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Nuclei Test - 48

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Nuclei Test - 48
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  • Question 1
    1 / -0

    Binding energy per nucleon versus mass number curve for nuclei is shown in fig.W, X, Y, and Z are four nuclei indicated on the curve. The process that would release energy is 

    Solution
    If it is mass A
    Y2ZY\rightarrow 2Z
    Reactant R=60×8.5=10meVR=60\times8.5=10meV
    Product P=2×30×5=300meVP=2\times30\times5=300meV
    ΔE=210meV\Delta E=-210meV
    Therefore Endothermic
    If it was C
    W2YW\rightarrow 2Y
    R=120×75=900meVR=120\times75=900meV
    P=2×60×8.5=1020meVP=2\times60\times8.5=1020meV
    ΔE=120meV\Delta E=120meV
    Therefore Exothermic
    If it was D
    XY+ZX\rightarrow Y+Z
    R=90×8=720meVP=60×8.5+30×5=66.0meVR=90\times8=720meV\\P=60\times8.5+30\times5=66.0meV
    ΔE=60meV\Delta E=-60meV
    Therefore Endothermic
  • Question 2
    1 / -0
    A stationary U238U^{238} nucleus undergoes α\alpha- decay. If the kinetic energy of product nucleus is EE, the total energy released  in the process is-
    Solution
    mass of alpha particle ma=4mm_a=4m, where mm is mass of a proton. 
    mass of Uranium nucleus is mu=238mm_u=238m after emission the residue will have mass mr=2384=234mm_r=238-4=234m

    We know KE=momentum2/2mKE=momentum^2/2m
    so for residue nucleus, it should be like this E=pr2/2mrE=p^2_r/2m_r

    or momentum of residue nucleus is  pr=2mrE2=468mE2p_r=\sqrt[2]{2m_rE}=\sqrt[2]{468mE}
    But during emission the alpha particle  should get same momentum in opposite direction to make net momentum as zero
     before and after emission so it should also be  pa=468mE2p_a=\sqrt[2]{468mE}

    If kinetic energy of alpha particle is EaE_a then Ea=pa2/2ma=468mE2×4m=58.5EE_a=p^2_a/2m_a=\dfrac{468mE}{2\times 4m}=58.5E

    So net energy released in the emission is the sum of above two i.e. 59.5E59.5E

  • Question 3
    1 / -0
    The binding energy per nucleon of deuteron is 1.2MeV1.2MeV and that of the helium atom is 7.1MeV7.1MeV. What is the energy released, if two deuteron atoms combine to form a single helium atom?  
    Solution
     Given - BE per nucleon of deuteron =1.2MeV and of Helium =7.1MeV12H+12H42He\begin{array}{c}\text { Given - } * B E \text { per nucleon of deuteron }=1.2 \mathrm{MeV} \\\text { and of Helium }=7.1\mathrm{MeV}\\{ }_{1}^{2} \mathrm{H}+{}_{1}^{2}\mathrm{H}\longrightarrow_{4}^{2}\mathrm{He}\end{array}


     Energy released = BE of Helium -2.BE of Deutron.  BE =BE ber nucleons X Num of nucleons. E=[4×7.1]2×[1.2×2] Mev E=23.6MeV\begin{array}{l}\Rightarrow \text { Energy released = BE of Helium -2.BE of Deutron. } \\\text { BE }=B E \text { ber nucleons } X \text { Num of nucleons. } \\\begin{aligned}\Rightarrow E &=[4 \times 7.1]-2\times[1.2 \times 2] \text { Mev } \\E &=23.6\mathrm{MeV}\end{aligned}\end{array}
  • Question 4
    1 / -0
    The number of atoms of the HeHe in 104 amu104\ amu is:
    Solution
    Since the mass of one atom of He = 4 a.m.u
    Therefore the number of atoms of the He in 104 amu is, 1044=26\dfrac { 104 }{ 4 } =26.
  • Question 5
    1 / -0
    What would be the energy required to dissociate completely 11 g of Ca (40)Ca \ (40) into its constituent particles? 
    Given: Mass of proton = 1.007277 amu,
    Mass of neutron = 1.00866 amu,
    Mass of Ca-40= 39.97545 amu
    (take amu = 931 MeV)
    Solution
    Given,
    mp=1.007277amum_p=1.007277 amu mass of proton
    mn=1.00866amum_n=1.00866amu mass of neutron
    mc(ca40)=39.97545amum_c(ca-40)=39.97545 amu
    and 1amu=931MeV/c21amu=931 MeV/c^2
    Z=20Z=20 No of proton for calcium
    N=20N=20 no. of neutron

    The energy required to dissociate completely is called binding energy.
    B.E=(Zmp+Nmnmc)c2B.E=(Zm_p+Nm_n-m_c)c^2 . . . .(1)

    B.E=(20×1.007277+20×1.0086639.97545)c2B.E=(20\times 1.007277+20\times 1.00866-39.97545)c^2
    B.E=0.34359×931MeVB.E=0.34359\times 931MeV

    B.E=319.6MeVB.E=319.6MeV This is for 1 atom of calcium-40

    No. of  atom in 1g1g of Ca-40 is,
    N=140×6.6022×1023N=\dfrac{1}{40}\times 6.6022\times 10^{23}

    N=1.65055×1022N=1.65055\times 10^{22} atom.
    The  total energy required is,
    E=N(B.E)E=N(B.E)
    E=1.65055×1022×319.6MeVE=1.65055\times 10^ {22}\times319.6MeV
    E=4.813×1024MeVE=4.813\times 10^{24}MeV
    The correct option is A.
  • Question 6
    1 / -0
    The binding energy of deutron is 2.2 MeV2.2 \space MeV and that of 24He^4_2 He is 28 MeV28 \space MeV. If two deutrons are fused to form one 24He^4_2 He then the energy released is:
    Solution
    1D22He4_1{D^2}{ \to _2}H{e^4}
    Energy released =282×2.2=23.6MeV=28 - 2 \times 2.2 = 23.6\,MeV
    Hence,
    option (B)(B) is correct answer.
  • Question 7
    1 / -0
    If x gx \ g  of AA (atomic mass 5050) contains nn atoms, how many atoms are there in 20 x g B20 \ x \ g \ B (atomic weight 100100)
    Solution

  • Question 8
    1 / -0
    The binding energy per nucleon of 37Li^7_3 Li and 24He^4_2 He nuclei are 5.60 MeV and 7.06 Me V, respectively. In the nuclear reaction 37Li+11He24H+24He+Q^7_3 Li + ^1_1 He \longrightarrow ^4_2 H + ^4_2 He + Q the value of energy Q released is
    Solution
    The binding energy for 1H1_1H^1 is around zero and also not give in the question so we can ignore it,
    Q=2(4×7.06)7×(5.60)Q=2(4\times 7.06)-7\times (5.60)
    =(56.4839.2)MeV=(56.48-39.2)MeV
    =17.28 MeV=17.28\ MeV
    =17.3 MeV=17.3\ MeV
  • Question 9
    1 / -0
    Neutron decay in free space is given as follows:
    en10H1+oe0+[]_{e}n^{1} \rightarrow_{0}H^{1}+_{-o}e^{0}+[]
    Then the parenthes is represents a 
    Solution
    en10H1+1e0+ some particles _{e}{n}^{1}\rightarrow_{0}H^{1}+_{-1}e^{0}+\text { some particles }
     The given decay is similar to βdecay.  To maintain the momentum, angular momentum  spin moment. The particle released is antineutrino. \begin{array}{l}\text { The given decay is similar to } \beta^{-} \text {decay. } \\\text { To maintain the momentum, angular momentum } \\\text { spin moment. The particle released is antineutrino. }\end{array}
  • Question 10
    1 / -0
    A slow neutron strikes a nucleus of 92235U^{235}_{92} {U} splitting it into lighter nuclei of 56141Ba^{141}_{56} {Ba} and 3692Kr^{92}_{36} {Kr} along with three neutrons. The energy released in this reaction is :
    (The masses of Uranium, Barium and Krypton in this reaction are 235.043933{235.043933} a.m.u, 140.917700{140.917700} a.m.u and 91.895400{91.895400} a.m.u respectively. The mass of a neutron is 1.0086651.008665 a.m.u)
    Solution

    92235U+01n56141Ba+3692Kr+3(01n)+Q{}_{92}^{235}U + {}_0^1n \to {}_{56}^{141}Ba + {}_{36}^{92}Kr + 3({}_0^1n) + Q

    Q is Energy released Q = ( Total mass of reactants – total mass ) ×C2 \times {C^2}

    =[Mass(92235U)+mass(01n)][mass56141Ba+mass3692Kr+mass301n]c2 = \left[ {Mass\left( {{}_{92}^{235}U} \right) + mass\left( {{}_0^1n} \right)} \right] - \left[ {mass{}_{56}^{141}Ba + mass{}_{36}^{92}Kr + mass3{}_0^1n} \right]{c^2}

    =(235043933140.91770091.89542×1.008665)u×c2 = \left( {235 \cdot 043933 - 140.917700 - 91.8954 - 2 \times 1.008665} \right)u \times {c^2}

    =198.9MeV = 198.9MeV

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