Nuclei


  1. A nucleus ZXA has mass represented by M(A, Z). If Mp and Mn denote the mass of proton and neutron respectively and B.E. the binding energy in MeV, then









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    The difference in mass of a nucleus and its constituents, ΔM, is called the mass defect and is given by
    ΔM = [ZMp + (A - Z)Mn] - M
    and binding energy = ΔMc2
    [{ZMp + (A - Z)Mn} - M]c2

    Correct Option: A

    The difference in mass of a nucleus and its constituents, ΔM, is called the mass defect and is given by
    ΔM = [ZMp + (A - Z)Mn] - M
    and binding energy = ΔMc2
    [{ZMp + (A - Z)Mn} - M]c2


  1. ​If M (A; Z), Mp and Mn denote the masses of the nucleus ZXA proton and neutron respectively  in units of u ( 1u = 931.5 MeV/c2) and  BE represents its bonding energy in MeV, then









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    Mass defect = ZMp + (A –Z)Mn–M(A,Z)

    or,
    B.E
    c2

    = ZMp + (A–Z) Mn–M(A,Z) ​
    ∴ M (A, Z) = ZMp + (A–Z)Mn
    B.E
    c2

    Correct Option: A

    Mass defect = ZMp + (A –Z)Mn–M(A,Z)

    or,
    B.E
    c2

    = ZMp + (A–Z) Mn–M(A,Z) ​
    ∴ M (A, Z) = ZMp + (A–Z)Mn
    B.E
    c2



  1. A nucleus of mass M emits a photon of frequency ν and the nucleus recoils. The recoil energy will be









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    Momentum

    Mu =
    E
    =
    hv
    cc

    Recoil energy

    Correct Option: B

    Momentum

    Mu =
    E
    =
    hv
    cc

    Recoil energy


  1. How does the binding energy per nucleon vary with the increase in the number of nucleons?









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    From the graph of BE/A versus mass number A  it is clear that, BE/A first increases and then decreases with increase in mass number.

    Correct Option: D


    From the graph of BE/A versus mass number A  it is clear that, BE/A first increases and then decreases with increase in mass number.



  1. The power obtained in a reactor using U235 disintegration is 1000 kW. The mass decay of U235 per hour is​​​









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    E = mc2

    m =
    E
    c2

    So, mass decay per second
    dm
    =
    1
    dE
    =
    1
    (Power in watt)
    dtc2dtc2

    =
    1
    × 1000 × 103
    (3 × 108)2

    and mass decay per hour
    =
    dm
    × 60 × 60
    dt

    =
    1
    × 106 × 3600
    (3 × 108)2

    = 4 × 10–8 kg ​= 40 micro gram

    Correct Option: C

    E = mc2

    m =
    E
    c2

    So, mass decay per second
    dm
    =
    1
    dE
    =
    1
    (Power in watt)
    dtc2dtc2

    =
    1
    × 1000 × 103
    (3 × 108)2

    and mass decay per hour
    =
    dm
    × 60 × 60
    dt

    =
    1
    × 106 × 3600
    (3 × 108)2

    = 4 × 10–8 kg ​= 40 micro gram