The fission properties of $^{239}_{94}\text{Pu}$ are very similar to those of $^{235}_{92}\text{U}$. The average energy released per fission is $180\ \text{MeV}$. How much energy, in MeV, is released if all the atoms in $1\ \text{kg}$ of pure $^{239}_{94}\text{Pu}$ undergo fission?

The fission properties of ²³⁹₉₄Pu are very similar to those of ²³⁵₉₂U. The average energy released per fission

Exercises · 13.17

^{239}_{94}\text{Pu}

Uttar Pradesh UpmspTextbookSubjective· 3mImportance★★★★★

34% · 17/50 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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" …

Start your 14-day free trial to unlock the full solution →

The total energy released is found by multiplying the number of atoms in 1 kg of $^{239}\text{Pu}$ by the energy per fission (180 MeV). Using Avogadro’s number and the molar mass, the result is approximately $4.54 \times 10^{26}\ \text{MeV}$ .

The core idea here is mass–energy equivalence — but not in the way you might first think. We aren’t directly converting the entire mass of plutonium into energy (that would be annihilation, not fission). Instead, each fission event converts a tiny fraction of the nucleus’s mass into kinetic energy of fragments and neutrons, which we measure as 180 MeV per fission. To find the total energy from 1 kg, we simply count how many fission events happen and multiply.

Let’s walk through it step by step.

Find the number of atoms in 1 kg of $^{239}\text{Pu}$ . The molar mass of $^{239}\text{Pu}$ is approximately $239\ \text{g/mol}$ (since the atomic mass number is 239). One mole contains Avogadro’s number of atoms, $N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}$ . For 1 kg = 1000 g, the number of moles is:

n = \frac{1000\ \text{g}}{239\ \text{g/mol}} \approx 4.184\ \text{mol}

So the number of atoms is:

N = n \times N_A = 4.184 \times 6.022 \times 10^{23} \approx 2.52 \times 10^{24}

Multiply by the energy per fission. Each fission releases 180 MeV. Therefore, total energy:

E = N \times 180\ \text{MeV} = 2.52 \times 10^{24} \times 180

Start your 14-day free trial to unlock the full solution →

Common Mistakes & How to Avoid Them

Mistake 1: Using the wrong mass number or atomic mass

Students often take the mass number (239) as the exact atomic mass in grams, or they confuse it with the molar mass in g/mol. The mass number is approximately the molar mass, but the real atomic mass of $^{239}\text{Pu}$ is 239.05216 u — close, but not exactly 239.

How to avoid: For exam problems, when the exact atomic mass is not given, use the mass number (239) as the molar mass in g/mol. This is the standard approximation in such questions. So 1 mole of $^{239}\text{Pu}$ has a mass of 239 g.

Mistake 2: Forgetting to convert kg to g

The problem gives mass in kg (1 kg), but the molar mass is in g/mol. Students sometimes plug 1 kg directly into the formula without converting.

How to avoid: Always check units. Convert 1 kg = 1000 g before using the molar mass.

Mistake 3: Confusing number of atoms with number of moles

Some students calculate the number of moles correctly but then forget to multiply by Avogadro's number to get the number of atoms.

How to avoid: Remember the chain:

\text{Number of atoms} = \frac{\text{mass in grams}}{\text{molar mass in g/mol}} \times N_A

Mistake 4: Using the wrong value of Avogadro's number

Using $6.022 \times 10^{23}$ is fine, but some students use $6.023 \times 10^{26}$ (which is for kg-mole) or forget the exponent entirely.

How to avoid: Stick to $N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}$ for gram-mole calculations.

Mistake 5: Arithmetic errors in the final multiplication

The numbers are large — $1000/239 \approx 4.184$ , multiplied by $6.022 \times 10^{23}$ , then by 180. Students often misplace decimal points or exponents.

How to avoid: Do the calculation step by step, and keep track of powers of 10 separately.

Correct Solution …

Start your 14-day free trial to unlock the full solution →

←Previous question13.16 — Write nuclear reaction equations for (i) α-decay of ²²⁶₈₈Ra (ii) α-dec…→Next question13.17 — A radioactive isotope has a half-life of T years. How long will it tak…

🎓Unlock everything free for 14 days

Unlock everything free for 14 days