Question

For nuclei with mass number close to 119 and 238 , the binding energies per nucleon are approximately \(7.6 \mathrm{MeV}\) and \(8.6 \mathrm{MeV}\) respectively. If a nucleus of mass number 238 breaks into two nuclei of nearly equal masses, what will be the approximate amount of energy released in the process of fission?
(A) \(214 \mathrm{MeV}\)
(B) \(119 \mathrm{MeV}\)
(C) \(2047 \mathrm{MeV}\)
(D) \(1142 \mathrm{MeV}\)

Answer 1

\begin{aligned}
&\text { Ans: (A) } \\
&\begin{array}{l}
\text { } 119 \rightarrow 7.6 \mathrm{MeV} \\
238 \rightarrow 8.6 \mathrm{MeV} \\
238 \rightarrow 119+119 \\
\begin{array}{l}
\mathrm{E}_{\mathrm{f}}=238 \times 7.6 \\
\therefore \mathrm{E}_{\mathrm{f}}-\mathrm{E}_{1}=-238 \mathrm{MeV}
\end{array} & \begin{array}{l}
\text { A negative value indicates the fission cannot take place. The data given are incorrect. } \\
\text { The BE per nucleon values are swapped in question, it should have been } 7.6 \text { MeV } \\
\text { for } 238 \text { and } 8.6 \mathrm{MeV} \text { for } 119 . \text { Thus } \Delta \mathrm{E}=+238 \mathrm{MeV} \text {. Thus } 214 \text { is closest and will } \\
\text { be taken as right answer }
\end{array}
\end{array}
\end{aligned}

Answer 2

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Answer 3

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