Q.How long can an electric lamp of 100 W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as
12H+12H→23He+n+3.27 MeV.
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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 — fusion energy comes from the mass defect, released as 3.27 MeV per D–D reaction.
Step 1: Energy per reaction: 3.27×1.602×10−13=5.24×10−13 J.
Step 2: Atoms of deuterium in 2.0 kg: 22000×6.022×1023=6.022×1026; each reaction uses 2 atoms, so reactions =3.011×1026.
Step 3: Total energy: 3.011×1026×5.24×10−13=1.577×1014 J. …
Converting the mass of deuterium into the number of D–D fusion reactions, then into total energy released, and finally into the time a 100 W lamp can run on it, gives a lamp lifetime of about 5.0×104 years.
Why this approach works
Deuterium nuclei fuse in pairs, each pair releasing 3.27 MeV. The lamp consumes 100 J every second. So the question reduces to: how many fusion events can 2.0 kg of deuterium supply, and how long will that energy last at 100 W? The only subtlety: each fusion event consumes two deuterium nuclei, so the number of reactions is half the number of atoms.
Step-by-step solution
1. Number of deuterium atoms in 2.0 kg
Deuterium (12H) has a molar mass of about 2.0 g/mol (mass number 2). So:
Moles=2.0 g/mol2000 g=1000 mol
Natoms=1000×NA=1000×6.022×1023=6.022×1026 atoms
2. Number of fusion reactions
Each reaction uses two deuterium nuclei:
Nreactions=26.022×1026=3.011×1026
3. Total energy released
Each reaction releases 3.27 MeV; using 1 MeV=1.602×10−13 J:
Ereaction=3.27×1.602×10−13=5.238×10−13 J
Etotal=(3.011×1026)×(5.238×10−13)=1.577×1014 J …
Method: Energy conservation via mass-energy equivalence (Einstein’s relation E=Δmc2).
Step 1 – Find the number of deuterium nuclei in 2.0 kg.
Deuterium (12H) has a molar mass of approximately 2.0 g/mol.
Number of moles in 2.0 kg:
n=2.0 g/mol2000 g=1000 mol
Number of nuclei:
N=n×NA=1000×6.022×1023=6.022×1026
Step 2 – Determine the energy released per fusion event.
Each fusion reaction involves two deuterium nuclei and releases 3.27 MeV.
So the number of fusion events possible from N nuclei is N/2 (since two nuclei are consumed per event).
Step 3 – Calculate total energy released.
Total energy in MeV:
Etotal=2N×3.27 MeV=26.022×1026×3.27
=3.011×1026×3.27≈9.846×1026 MeV
Convert to joules (1 MeV=1.602×10−13 J):
Etotal=9.846×1026×1.602×10−13≈1.577×1014 J
Step 4 – Relate energy to lamp power. …
Common Mistakes & How to Avoid Them
1. Forgetting to convert MeV to joules
The energy released per fusion event is given in MeV, but power is in watts (J/s). Many students plug 3.27 MeV directly into the calculation without converting to joules.
How to avoid: Always write the conversion factor explicitly before starting:
1 MeV=1.6×10−13 J.
So 3.27 MeV=3.27×1.6×10−13 J.
Never mix MeV and J in the same equation — convert everything to SI units first.
2. Using the wrong number of nuclei in 2 kg of deuterium
Deuterium is 12H, so its molar mass is 2 g/mol, not 1 g/mol. A common error is to treat it as ordinary hydrogen.
How to avoid:
Number of moles = molar massmass=2 g/mol2000 g=1000 mol.
Number of nuclei = 1000×NA=1000×6.02×1023=6.02×1026.
Each fusion reaction consumes two deuterium nuclei. So the number of reactions possible is half the number of deuterium nuclei.
3. Forgetting that each reaction uses two deuterium atoms
Students sometimes take the total number of deuterium nuclei as the number of reactions. But the reaction 12H+12H needs two per event.
How to avoid:
Number of reactions = 2total deuterium nuclei=26.02×1026=3.01×1026.
4. Confusing power with energy
Power is energy per second. The lamp consumes 100 J every second. The total energy from fusion gives the total time directly as t=PEtotal.
How to avoid:
Write the relation clearly:
P=tE⇒t=PE.
Do not multiply power and time — that gives energy, not time.
5. Arithmetic errors in powers of ten …
Showing the 12 most recent of 14 on this concept.
- AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQQ.What is the energy released in the following fission reaction? z236U→a117X+(z−a)117Y+2(01n) Given that the binding energy per nucleon of X and Y is 8.8 MeV and that of z236U is 7.5 MeV (A) 256.4 MeV (B) 248.6 MeV (C) 274.2 MeV (D) 289.2 MeV
›Reveal solutionSolution
Energy released in fission equals the increase in total nuclear binding energy from reactant to products; computing this gives 289.2 MeV.
Concept and Intuition
Nuclear fission releases energy because the fragment nuclei are, per nucleon, more tightly bound than the original heavy nucleus (binding energy per nucleon peaks around mass number ~56 and is higher for the mid-mass fragments than for very heavy nuclei like uranium). The energy released is exactly the difference between the total binding energy of the products and that of the reactant — free neutrons contribute zero binding energy since they are not bound in any nucleus.
Step-by-Step Solution
- Total binding energy of the reactant, Z236U: BEreactant=236×7.5=1770 MeV.
- The two product nuclei X and Y together have 117+117=234 nucleons, each bound at 8.8 MeV/nucleon: BEproducts=234×8.8=2059.2 MeV.
- The 2 free neutrons emitted carry no binding energy (they are not part of a nucleus). …
- AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQQ.Energy released in the fission of a single 92U235 nucleus is 200 MeV. The fission rate of 92U235 fueled reactor operating at a power level of 5W is (A) 1.56×1010 s−1 (B) 1.56×1011 s−1 (C) 1.56×1016 s−1 (D) 1.56×10−17 s−1
›Reveal solutionSolution
Divide reactor power by energy released per fission event to get the fission rate: ≈1.56×1011 fissions per second.
Concept and Intuition
A nuclear reactor's power output is the total energy released per second, which equals (energy released per fission) × (number of fissions per second). So the fission rate is just power divided by per-fission energy, after converting MeV to joules using 1 MeV=1.6×10−13 J.
Step-by-Step Solution
- Convert energy per fission to SI: E=200 MeV=200×1.6×10−13 J=3.2×10−11 J.
- Reactor power P=5 W=5 J/s.
- Fission rate R=EP=3.2×10−115=1.5625×1011 fissions per second.
Common Mistakes …
- AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQQ.Each nuclear fission of 235U releases 200 MeV of energy. If a reactor generates 1 MW power, then the rate of fission in the reactor is (A) 3.125×106 (B) 3.125×108 (C) 3.125×1010 (D) 3.125×1016
›Reveal solutionSolution
This tests converting a reactor's power output into a fission rate using the known energy released per fission event. The rate comes out to 3.125×1016 fissions/second.
Concept and Intuition
Power is energy delivered per unit time. If each fission event reliably releases a fixed amount of energy, then the number of fission events needed per second to sustain a given power output is simply that power divided by the energy released per event — exactly like asking how many coins of a given value you need per second to make up a certain rate of payment.
Step-by-Step Solution
- Energy released per fission: 200 MeV. Convert to joules using 1 eV=1.6×10−19 J:
Efission=200×106×1.6×10−19 J=3.2×10−11 J
- Reactor power: P=1 MW=106 W=106 J/s.
- Rate of fission (fissions per second): …
- AP EAPCET 2024Set ap-2024-05-17-AN1 markMCQQ.The energy released in the fission of one 92U235 nucleus is 200 MeV. The energy released in the fission of 235 g mass of 92U235 is nearly (A) 15.84×1012 J (B) 19.27×1012 J (C) 13.59×1012 J (D) 17.73×1012 J
›Reveal solutionSolution
235 g of U-235 is exactly one mole of nuclei; multiplying the per-nucleus fission energy by Avogadro's number gives 19.27×1012 J.
Concept and Intuition
The atomic mass of U-235 in grams equals one mole of that isotope (this is precisely why 235 g is chosen in the problem — it's a clean way to get exactly NA nuclei). So the total fission energy from 235 g is just the per-nucleus energy times Avogadro's number.
Step-by-Step Solution
- Convert per-nucleus energy to joules: 200 MeV=200×106×1.6×10−19 J=3.2×10−11 J.
- Number of nuclei in 235 g: N=NA=6.022×1023.
- Total energy: E=3.2×10−11×6.022×1023.
- 3.2×6.022=19.27, and the powers of ten give 10−11+23=1012, so E≈19.27×1012 J. …
- AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQQ.The decrease in each day in the Uranium mass of the material in a Uranium reactor operating at a power of 12 MW is (Energy released in one 92U235 fission is about 200 MeV) (A) 12.64×10−2 kg (B) 11.50×10−2 g (C) 12.64 kg (D) 12.64 g
›Reveal solutionSolution
This tests converting reactor power output into a mass of fissile fuel consumed per day, via the energy released per fission event.
Concept and Intuition
A nuclear reactor's power output comes from a huge number of individual fission events, each releasing a fixed amount of energy (~200 MeV for U-235). Knowing the total energy needed per day and the energy per fission tells us how many nuclei fission per day; multiplying by the mass per nucleus (via Avogadro's number) gives the mass consumed.
Step-by-Step Solution
- Total energy released per day: E=P×t=12×106 W×(24×3600 s)=12×106×86400=1.0368×1012 J.
- Energy released per fission: 200 MeV=200×106×1.6×10−19 J=3.2×10−11 J.
- Number of fissions per day: N=EfissionE=3.2×10−111.0368×1012≈3.24×1022. …
- AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQQ.If the energy released per fission of a 92235U nucleus is 200 MeV, the energy released in the fission of 0.1 kg of 92235U in kilowatt-hour is. (A) 22.8×105 (B) 22.8×107 (C) 11.4×105 (D) 820×1010
›Reveal solutionSolution
Count the U-235 atoms in the given mass, multiply by the per-fission energy, then convert joules to kilowatt-hours.
Concept and Intuition
Each fission event releases a fixed amount of energy (given as 200 MeV). To get the total energy from a macroscopic mass, we need the number of atoms/nuclei present, found via moles and Avogadro's number. The final conversion from joules to kWh uses the standard 1kWh=3.6×106 J.
Step-by-Step Solution
- Moles in 0.1 kg = 100 g: n=235100=0.4255 mol.
- Number of atoms: N=0.4255×6.022×1023=2.562×1023.
- Energy per fission in joules: 200 MeV=200×1.6×10−13 J=3.2×10−11 J.
- Total energy: E=2.562×1023×3.2×10−11=8.20×1012 J. …
- AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQQ.In a nuclear reactor, the fuel is consumed at the rate of 1×10−3 gs−1. The power generated in kW is (A) 9×1014 (B) 9×107 (C) 9×108 (D) 9×1012
›Reveal solutionSolution
Converting the mass-consumption rate to a power via E=mc2 gives 9×107 kW.
Concept and Intuition
In a nuclear reactor, mass is converted into energy according to Einstein's mass-energy equivalence, E=mc2. If a certain mass of fuel is "consumed" (i.e., converted to energy) every second, then the energy released per second — which is exactly the power output — follows directly by plugging the mass rate into this formula.
Step-by-Step Solution
- Given mass consumption rate: dtdm=1×10−3 g/s=1×10−6 kg/s.
- Power (energy released per second): P=dtdmc2.
- c2=(3×108)2=9×1016 m2/s2.
- P=1×10−6×9×1016=9×1010 W. …
- AP EAPCET 2023Set ap-2023-05-23-AN1 markMCQQ.If the speed of the light was half the present value, the energy released in a nuclear reaction decreases by (A) 100% (B) 75% (C) 50% (D) 25%
›Reveal solutionSolution
Since nuclear energy release follows E=mc2, halving c reduces c2 (and hence E) to one-quarter of its original value — a 75% decrease.
Concept and Intuition
The mass–energy equivalence relation E=mc2 shows energy scales with the square of the speed of light, not linearly with it. So even a modest change in c produces an amplified (squared) change in the energy released for the same mass defect.
Step-by-Step Solution
- Original energy: E=mc2.
- New speed: c′=c/2, so new energy E′=m(c′)2=m4c2=4E. …
- AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQQ.The nucleus having highest binding energy per nucleon is (A) 816O (B) 2656Fe (C) 84208Pb (D) 24He
›Reveal solutionSolution
The binding-energy-per-nucleon curve peaks around iron (A ≈ 56); among the given nuclei, Fe-56 sits closest to that peak.
Concept and Intuition
The famous binding energy per nucleon vs. mass number curve rises steeply for light nuclei, peaks around A=56 (iron group), then falls slowly for heavier nuclei. This is why fusion releases energy for light elements and fission releases energy for heavy elements — both processes move nuclei toward the iron peak.
Step-by-Step Solution
- Recall approximate binding energy per nucleon values: 24He≈7.1 MeV, 816O≈8.0 MeV, 2656Fe≈8.8 MeV, 82208Pb≈7.9 MeV.
- Comparing these, Fe-56 has the maximum value among the choices. …
- AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQQ.The process that mainly takes place in stars to produce energy (A) nuclear fission (B) nuclear fusion (C) ionization (D) annihilation
›Reveal solutionSolution
This tests basic astrophysics: the source of a star's energy. Answer: nuclear fusion.
Concept and Intuition
Stars like our Sun are powered by thermonuclear fusion reactions in their cores, where
extreme temperature and pressure force light nuclei (mainly hydrogen) to fuse into
heavier nuclei (helium), converting a tiny fraction of mass into a huge amount of
energy (E=mc2) that is radiated as heat and light.
Step-by-Step Solution
- Compare the options: nuclear fission is the splitting of heavy nuclei (used in reactors on Earth, not the dominant stellar process); ionization and annihilation don't release nearly enough sustained energy to power a star for billions of years.
- In stars, hydrogen nuclei fuse (proton-proton chain, or CNO cycle in heavier stars) …
- AP EAPCET 2022Set ap-2022-07-12-AN1 markMCQQ.An experimental train uses 1 g of nuclear material to run. 90 % of the produced energy is wasted to overcome the frictional force between the wheels and the track. If the weight of each coach is 18 ton (18000 kg) and it runs at a speed of 100 m s−1, the number of coaches the engine can drag at a time is (Speed of light in vacuum is =3×108 m s−1) (A) 100 (B) 1000 (C) 10000 (D) 100000
›Reveal solutionSolution
Mass-energy from the fuel splits 90%-wasted / 10%-useful; the useful 10% supplies the kinetic energy of the coaches at their running speed, from which the number of coaches follows.
Concept and Intuition
By E=mc2, even a tiny mass of nuclear fuel releases enormous energy. Here, 90% of that energy is dissipated as heat overcoming friction between wheels and track (this is the "wasted" part), leaving 10% as the useful energy that actually goes into moving the train — i.e. giving the coaches their kinetic energy at running speed.
Step-by-Step Solution
- Total energy released: E=mc2=(10−3 kg)×(3×108 m/s)2=9×1013 J.
- Useful (non-wasted) energy: 10% of E = 9×1012 J.
- This equals the total kinetic energy of n coaches, each of mass M=18000 kg, moving at v=100 m/s: …
- AP EAPCET 2021Set eng-2021-08-19-AN1 markMCQQ.Which of the following statement is correct? (A) The rest mass of the stable nucleus is less than the sum of the rest masses of its separated nucleons (B) The rest mass of the stable nucleus is greater than the sum of the rest masses of its separated nucleons (C) In nuclear fission, energy is released by fusion is of two nuclei of medium mass (approximate 100amu) (D) In nuclear fission, energy is released by fragmentation is very low atomic mass nucleus
›Reveal solutionSolution
A stable nucleus's rest mass is less than the sum of its free nucleons' rest masses — the mass defect that accounts for its binding energy — making option (A) correct.
Concept and Intuition
Binding a set of nucleons together into a nucleus releases energy (the binding energy), and by mass-energy equivalence, that released energy corresponds to a loss of rest mass. So a bound (stable) nucleus always weighs less than its separated, free constituent nucleons — this "mass defect" is a foundational nuclear-physics fact, and it is precisely why nuclear reactions that increase binding energy per nucleon (fission of heavy nuclei, or fusion of light nuclei) release energy.
Step-by-Step Solution
- Binding energy: BE=(sum of rest masses of free nucleons−rest mass of nucleus)c2>0 for any stable, bound nucleus.
- This directly means: rest mass of nucleus < sum of rest masses of separated nucleons — confirming option (A) and ruling out (B).
- Nuclear fission is the splitting of a heavy nucleus (like U-235) into medium-mass fragments (~100 amu each), releasing energy because the fragments are more tightly bound per nucleon than the parent — this is fission, not "fusion of two nuclei of medium mass" as option (C) mislabels it. …
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