NEET-XII-Physics
46: The Nucleus
- #7Find the energy liberated in the reaction
223Ra → 209Pb + 14C.
The atomic masses needed are as follows.
223Ra 209Pb 14C
22..018 u 208.981 u 14.003 uAns : Given:
Atomic mass of 223Ra, m(223Ra) = 223.018 u
Atomic mass of 209Pb, m(209Pb) = 208.981 u
Atomic mass of 14C, m(14C) = 14.003 u
Reaction:
`` {}^{223}\,\mathrm{\,Ra\,}{\to }^{209}\,\mathrm{\,Pb\,}+{}^{14}\,\mathrm{\,C\,}``
`` ``
`` \,\mathrm{\,Energy\,},E=\left[m\left({}^{223}Ra\right)-\left(m\left({}^{209}Pb\right)+m\left({}^{14}\,\mathrm{\,C\,}\right)\right)\right]{c}^{2}``
`` =\left[223.018\,\mathrm{\,u\,}-\left(208.981+14.003\right)\,\mathrm{\,u\,}\right]{c}^{2}``
`` =0.034\times 931\,\mathrm{\,MeV\,}``
`` =31.65\,\mathrm{\,MeV\,}``
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