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NEET-XII-Physics

46: The Nucleus

with Solutions - page 5
Qstn# iv-14 Prvs-QstnNext-Qstn
  • #14
    Potassium-40 can decay in three modes. It can decay by β--emission, B*-emission of electron capture. (a) Write the equations showing the end products. (b) Find the Q-values in each of the three cases. Atomic masses of
    Ar1840, K1940 and Ca2040are 39.9624 u, 39.9640 u and 39.9626 u respectively.
    Ans : (a) Decay of potassium-40 by β-emission is given by
    `` {}_{19}\,\mathrm{\,K\,}^{40}{\to }_{20}{\,\mathrm{\,Ca\,}}^{40}+{\,\mathrm{\,\beta \,}}^{-}+\stackrel{¯}{\,\mathrm{\,v\,}}``
    Decay of potassium-40 by β+ emission is given by
    `` {}_{19}\,\mathrm{\,K\,}^{40}{\to }_{18}{\,\mathrm{\,Ar\,}}^{40}+{\,\mathrm{\,\beta \,}}^{+}+\,\mathrm{\,v\,}``
    Decay of potassium-40 by electron capture is given by
    `` {}_{19}\,\mathrm{\,K\,}^{40}+{\,\mathrm{\,e\,}}^{-}{\to }_{18}{\,\mathrm{\,Ar\,}}^{40}+\,\mathrm{\,v\,}`` (b) Qvalue in the β- decay is given by
    Qvalue = [m(19K40) - m(20Ca40)]c2
    = [39.9640 u - 39.9626 u]c2
    = 0.0014 `` \times `` 931 MeV
    = 1.3034 MeV
    Qvalue in the β+ decay is given by
    Qvalue = [m(19K40) - m(20Ar40) - 2me]c2
    = [39.9640 u - 39.9624 u - 0.0021944 u]c2
    = (39.9640 - 39.9624) 931 MeV - 1022 keV
    = 1489.96 keV - 1022 keV
    = 0.4679 MeV
    Qvalue in the electron capture is given by
    Qvalue = [ m(19K40) - m(20Ar40)]c2
    = (39.9640 - 39.9624)uc2
    = 1.4890 = 1.49 MeV
    Page No 442:
  • #14-a
    Write the equations showing the end products.
    Ans : Decay of potassium-40 by β-emission is given by
    `` {}_{19}\,\mathrm{\,K\,}^{40}{\to }_{20}{\,\mathrm{\,Ca\,}}^{40}+{\,\mathrm{\,\beta \,}}^{-}+\stackrel{¯}{\,\mathrm{\,v\,}}``
    Decay of potassium-40 by β+ emission is given by
    `` {}_{19}\,\mathrm{\,K\,}^{40}{\to }_{18}{\,\mathrm{\,Ar\,}}^{40}+{\,\mathrm{\,\beta \,}}^{+}+\,\mathrm{\,v\,}``
    Decay of potassium-40 by electron capture is given by
    `` {}_{19}\,\mathrm{\,K\,}^{40}+{\,\mathrm{\,e\,}}^{-}{\to }_{18}{\,\mathrm{\,Ar\,}}^{40}+\,\mathrm{\,v\,}``
  • #14-b
    Find the Q-values in each of the three cases. Atomic masses of
    Ar1840, K1940 and Ca2040are 39.9624 u, 39.9640 u and 39.9626 u respectively.
    Ans : Qvalue in the β- decay is given by
    Qvalue = [m(19K40) - m(20Ca40)]c2
    = [39.9640 u - 39.9626 u]c2
    = 0.0014 `` \times `` 931 MeV
    = 1.3034 MeV
    Qvalue in the β+ decay is given by
    Qvalue = [m(19K40) - m(20Ar40) - 2me]c2
    = [39.9640 u - 39.9624 u - 0.0021944 u]c2
    = (39.9640 - 39.9624) 931 MeV - 1022 keV
    = 1489.96 keV - 1022 keV
    = 0.4679 MeV
    Qvalue in the electron capture is given by
    Qvalue = [ m(19K40) - m(20Ar40)]c2
    = (39.9640 - 39.9624)uc2
    = 1.4890 = 1.49 MeV
    Page No 442: