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  • <img alt="{ }_{92}^{238} \mathrm{~A} \rightarrow{ }_{90}^{234} \mathrm{~B}+{ }_2^4 \mathrm{D}+\mathrm{Q}" src="https://entrancecorner.oncodecogs.com/gif.latex?%7B%20%7D_%7B92%7D%5E%7B238%7D%20%5Cma

{ }_{92}^{238} \mathrm{~A} \rightarrow{ }_{90}^{234} \mathrm{~B}+{ }_2^4 \mathrm{D}+\mathrm{Q}

In the given nuclear reaction, the approximate amount of energy released will be:

\text { [Given, mass of }{ }_{92}^{238} \mathrm{~A}=238.05079 \times 931.5 \mathrm{MeV} / \mathrm{c}^2

                                                     \\\text{mass of } { }_{90}^{234} \mathrm{~B}=234.04363 \times 931.5 \mathrm{MeV} / \mathrm{c}^2, \\\\\text{mass of } \left.{ }_2^4 \mathrm{D}=4.00260 \times 931.5 \mathrm{MeV} / \mathrm{c}^2\right]

Option: 1

4.25 MeV


Option: 2

5.9 MeV


Option: 3

3.82 MeV

 


Option: 4

2.12 MeV


Answers (1)

best_answer

\begin{aligned} & \mathrm{Q}=\Delta \mathrm{mC}^2 \\ & \mathrm{Q}=(238.05079-234.04363-4.00260) \times 931.5 \mathrm{MeV} \\ & \mathrm{Q}=0.00456 \times 931.5 \mathrm{MeV} \\ & \mathrm{Q}=4.25 \mathrm{MeV} \end{aligned}

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HARSH KANKARIA

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