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NEET-XII-Physics

46: The Nucleus

with Solutions - page 5

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  • Qstn #11
    A free neutron beta-decays to a proton with a half-life of 14 minutes. (a) What is the decay constant? (b) Find the energy liberated in the process.
    Ans : Given:
    Half-life period of free neutron beta-decays to a proton, `` {T}_{1/2}`` = 14 minutes
    Half-life period, T1/2 = `` \frac{0.6931}{\lambda }``
    Here, `` \lambda `` = Decay constant
    `` \therefore \lambda =\frac{0.693}{14\times 60}``
    `` =8.25\times {10}^{-4}{\,\mathrm{\,s\,}}^{-1}``
    If mp is the mass of proton, let mn and me be the mass of neutron and mass of electron, respectively.
    `` \therefore \; Energ\,\mathrm{\,y\,}\,\mathrm{\,liberated\,},E=[{m}_{n}-\left({m}_{p}+{m}_{e}\right)]{c}^{2}``
    `` =[1.008665\,\mathrm{\,u\,}-\left(1.007276+0.0005486\right)\,\mathrm{\,u\,}]{c}^{2}``
    `` =0.0008404\times 931\,\mathrm{\,MeV\,}``
    `` =782\,\mathrm{\,keV\,}``
    `` ``
    Page No 442:
  • Qstn #12
    Complete the following decay schemes.
    Ans : `` \left(\,\mathrm{\,a\,}\right){}_{88}{}^{226}\,\mathrm{\,Ra\,}\to {}_{2}{}^{4}\,\mathrm{\,\alpha \,}+{}_{86}{}^{222}\,\mathrm{\,Rn\,}``
    `` \left(\,\mathrm{\,b\,}\right){}_{8}{}^{19}\,\mathrm{\,O\,}\to {}_{9}{}^{19}\,\mathrm{\,F\,}+\stackrel{¯}{e}+\stackrel{¯}{v}``
    `` \left(\,\mathrm{\,c\,}\right){}_{13}{}^{25}\,\mathrm{\,Ar\,}\to {}_{12}{}^{25}\,\mathrm{\,Mg\,}+{e}^{+}+v``
    Page No 442:
  • Qstn #13
    In the decay 64Cu → 64Ni + e+ + v, the maximum kinetic energy carried by the positron is found to be 0.650 MeV.
    Ans : Given:
    Maximum kinetic energy of the positron, K = 0.650 MeV
  • #13-a
    What is the energy of the neutrino which was emitted together with a positron of kinetic energy 0.150 MeV?
    Ans : Neutrino and positron are emitted simultaneously.
    ∴ Energy of neutrino = 0.650 - Kinetic energy of the given positron
    = 0.650 - 0.150
    = 0.5 MeV = 500 keV
  • #13-b
    What is the momentum of this neutrino in kg m s-1?
    Use the formula applicable to a photon.
    Ans : Momentum of the neutrino, `` P=\frac{E}{c}``
    Here, E = Energy of neutrino
    c = Speed of light
    `` \Rightarrow P=\frac{500\times 1.6\times {10}^{-19}}{3\times {10}^{8}}\times {10}^{3}``
    `` =2.67\times {10}^{-22}{\,\mathrm{\,Kgms\,}}^{-1}``
    Page No 442:
  • Qstn #14
    Potassium-40 can decay in three modes. It can decay by β--emission, B*-emission of electron capture.
  • #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:
  • Qstn #15
    Lithium (Z = 3) has two stable isotopes 6Li and 7Li. When neutrons are bombarded on lithium sample, electrons and α-particles are ejected. Write down the nuclear process taking place.
    Ans : The nuclear process taking place is shown below.
    `` {}_{8}{}^{6}\,\mathrm{\,Li\,}+n\to {}_{3}{}^{7}\,\mathrm{\,Li\,}``
    `` {}_{3}{}^{7}\,\mathrm{\,Li\,}+n\to {}_{3}{}^{8}\,\mathrm{\,Li\,}\to {}_{4}{}^{8}\,\mathrm{\,Be\,}+\stackrel{¯}{v}+{e}^{-}``
    `` {}_{4}{}^{8}\,\mathrm{\,Be\,}\to {}_{2}{}^{4}\,\mathrm{\,He\,}+{}_{2}{}^{4}\,\mathrm{\,He\,}``
    Page No 442:
  • Qstn #16
    The masses of 11C and 11B are respectively 11.0114 u and 11.0093 u. Find the maximum energy a positron can have in the β*-decay of 11C to 11B.
    Ans : Given:
    Mass of 11C, m(11C) = 11.0114 u
    Mass of 11B, m(11B) = 11.0093 u
    Energy liberated in the β+ decay (Q) is given by
    `` Q=\left[m\left({}^{11}C\right)-m\left({}^{11}B\right)-2{m}_{e}\right]{c}^{2}``
    = (11.0114 u - 11.0093 u `` -`` 2`` \times ``0.0005486 u)c2
    = 0.0010028 `` \times `` 931 MeV
    = 0.9336 MeV = 933.6 keV
    For maximum KE of the positron, energy of neutrino can be taken as zero.
    ∴ Maximum KE of the positron = 933.6 keV
    Page No 442:
  • Qstn #17
    228Th emits an alpha particle to reduce to 224Ra. Calculate the kinetic energy of the alpha particle emitted in the following decay:
    Th228→Ra224*+αRa224*→224Ra+Υ(217 keV).Atomic mass of 228Th is 228.028726 u, that of 224Ra is 224.020196 u and that of
    H24is 4.00260 u.
    Ans : Given:
    Atomic mass of 228Th, m(228Th) = 228.028726 u
    Atomic mass of 224Ra, m(224Ra) = 224.020196 u
    Atomic mass of `` {}_{2}{}^{4}\,\mathrm{\,H\,}``, m(`` {}_{2}{}^{4}\,\mathrm{\,H\,}``) = 4.00260 u
    Mass of 224Ra = 224.020196 × 931 + 0.217 MeV = 208563.0195 MeV
    Kinetic energy of alpha particle, K = `` \left[m\left({}^{228}\,\mathrm{\,Th\,}\right)-\left[m\left({}^{224}\,\mathrm{\,Ra\,}\right)+m\left({}_{2}{}^{4}\,\mathrm{\,H\,}\right)\right]\right]``c2
    = (228.028726 × 931) - [(208563.0195 + 4.00260 × 931]
    = 5.30383 MeV = 5.304 MeV
    Page No 442:
  • Qstn #18
    Calculate the maximum kinetic energy of the beta particle emitted in the following decay scheme:
    12N → 12C* + e+ + v
    12C* → 12C + γ (4.43MeV).
    The atomic mass of 12N is 12.018613 u.
    Ans : Given:
    Atomic mass of 12N, m(12N) = 12.018613 u
    12N → 12C* + e+ + v
    12C* → 12C + γ (4.43 MeV)
    Net reaction is given by
    12N → 12C + e+ + v + γ (4.43 MeV)
    Qvalue of the `` {\beta }^{+}`` decay will be
    Qvalue= [ m(12N) `` -`` (m(12C*) + 2me)]c2
    = [12.018613 `` \times ``931 MeV `` -`` (12`` \times ``931 + 4.43) MeV `` -`` (2`` \times ``511) keV]
    = [11189.3287 `` -`` 11176.43 `` -`` 1.022] MeV
    = 11.8767 MeV = 11.88 MeV
    The maximum kinetic energy of beta particle will be 11.88 MeV, assuming that neutrinos have zero energy.
    Page No 442:
  • Qstn #19
    The decay constant of
    Hg80197(electron capture to
    Au79197) is 1.8 × 10-4 S-1.
    Ans : Given:
    Decay constant of `` {}_{80}{}^{197}\,\mathrm{\,Hg\,}``, `` \lambda `` = 1.8 × 10-4 s`` -1``
  • #19-a
    What is the half-life?
    Ans : Half-life, `` {T}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{0.693}{\lambda }``
    `` \Rightarrow {T}_{1/2}=\frac{0.693}{1.8\times {10}^{-4}}``
    `` =3850\,\mathrm{\,s\,}=64\,\mathrm{\,minutes\,}``
    `` ``
  • #19-b
    What is the average-life?
    Ans : `` \,\mathrm{\,Average\,}\,\mathrm{\,life\,},{T}_{av}=\frac{{T}_{1/2}}{0.693}``
    `` =\frac{64}{0.693}``
    `` =92\,\mathrm{\,minutes\,}``
    `` ``
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