COMEDK · Physics · 29. Nuclear Physics
In a nuclear reaction \({ }_4^{10} \mathrm{Be}+{ }_1^1 p \rightarrow{ }_5^{10} \mathrm{~B}+{ }_0^1 n\), what is the total energy released if the binding energy of beryllium -9 is 58.2 MeV and that of boron- 10 is 64.7 MeV ?
- A \(5.78 \times 10^{12} J\)
- B \(6.04 \times 10^{12} J\)
- C \(9.61 \times 10^{-12} J\)
- D \(1.04 \times 10^{-12} J\)
Answer & Solution
Correct Answer
(D) \(1.04 \times 10^{-12} J\)
Step-by-step Solution
Detailed explanation
The nuclear reaction is given by \({}_{4}^{10}\text{Be} + {}_{1}^{1}\text{p} \rightarrow {}_{5}^{10}\text{B} + {}_{0}^{1}\text{n}\). The energy released in a nuclear reaction is given by the difference in the total binding energy of the products and the reactants.…
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