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).

Answer 1

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

Answer 2

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