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 $62.92960\ \text{u}$).

A given coin has a mass of 3.0g. Calculate the nuclear energy that would be required to separate all the neutr

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3.0\ \text{g}

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Concept understanding — Mass Energy Equivalence

Mass Energy Equivalence: From Intuition to the Formula

Imagine you have a lump of coal. You know you can burn it to get heat, and that heat can run a steam engine. The energy you get out seems to come from the chemical bonds in the coal. But what if I told you that the coal itself — just sitting there, not burning — already contains a staggering amount of energy locked inside its very mass? That is the core idea of mass-energy equivalence.

The Intuition: Mass is Frozen Energy

Think of mass as a kind of "frozen" or "stored" energy. When you burn coal, you are only releasing a tiny fraction of this stored energy — the energy in the chemical bonds. The rest of the mass remains as matter. But if you could somehow completely convert that lump of coal into pure energy, you would get an unimaginable amount — enough to power a city for years.

This is not a metaphor. Mass and energy are not two separate things that can be converted into each other like dollars and rupees. They are the same fundamental thing, just in different forms. Mass is a highly concentrated form of energy. Energy, when concentrated enough, behaves like mass.

The Precise Statement

The relationship is given by the most famous equation in physics:

E = mc^2

Where:

$E$ is the energy equivalent of the mass (in joules, J)
$m$ is the mass (in kilograms, kg)
$c$ is the speed of light in vacuum ( $3 \times 10^8$ m/s)

The speed of light is a huge number. Squaring it makes it enormous. This is why a tiny amount of mass corresponds to a colossal amount of energy.

What This Equation Actually Means

The equation tells you exactly how much energy is "stored" inside any object with mass $m$ . If you could annihilate that mass completely, you would get $E$ joules of energy.

Example: A 1 kg mass (like a litre of water) contains:

E = 1 \times (3 \times 10^8)^2 = 9 \times 10^{16} \text{ J}

That is 90 quadrillion joules — roughly the energy released by a 20-megaton nuclear bomb. This is not energy you can normally access; it is locked inside the nucleus of atoms.

Where Does This Show Up in Real Life?

Nuclear Reactions: In nuclear fission (splitting atoms) or fusion (joining atoms), a tiny fraction of the mass of the nucleus is converted into energy. The mass of the products is slightly less than the mass of the reactants. The "missing" mass has become energy — exactly as $E=mc^2$ predicts. This is how the Sun works and how nuclear power plants generate electricity.
Particle Physics: When a particle and its antiparticle meet, they annihilate completely into pure energy (usually gamma rays). The energy produced equals $mc^2$ for the two particles. …

Why this formula?

Why $E = mc^2$ — The Reasoning Behind Mass-Energy Equivalence

The formula $E = mc^2$ is not a random guess. Einstein arrived at it by thinking deeply about what happens to energy when you move an object. The core insight: if an object gains energy, it must behave as if it has gained mass.

The Starting Point: Relativistic Momentum

In special relativity, the momentum of a particle is not simply $p = mv$ . Instead, it is:

p = \frac{m_0 v}{\sqrt{1 - v^2/c^2}}

where $m_0$ is the rest mass (mass measured when the object is at rest). This formula already tells us something strange: as speed approaches $c$ , momentum shoots toward infinity — no object with mass can reach the speed of light.

The Energy-Momentum Relation

Einstein then asked: what is the correct expression for kinetic energy that matches this new momentum? In classical physics, kinetic energy is $K = \frac{1}{2}mv^2$ . But that formula fails at high speeds.

The relativistic kinetic energy turns out to be:

K = \frac{m_0 c^2}{\sqrt{1 - v^2/c^2}} - m_0 c^2

This looks odd — why subtract $m_0 c^2$ ? Because when $v = 0$ , the first term becomes $m_0 c^2$ , and we want $K = 0$ at rest. So the subtraction gives zero kinetic energy when the object is stationary.

Note

The term $m_0 c^2$ appears naturally as a rest energy — energy that an object has simply because it has mass, even when completely at rest.

The Crucial Step: What Happens When You Add Energy?

Now consider a box that emits light (photons) in opposite directions. The light carries away energy. Classical physics says the box loses energy but its mass stays the same. Einstein showed this cannot be true.

The argument (simplified): if the box emits a pulse of light with energy $E$ , the light carries momentum $p = E/c$ . By conservation of momentum, the box recoils. But after the light is absorbed by the opposite wall, the box stops. The net effect: the box has moved slightly. Its center of mass has shifted — unless the energy carried by the light also carried mass.

Important

For the center of mass of the entire system (box + light) to remain stationary, the light must behave as if it has an effective mass $m = E/c^2$ . Therefore, energy itself has inertia.

The Full Formula

The total energy of any object — moving or at rest — is:

E = \frac{m_0 c^2}{\sqrt{1 - v^2/c^2}}

For an object at rest ( $v = 0$ ), this reduces to:

E = m_0 c^2

For a moving object, the total energy is the sum of rest energy and kinetic energy:

E = m_0 c^2 + K

E = mc^2

Why It's Not Just a "Conversion" …

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Separating every nucleon in the coin means supplying the coin's total nuclear binding energy: the mass defect per $^{63}_{29}\text{Cu}$ atom ( $0.5919\ \text{u}\to551.4\ \text{MeV}$ ) times the $2.87\times10^{22}$ atoms in $3.0\ \text{g}$ , giving $\approx2.5\times10^{12}\ \text{J}$ .

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:

\Delta m = 29\,m(^1\text{H}) + 34\,m_n - m(^{63}\text{Cu})

\Delta m = 29(1.007825) + 34(1.008665) - 62.92960

\Delta m = 29.226925 + 34.294610 - 62.92960 = 0.591935\ \text{u}

2. Binding energy per atom

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

E_{\text{atom}} = 0.591935\times931.5 = 551.4\ \text{MeV}

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

E_{\text{atom}} = 551.4\times1.602\times10^{-13} = 8.833\times10^{-11}\ \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 …

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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:

n = \frac{3.0}{62.92960} \approx 0.04767\ \text{mol}

Number of atoms:

N = n \times N_A = 0.04767 \times 6.022 \times 10^{23} \approx 2.87 \times 10^{22}\ \text{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:

29(1.007825) + 34(1.008665) = 29.226925 + 34.29461 = 63.521535\ \text{u}

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

Mass defect for one atom:

\Delta m = 63.521535 - 62.92960 = 0.591935\ \text{u} \approx 0.59193\ \text{u}

Step 3 — Convert mass defect to energy per atom

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

E_{\text{per atom}} = 0.59193 \times 931.5 \approx 551.4\ \text{MeV}

Step 4 — Total energy for the coin …

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

Watch out

A shortcut that causes errors: some students divide $3.0$ by $62.92960$ directly, forgetting the conversion factor. This gives a meaningless number.

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)

Important

Always check whether you're given atomic masses (include electrons) or nuclear masses. Here, $62.92960\ \text{u}$ is the atomic mass of $^{63}\text{Cu}$ , so you must account for electrons when computing the mass defect.

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:

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

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1. Mass defect of one 2963Cu^{63}_{29}\text{Cu}2963​Cu atom

2. Binding energy per atom

3. Number of atoms in the coin

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

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