Q.A given coin has a mass of $3.0\ \text{g}$. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of $^{63}_{29}\text{Cu}$ atoms (of mass $62.92960\ \text{u}$).

The energy to pull every nucleon apart is the total nuclear binding energy of the coin, from the mass defect ($E=\Delta m\,c^2$).

Mass defect per $^{63}_{29}\text{Cu}$ atom (29 protons + 34 neutrons, comparing atomic masses so electrons cancel):

Binding energy per atom:

Number of atoms in $3.0\ \text{g}$:

$N = \frac{3.0}{62.92960}\times6.022\times10^{23} = 2.871\times10^{22}$ …

To pull apart every proton and neutron in the coin, we must supply the total nuclear binding energy of every atom in it. Binding energy is what holds each nucleus together; by mass–energy equivalence it equals the mass defect converted to energy, $E=\Delta m\,c^2$.

1. Mass defect of one $^{63}_{29}\text{Cu}$ atom

Copper-63 has $Z=29$ protons and $N=63-29=34$ neutrons. Since the quoted mass $62.92960\ \text{u}$ is the atomic mass (it already includes the 29 electrons), we compare it against 29 hydrogen atoms plus 34 free neutrons, so the electron masses cancel exactly:

2. Binding energy per atom

Using $1\ \text{u}=931.5\ \text{MeV}/c^2$:

In SI units ($1\ \text{MeV}=1.602\times10^{-13}\ \text{J}$):

3. Number of atoms in the coin

An atomic mass of $62.92960\ \text{u}$ means a molar mass of $62.92960\ \text{g/mol}$, so …

Method: Mass Defect → Binding Energy (Mass-Energy Equivalence)

The idea is simple: the nucleus is held together by the strong force, and to pull it apart you must supply energy equal to the binding energy. That binding energy comes from the mass defect — the difference between the mass of the separated nucleons and the actual mass of the nucleus. Einstein’s $E = mc^2$ converts that missing mass into energy.

Step 1 — Find the number of atoms in the coin

Mass of coin = $3.0\ \text{g} = 3.0 \times 10^{-3}\ \text{kg}$

Molar mass of $^{63}\text{Cu}$ = $62.92960\ \text{g/mol}$ (since 1 u ≈ 1 g/mol for this purpose)

Number of moles:

Number of atoms:

Step 2 — Find the mass defect for one atom

For $^{63}_{29}\text{Cu}$:

• Protons = 29, neutrons = $63 - 29 = 34$

The given mass ($62.92960\ \text{u}$) is an atomic mass, so pair it with the hydrogen ATOM mass (not the bare proton mass — that would silently drop 29 electron masses):

• Mass of a hydrogen atom = $1.007825\ \text{u}$
• Mass of a neutron = $1.008665\ \text{u}$

Total mass of 29 hydrogen atoms + 34 neutrons:

Actual atomic mass of one $^{63}\text{Cu}$ atom = $62.92960\ \text{u}$

Mass defect for one atom:

Step 3 — Convert mass defect to energy per atom

Using $1\ \text{u} = 931.5\ \text{MeV}/c^2$:

Step 4 — Total energy for the coin …

Common Mistakes on Mass–Energy Equivalence Problems

Mistake 1: Forgetting to convert mass from grams to atomic mass units

The coin's mass is given as $3.0\ \text{g}$, but the copper atom's mass is in atomic mass units (u). Many students directly use the gram value in Einstein's equation without converting.

How to avoid: First find how many copper atoms are in the coin. The mass of one $^{63}\text{Cu}$ atom is $62.92960\ \text{u}$. Convert this to grams using $1\ \text{u} = 1.660539 \times 10^{-27}\ \text{kg}$, or better, use the fact that $1\ \text{u} = 1.660539 \times 10^{-24}\ \text{g}$.

Number of atoms $N = \frac{\text{mass of coin}}{\text{mass of one atom}} = \frac{3.0}{62.92960 \times 1.660539 \times 10^{-24}}$

Mistake 2: Using the atomic mass instead of the mass defect

The energy required to separate all nucleons is the binding energy, which comes from the mass defect — not from the atomic mass itself.

Students often plug $62.92960\ \text{u}$ into $E = mc^2$ and get a huge number, not realising that this is the mass of the whole atom, not the missing mass.

How to avoid: The mass defect $\Delta m$ is the difference between the mass of the separated nucleons and the actual mass of the nucleus.

For $^{63}_{29}\text{Cu}$:

• Number of protons = 29
• Number of neutrons = $63 - 29 = 34$
• Mass of 29 protons = $29 \times 1.007276\ \text{u}$
• Mass of 34 neutrons = $34 \times 1.008665\ \text{u}$
• Mass of 29 electrons = $29 \times 0.000548\ \text{u}$ (if using atomic masses)

Mistake 3: Forgetting to multiply by the number of atoms

After finding the binding energy for one copper nucleus, students sometimes stop there. The question asks for the energy to separate all nucleons in the entire coin.

How to avoid: Once you have the binding energy per nucleus (in J or MeV), multiply by the number of atoms $N$ found in Mistake 1.

Mistake 4: Unit confusion — mixing MeV, J, and u

Students often compute $\Delta m$ in u, then use $E = \Delta m c^2$ but forget that $1\ \text{u} \cdot c^2 = 931.5\ \text{MeV}$. They might try to convert to joules incorrectly.

How to avoid: Use the conversion directly:

$E_{\text{per nucleus}} = \Delta m\ (\text{in u}) \times 931.5\ \text{MeV/u}$ …