Q.The fission properties of 94239Pu are very similar to those of 92235U. The average energy released per fission is 180 MeV. How much energy, in MeV, is released if all the atoms in 1 kg of pure 94239Pu undergo fission?
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🔒 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=mc2
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 (3×108 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×(3×108)2=9×1016 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?
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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=mc2 predicts. This is how the Sun works and how nuclear power plants generate electricity.
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Particle Physics: When a particle and its antiparticle meet, they annihilate completely into pure energy (usually gamma rays). The energy produced equals mc2 for the two particles. …
Why this formula?
Why E=mc2 — The Reasoning Behind Mass-Energy Equivalence
The formula E=mc2 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=1−v2/c2m0v
where m0 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=21mv2. But that formula fails at high speeds.
The relativistic kinetic energy turns out to be:
K=1−v2/c2m0c2−m0c2
This looks odd — why subtract m0c2? Because when v=0, the first term becomes m0c2, and we want K=0 at rest. So the subtraction gives zero kinetic energy when the object is stationary.
The term m0c2 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.
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/c2. Therefore, energy itself has inertia.
The Full Formula
The total energy of any object — moving or at rest — is:
E=1−v2/c2m0c2
For an object at rest (v=0), this reduces to:
E=m0c2
For a moving object, the total energy is the sum of rest energy and kinetic energy:
E=m0c2+K
E=mc2
where m is the relativistic mass m=1−v2/c2m0, or equivalently:
E2=(pc)2+(m0c2)2
Why It's Not Just a "Conversion" …
Concept: Mass–Energy Equivalence – the energy released is the number of fissions multiplied by the energy per fission.
Step 1: Number of atoms in 1 kg of 94239Pu
Molar mass M=239 g/mol=0.239 kg/mol.
Number of moles n=0.239 kg/mol1 kg≈4.184 mol.
Avogadro’s number NA=6.022×1023 atoms/mol.
Total atoms N=n×NA≈4.184×6.022×1023≈2.52×1024.
Step 2: Total energy released …
The total energy released is found by multiplying the number of atoms in 1 kg of 239Pu by the energy per fission (180 MeV). Using Avogadro’s number and the molar mass, the result is approximately 4.54×1026 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 239Pu. The molar mass of 239Pu is approximately 239 g/mol (since the atomic mass number is 239). One mole contains Avogadro’s number of atoms, NA=6.022×1023 mol−1. For 1 kg = 1000 g, the number of moles is:
n=239 g/mol1000 g≈4.184 mol
So the number of atoms is:
N=n×NA=4.184×6.022×1023≈2.52×1024
- Multiply by the energy per fission. Each fission releases 180 MeV. Therefore, total energy:
E=N×180 MeV=2.52×1024×180
Method: Direct calculation using Avogadro’s number and mass–energy equivalence.
The idea is simple: find the number of atoms in 1 kg of plutonium-239, then multiply by the energy released per fission.
Step 1 – Find the number of moles in 1 kg of 239Pu.
The molar mass of 239Pu is 239 g/mol (since the mass number is 239).
1 kg=1000 g, so
n=2391000 mol.
Step 2 – Find the number of atoms.
Avogadro’s number NA=6.022×1023 atoms/mol.
N=n×NA=2391000×6.022×1023.
Step 3 – Multiply by the energy per fission.
Each fission releases 180 MeV.
E=N×180=2391000×6.022×1023×180.
Step 4 – Calculate.
First, 2391000≈4.1841. …
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 239Pu 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 239Pu 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:
Number of atoms=molar mass in g/molmass in grams×NA
Mistake 4: Using the wrong value of Avogadro's number
Using 6.022×1023 is fine, but some students use 6.023×1026 (which is for kg-mole) or forget the exponent entirely.
How to avoid: Stick to NA=6.022×1023 mol−1 for gram-mole calculations.
Mistake 5: Arithmetic errors in the final multiplication
The numbers are large — 1000/239≈4.184, multiplied by 6.022×1023, 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 …
Showing the 12 most recent of 15 on this concept.
- CBSE 2026Set A1 markMCQQ.How many joules of energy will be released due to mass defect of 1 mg? (A) 9 × 10^6 J (B) 3 × 10^8 J (C) 9 × 10^10 J (D) 3 × 10^2 J
›Reveal solutionSolution
E = mc² with m = 1 mg = 1 × 10⁻⁶ kg gives 9 × 10¹⁰ J.
Using Einstein's mass–energy relation E=mc2:
Mass defect m=1 mg=1×10−6 kg.
Speed of light c=3×108 m/s.
…
- CBSE 2026Set ANNUAL1 markMCQQ.The equivalent energy of 10 g of substance is(a) 9 x 10^14 J(b) 9 x 10^13 J(c) 3 x 10^16 J(d) 6 x 10^13 J
›Reveal solutionSolution
Mass-energy equivalence E = m*c^2 converts a mass of 10 g directly into an energy of 9 x 10^14 J.
Einstein's mass-energy equivalence relation states that mass and energy are interconvertible, related by
E = m * c^2
where m is the mass (in kg) and c = 3 x 10^8 m/s is the speed of light in vacuum. Here m = 10 g = 0.01 kg. Substituting:
E = 0.01 * (3 x 10^8)^2 = 0.01 * 9 x 10^16 = 9 x 10^14 J
…
- CBSE 2026Set ANNUAL1 markMCQQ.The energy equivalent of 1 mg of substance is(a) 9 × 10^10 J(b) 9 × 10^13 J(c) 1.35 × 10^14 J(d) 3 × 10^13 J
›Reveal solutionSolution
Einstein's mass–energy relation E=mc2 shows even a tiny mass carries an enormous amount of energy.
Given: m=1 mg=1×10−6 kg, c=3×108 m/s.
E=mc2=(1×10−6)(3×108)2=(1×10−6)(9×1016)
E=9×1010 J
…
- CBSE 2025Set IMPROVEMENT1 markQ.Write the energy equivalent of one atomic mass unit in joules.
›Reveal solutionSolution
Using Einstein's mass–energy relation E=mc2 with m=1u=1.6605×10−27kg gives the energy equivalent of 1 amu.
The energy equivalent of a mass m is E=mc2. For one atomic mass unit, m=1.6605×10−27kg and c=3×108m/s:
E=(1.6605×10−27)×(3×108)2=1.6605×10−27×9×1016 …
- CBSE 2025Set D1 markMCQQ.Equivalent energy of 1 amu is (A) 190 MeV (B) 139 MeV (C) 913 MeV (D) 931 MeV
›Reveal solutionSolution
Mass–energy equivalence gives 1 u ≈ 931 MeV.
Using Einstein's relation E = mc² with 1 u = 1.6605×10⁻²⁷ kg:
E=(1.6605×10−27)(3×108)2=1.4924×10−10 J
Converting to MeV (1 MeV = 1.602×10⁻¹³ J): …
- CBSE 2024Set A1 markQ.Match Column 'A' with Column 'B' and write the correct pair. Column 'A' item: 'Mass-energy equivalence relation'. Column 'B' options:(i) De-Broglie(ii) Maxwell(iii) Ohm(iv) Einstein(v) Coulomb(vi) Lenz(vii) Young.
›Reveal solutionSolution
Mass-energy equivalence, E = mc², was given by Albert Einstein.
Albert Einstein, through his theory of relativity, showed that mass and energy are equivalent and interconvertible, related by:
E=mc2 …
- CBSE 2024Set ANNUAL1 markQ.Write Einstein's mass-energy equivalent relation.
›Reveal solutionSolution
Einstein's mass-energy equivalence states that mass itself is a form of energy, related by the constant c².
Einstein's special theory of relativity showed that mass and energy are two forms of the same physical quantity and are interconvertible. The mass-energy equivalence relation is:
E=mc2
…
- CBSE 2023Set ANNUAL1 markQ.Answer in one word/sentence: Give the mass-energy equivalence equation.
›Reveal solutionSolution
Einstein's equation E = m*c^2 shows that mass and energy are equivalent and interconvertible, with c^2 (a very large number) as the conversion factor.
According to Einstein's theory, any mass m has an associated rest energy E given by:
E = m*c^2
…
- CBSE 2022Set ANNUAL1 markQ.Calculate the energy equivalent of 1 gm. of substance.
›Reveal solutionSolution
Use E=mc2 with m=1g=10−3kg.
By Einstein's mass-energy equivalence relation, E=mc2.
Here m=1gm=1×10−3kg and c=3×108m/s (given constant).
…
- CBSE 2022Set I1 markMCQQ.Which of the following relations is correct for mass and energy? (A) m = E (B) m^2 = E (C) mc^2 = E (D) m = √E / 2
›Reveal solutionSolution
Mass and energy are equivalent: E = mc².
Einstein's special theory of relativity established that mass and energy are interchangeable. A mass m is equivalent to an energy:
E=mc2
…
- CBSE 2022Set ANNUAL1 markMCQQ.Match Column A item 'Energy – mass equality' with the correct item in Column B.(a) kg m^2(b) Poise(c) Cp - Cv = R(d) E = mc^2(e) 1/frequency(f) distance(g) 24 hours
›Reveal solutionSolution
The mass-energy equivalence relation, discovered by Einstein, is E = mc^2, where c is the speed of light in vacuum.
This relation states that a mass m has an equivalent rest energy E = mc^2, and conversely that energy has an equivalent mass. It underlies nuclear energy release (a small mass defect converts into a large a …
- CBSE 2019Set HE2341 markMCQQ.The energy equivalent of 1 gram of substance is:(i) 9×10^6 Joule(ii) 3×10^13 Joule(iii) 3×10^6 Joule(iv) 9×10^13 Joule
›Reveal solutionSolution
E = mc² with m = 1 g = 10⁻³ kg and c = 3×10⁸ m/s gives E = 9×10¹³ J.
Einstein's mass-energy equivalence relation states that mass and energy are interconvertible:
E=mc2
Here m=1 g=1×10−3 kg and c=3×108 m/s (speed of light).
E=(1×10−3)×(3×108)2=(1×10−3)×(9×1016)=9×1013 J …
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