is Δm = 236.045563 u - 235.849115 u = 0.196448 u

Now, we can calculate the reaction energy using Einstein's famous equation, E=Δmc^2, where c is the speed of light:

E = (0.196448 u) x (931.5 MeV/c²) = 183.2 MeV

Therefore, the correct choice is D) 180 MeV since it is the closest option to the calculated energy value.

An Excited 92236^{236} _{92}92236​ U∗ Nucleus Undergoes Fission into Two Fragments as Follows