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  • In the following reaction, the energy released is <img alt="4_1^1 \mathrm{H} \rightarrow{ }_2^4 \mathrm{He}+2 \mathrm{e}^{+}+\text {Energy }" src="/latex-image/?4_1%5E1%20%5Cmathrm%7BH%7D

In the following reaction, the energy released is

4_1^1 \mathrm{H} \rightarrow{ }_2^4 \mathrm{He}+2 \mathrm{e}^{+}+\text {Energy }

Given:

\text { Mass of }{ }_1^1 \mathrm{H}=1.007825 \mathrm{u}

\text { Mass of }{ }_2^4 \mathrm{H}=4.002603 \mathrm{u}

\text { Mass of } \mathrm{e}^{+}=0.000548 \mathrm{u}

Option: 1

12.33 MeV 


Option: 2

24.67 MeV


Option: 3

25.7 MeV 


Option: 4

49.34 MeV


Answers (1)

best_answer

The given nuclear reaction is

4{ }_1^1 \mathrm{H} \rightarrow{ }_2^4 \mathrm{He}+2 \mathrm{e}^{+}+\text {Energy }

The energy released during the process is

\mathrm{Q}=\left[4 \mathrm{~m}_1\left({ }_1^1 \mathrm{H}\right)-\mathrm{m}\left({ }_2^4 \mathrm{He}\right)-2\left(\mathrm{me}^{+}\right)\right] \mathrm{c}^2

=[(4 \times 1.007825-4.002603)-2 \times 0.000548] \mathrm{u} \times \mathrm{c}^2

=[4.0313-4.002603-0.001096] \mathrm{u} \times \mathrm{c}^2

=(0.027601 \mathrm{u}) \mathrm{c}^2=(0.027601 \mathrm{u})(931.5 \mathrm{MeV} / \mathrm{u})=25.7 \mathrm{MeV}

Posted by

vinayak

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