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Nuclei Test - 41

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Nuclei Test - 41
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  • Question 1
    1 / -0
    How many grams are there in a 7.07.0 mole sample of sodium hydroxide?
    Solution
    The formula of sodium hydroxide is NaOHNaOH
    Atomic weight of NaNa = 23 g/mol23\ {g}/{mol}
    Atomic weight of O O  16 g/mol16\ {g}/{mol}
    Atomic weight of HH = 1 g/mol 1\ {g}/{mol}
    \therefore molecular weight of NaOHNaOH = atomic mass of NaNa + atomic mass of OO + atomic mass of HH
                                                        = 23+16+1=40 g/mol23 + 16 + 1 = 40\ {g}/{mol}
    Therefore, 1 mole of NaOH=40 g/molNaOH = 40\ g/mol
    7 moles of  NaOH= 7 mole×40 g/mol=280g NaOH =  7 \ mole \times 40\ g/mol= 280g

    Hence, option BB is correct.
  • Question 2
    1 / -0
    The mass defect in a nucleus is 3.53.5 amu. Then the binding energy of the nucleus is.
    Solution
    A
    Binding energy  =(mass defect) c2=(Δm)c2c^{2} = (\Delta m) c^{2}
    =Δ(931)MeV=\Delta(931)MeV
    =3.5×931=3258.5MeV=3.5×931 = 3258.5 MeV
  • Question 3
    1 / -0
    Calculate the mass defect.
    Solution
    Possible fusion reaction
    12H+13HxyHe+01n1+1=x+0x=22+3=y+1y=412H+13H24He+01n_{ 1 }^{ 2 }{ H }+_{ 1 }^{ 3 }{ H }\rightarrow _{ x }^{ y }{ He }+_{ 0 }^{ 1 }{ n }\\ 1+1=x+0\Rightarrow x=2\\ 2+3=y+1\Rightarrow y=4\\ _{ 1 }^{ 2 }{ H }+_{ 1 }^{ 3 }{ H }\rightarrow _{ 2 }^{ 4 }{ He }+_{ 0 }^{ 1 }{ n }
    Mass defect =m=(m(12H)+m(13H) )(m(24He)+m(01n) )=(2.014012+3.0160504.0026031.0087)u=0.018759u0.01876u=\triangle m=\left( m\left( _{ 1 }^{ 2 }{ H } \right) +m\left( _{ 1 }^{ 3 }{ H } \right)  \right) -\left( m\left( _{ 2 }^{ 4 }{ He } \right) +m\left( _{ 0 }^{ 1 }{ n } \right)  \right) \\ =(2.014012+3.016050-4.002603-1.0087)u\\ =0.018759u\approx 0.01876u
  • Question 4
    1 / -0
    The Q value is the ________ energy released in the decay at rest.
    Solution
    The Q value is the amount of energy released by a nuclear reaction. It is based on the mass-energy equivalence. 
    The Q value is defined as change in kinetic energy, i.e Q=KEfinalKEinitial=(mfinalminitial)c2Q=KE_{final}-KE_{initial}=(m_{final}-m_{initial})c^2.
  • Question 5
    1 / -0
    The binding energy/nucleon of deuteron (1H2)(_{1}H^{2}) and the helium atom (2He4)(_{2}He^{4}) are 1.11.1 MeV and 77MeV respectively. If the two deuteron atoms fuse to form a single helium atom, then the energy released is.
    Solution
    The nuclear reaction is:  1H2+1H22He4_1H^2 + _1H^2 \rightarrow _2He^4
    Binding energy of deutron, Ed=1.1×2=2.2E_d = 1.1\times 2 = 2.2 MeV
    Binding energy of helium, EHe=7×4=28E_{He} = 7\times 4 = 28 MeV
    Energy released, Q=EHe2EdQ = E_{He} - 2E_d
    \therefore  Q=282(2.2)=23.6Q = 28 - 2(2.2) = 23.6 MeV
  • Question 6
    1 / -0
    Which of the following is a nuclear reaction?
    Solution
    Isotopes of hydrogen nuclei combine to form helium nuclei 
    1H3+1H22He4+n1_{1}H^{3}+_{1}H^{2}\rightarrow_{2}He^{4}+n^{1}+ energy 
  • Question 7
    1 / -0
    An alpha particle (4He^4He)has a mass of 4.00300 amu. A proton has mass of 1.00783 amu and a neutron has mass of 1.00867 amu respectively. The binding energy of alpha particle estimated from these data is the closest to
    Solution
    An alpha particle (24He)(_2^4He) has 2 protons and two neutrons.
    Mass defect      ΔM=2mp+2mnmHe\Delta M = 2m_p + 2m_n - m_{He}
    \therefore   ΔM=2(1.00783)+2(1.00867)4.00300\Delta M=2(1.00783) + 2(1.00867)-4.00300  =0.03000 = 0.03000 amu
    Binding energy of alpha particle    B.E=ΔMc2 = \Delta M c^2
    \therefore   B.E =0.03000×931.5=27.9 = 0.03000 \times 931.5 =27.9 MeV                  (1(\because 1 amu c2=931.5MeVc^2=931.5 MeV ))
  • Question 8
    1 / -0
    The equation E=mc2E=mc^2 was theoretical. It received experimental proof from.
    Solution
    The phenomenon of radioactivity 
    The equation E=mc2E=m{c}^{2} was found therotically and it is not proved initially. Later in the theorem received the proof from the phenomenon of radioactivity. 
  • Question 9
    1 / -0
    41H1  2He4+2e+10+264_1H^1\rightarrow  \ _2He^4+2e^0_{+1}+26 MeV represents
    Solution

  • Question 10
    1 / -0
    When a radioactive isotope 88Ra228_{88} Ra^{228}  decays in series by the emission of three α\alpha -particles and a β\beta  particle the isotope finally formed is :
    Solution
    In each alpha decay, mass number of the parent nucleus decreases by 44 units while atomic number decreases by 22 units. In each beta decay,  atomic number increases by 11 unit but mass number remains the same.
    So, mass number of the isotope formed  A=2283×4=216A = 228-3\times 4 = 216
    Atomic number  Z=883×2+1=83Z = 88-3\times 2+1 = 83
    So correct answer is option C.
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