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).

Question

Physics

Byrd

16 July, 06:01

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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Answers (2)

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Phoenix Kemp · Answer 1

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

Jerimiah Mooney · Answer 2

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