A Level Physics 9702: Binding Energy per Nucleon of Iron-56 and the Stability Curve

Iron-56, $^{56}_{26}\text{Fe}$ , is one of the most tightly bound nuclides found in nature. Its nucleus contains 26 protons and 30 neutrons. The mass of a free proton is $m_p=1.007276\text{ u}$ , the mass of a free neutron is $m_n=1.008665\text{ u}$ , and the mass of the assembled iron-56 nucleus is $m_{nuc}=55.920800\text{ u}$ .

Take $1\text{ u} = 931.5\text{ MeV}$ .

(a) Calculate the mass defect $\Delta m$ of the iron-56 nucleus, in u. [2]

(b) Calculate the binding energy of the nucleus, in MeV. [2]

(d) With reference to the shape of the graph of binding energy per nucleon against nucleon number, explain why iron-56 is one of the most stable nuclides, and why both the fission of very heavy nuclei and the fusion of very light nuclei can release energy. [3]

Part (a): Mass defect

The nucleus is made of 26 separate protons and 30 separate neutrons, so their total mass if they were unbound would be: $26m_p+30m_n=26(1.007276)+30(1.008665)=26.189176+30.259950=56.449126\text{ u}$

The mass defect is the difference between this total and the actual mass of the bound nucleus: $\Delta m=(26m_p+30m_n)-m_{nuc}=56.449126-55.920800=0.528326\text{ u}$

Recompute as a check, grouping the terms differently: $26(m_p+m_n)+4m_n=26(2.015941)+4(1.008665)=52.414466+4.034660=56.449126\text{ u}$ , the same total, and $56.449126-55.920800=0.528326\text{ u}$ .

$\Delta m=\boxed{0.528}\text{ u (3 s.f.)}$

Part (b): Binding energy in MeV

Using the given conversion $1\text{ u}=931.5\text{ MeV}$ directly on the mass defect from (a): $E=\Delta m\times931.5=0.528326\times931.5$

$E=0.528326\times900+0.528326\times31.5=475.49+16.64=492.14\text{ MeV}$

Recompute as a check by multiplying directly: $0.528\times931.5=491.8\text{ MeV}$ , consistent with the split-sum method above (small difference only from rounding $\Delta m$ to 3 s.f. first).

$E=\boxed{492}\text{ MeV (3 s.f.)}$

Part (c): Binding energy per nucleon

Iron-56 has $56$ nucleons in total (26 protons $+$ 30 neutrons), so dividing the full (unrounded) binding energy from (b) by 56: $\text{BE per nucleon}=\frac{492.14}{56}=8.788\text{ MeV}$

Recompute as a check: $56\times8.788=492.1\text{ MeV}$ , consistent with the binding energy found in (b).

$\text{BE per nucleon}=\boxed{8.79}\text{ MeV (3 s.f.)}$

Part (d): Why iron-56 is so stable

The graph of binding energy per nucleon against nucleon number rises steeply for light nuclei, reaches a broad maximum around nucleon number $56$ (iron and its neighbours), and then falls slowly for heavier nuclei. Iron-56 lies essentially at this peak, so on average each of its nucleons is more strongly bound than the nucleons in almost any other nuclide. This is why iron-56 is exceptionally stable and needs neither to split apart nor to join with another nucleus to become more tightly bound.

This same curve explains why energy is released by moving toward the peak from either side:

Fission splits a very heavy nucleus (far down the right-hand, gently-falling side of the curve) into two medium-mass fragments that sit closer to the peak, so the fragments have a higher binding energy per nucleon than the original nucleus.
Fusion joins very light nuclei (far down the left-hand, steeply-rising side of the curve) into a single heavier nucleus that also sits closer to the peak, again increasing the binding energy per nucleon.

In both cases the total binding energy of the products exceeds that of the starting nuclei, so by $E=c^2\Delta m$ this increase in binding energy corresponds to a decrease in total mass, and the “missing” mass is released as energy.

Final answers

(a) $\Delta m=\boxed{0.528}\text{ u}$
(b) $E=\boxed{492}\text{ MeV}$
(c) Binding energy per nucleon $=\boxed{8.79}\text{ MeV}$
(d) Iron-56 sits at the peak of the binding energy per nucleon curve; fission (from the heavy side) and fusion (from the light side) both move nucleons toward this peak, releasing energy