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NEET-XII-Physics

46: The Nucleus

with Solutions - page 4
Qstn# iv-4 Prvs-QstnNext-Qstn
  • #4
    How 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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