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NEET-XII-Physics
46: The Nucleus
- #4How much energy is released in the following reaction:
7Li + p → α + α.
Atomic mass of 7Li = 7.0160 u and that of 4He = 4.0026 u.Ans : Given:
Mass of 7Li = 7.0160 u
Mass of 4He = 4.0026 u.
Reaction:
`` \,\mathrm{\,L\,}{i}^{7}+p\mathit{\to }\alpha \mathit{+}\alpha \mathit{+}E\mathit{,}``
`` ``
`` ``
Energy release `` \left(E\right)`` is given by
`` E=\left[m\left({}^{7}\,\mathrm{\,Li\,}\right)+\left({m}_{p}\right)-2\times m\left({}^{4}\,\mathrm{\,He\,}\right)\right]{c}^{2}``
`` =\left[\left(7.0160\,\mathrm{\,u\,}+1.007276\,\mathrm{\,u\,}\right)-2\left(4.0026\,\mathrm{\,u\,}\right)\right]{c}^{2}``
`` =(8.023273\,\mathrm{\,u\,}-8.0052\,\mathrm{\,u\,}){\,\mathrm{\,c\,}}^{2}``
`` =0.018076\times 931\,\mathrm{\,MeV\,}``
`` =16.83\,\mathrm{\,MeV\,}``
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