# Negative or positive binding energy?

I'm working with modern physics atm, and can't seem to wrap my head around the binding energies of some molecules.

At first, I thought that a negative binding energy = unstable molecule, and a positive binding energy = stable molecule.

But for example, Cobalt-60 is a radioactive (unstable) isotope and has a positive binding energy. But deuterium, which is a stable isotope of hydrogen, also has a positive binding energy. How is this possible? If both a stable and an unstable molecule can have a positive binding energy, what does it mean if a binding energy is negative/positive? What's the physical meaning? See calculations below:

$$Mass[27 protons] + Mass[33 neutrons] - Mass[Cobalt-60] = 27*1.00727647+33*1.00866501-59.9338222=0.5485...$$ $$Mass[1 proton] + Mass[1 neutron] - Mass[Deuterium] = 1*1.00727647+1*1.00866501-2.01410178=0.0018...$$

## 1 Answer

Nature tends to the lowest energy state possible. Therefore stable bound states have negative binding energy.

As an example, deuterium is stable against breaking up into a proton and a neutron since

$$m_d-m_p-m_n = 1876.12 \textrm{ MeV } - 938.27 \textrm{ MeV } - 939.57 \textrm{ MeV } = -1.72 \textrm{ MeV}.$$

Similarly, $^{60}$Co is stable against breaking up into its constituent protons and neutrons,

$$m_{^{60}\mathrm{Co}}-27m_p-33m_n=-511\textrm{ MeV}.$$

However, $^{60}$Co is unstable against decay to $^{60}$Ni

$$m_{^{60}\mathrm{Co}}-m_{^{60}\mathrm{Ni}}=2.83\textrm{ MeV};$$

i.e. every $^{60}$Co nucleus that decays to $^{60}$Ni releases 2.83 MeV of energy.