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In the fusion reaction, ${ }_1 \mathrm{H}^2+{ }_1 \mathrm{H}^2 \rightarrow{ }_2 \mathrm{He}^4+\mathrm{Q}, \mathrm{Q}$ is energy released. If ' $\mathrm{c}$ ' is the speed of light and ' $\mathrm{m}$ ' is the mass of each deuterium nucleus then the mass of helium nucleus formed is
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$2 \mathrm{~m}-\frac{\mathrm{Q}}{\mathrm{c}^2}$
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