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When two dentrons (.(1)H^(2)) fuse to fr...

When two dentrons `(._(1)H^(2))` fuse to from a helium nucleus `._(2)He^(4), 23.6 MeV` energy is released. Find the binding energy of helium if it is 1.1 MeV for each nucleon of deutrim.

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The correct Answer is:
28 meV
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In fusion reaction ._(1)H^(2) + ._(1)H^(2) to ._(2)He^(4) + Q Find the energy released while binding energy per nucleon for deuterium is 1.12 MeV and for helium nucleus is 7.07 MeV

What do you understand by the fact that the binding energy of helium nucleus is 28.17MeV?

Knowledge Check

  • If two deuterons are fused to form one Helium than what will be amount of energy released, if the binding energy of deuteron is 2.2MeV and that of Helium is 28MeV

    A
    25.8MeV
    B
    23.6MeV
    C
    19.2MeV
    D
    30.2MeV

  • Two deuterons udnergo nuclear fusion to form a Helium nucleus. Energy released in this process is : (given binding energy per nucleon for deuteron = 1.1 MeV and for helium = 7.0 MeV)

    A
    23.6 MeV
    B
    25.8 MeV
    C
    30.2 MeV
    D
    32.4 MeV

  • The mass defect of He_(2)^(4) He is 0.03 u. The binding energy per nucleon of helium (in MeV) is

    A
    6.9825
    B
    27.93
    C
    2.793
    D
    69.825

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    The binding energy of ._(2)^(4)He is 28.57 MeV. What shall be the binding energy per nucleon of this element?

    What is the binding energy per nucleon for Helium in MeV if the mass defect for the nucleus of Helium is 0.0303 amu.

    The mass defect for the nucleus of helium is 0.0303 a.m.u. What is the binding energy per nucleon for helium in MeV ?

    The binding energy per nucleon of deuteron (_(1)^(2)H) and helium nucleus (_(2)^(4)He) is 1.1 MeV and 7 MeV respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is

    What is the binding energy per nucleon for Helium( in MeV) if the mass defect for the nucleus of Helium is 0.0303 amu?

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