Find the energy (in MeV) that binds the neutron to the _7^14 text(N) nucleus by considering the mass of _7^13 text(N) (atomic mass = 13.005

Question

Find the energy (in MeV) that binds the neutron to the _7^14 text(N) nucleus by considering the mass of _7^13 text(N) (atomic mass = 13.005 738 u) and the mass of _1^0 text(n) (atomic mass = 1.008 665 u), as compared to the mass of _7^14 text(N) (atomic mass = 14.003 074 u).

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Huyền Thanh 4 years 2021-07-24T03:47:59+00:00 2 Answers 13 views 0

Answers ( )

  1. thuthuy
    0
    2021-07-24T03:49:47+00:00
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    Answer:

    Binding energy = 10.55 MeV

    Explanation:

    The mass defect is;

    Δm = 13.005738u + 1.008665u – 14.003074u = 0.011329u

    In mass defect, a mass of 1 u is equivalent to an energy of 931.5 MeV.

    Thus, the binding energy of the neutron in MeV will be;

    0.011329 x 931.5 = 10.55 MeV

  2. kiet
    0
    2021-07-24T03:49:48+00:00
    Reply

    Answer:

    Binding energy is 10.55 MeV

    Explanation:

    Binding energy is the minimum energy required to separate a particle into separate smaller parts

    The mass defect (Δm) is given as:

    Δm = (13.005738 u + 1.008665 u) – 14.003074 = 14.014403 – 14.003074 = 0.011329

    Δm = 0.011329 u

    Binding energy = Δm × (931.5 v / 1u)

    Therefore: Binding energy = 0.011329 u × (931.5 MeV / 1u) = 10.55 MeV

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