Q.Suppose India had a target of producing by 2020 AD, 200,000 MW of electric power, ten percent of which was to be obtained from nuclear power plants. Suppose we are given that, on an average, the efficiency of utilization (i.e. conversion to electric energy) of thermal energy produced in a reactor was 25%. How much amount of fissionable uranium would our country need per year by 2020? Take the heat energy per fission of 235U to be about 200MeV.
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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" …
Working backward from the required electrical output, through the stated conversion efficiency to get thermal power, then dividing by the energy released per fission, gives the number of fissions — and hence the mass of U-235 — needed each year. …
Find the required nuclear electric output (10% of 200,000 MW), convert to thermal power using the 25% efficiency, find the annual thermal energy, divide by 200 MeV/fission to get the number of fissions, and convert to mass. Result: about 3.08 × 10⁴ kg (roughly 30.8 tonnes) of U-235 per year.
Step 1 — Required nuclear electric power
Pelec=10%×200,000 MW=20,000 MW=2×1010 W
Step 2 — Required thermal power
Since only 25% of thermal energy converts to electricity:
Pthermal=0.25Pelec=0.252×1010=8×1010 W
Step 3 — Thermal energy needed per year
Eyear=Pthermal×(1 yr in seconds)=8×1010×3.154×107=2.5232×1018 J
Step 4 — Number of fissions needed
Efission=200 MeV=200×1.6×10−13=3.2×10−11 J
N=3.2×10−112.5232×1018=7.885×1028 fissions/year
Step 5 — Mass of U-235 needed …
- GUJCET 2026Set x1 markMCQQ.The energy equivalent of 1.0 kg of substance is ______. (A) 9×1013 J (B) 3×1013 J (C) 9×1016 J (D) 9×1014 J
›Reveal solutionSolution
Mass–energy equivalence: E=mc2=9×1016 J for 1 kg.
Einstein's mass–energy relation:
E=mc2
With m=1.0 kg and c=3×108 m/s: …
- GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQQ.Equivalent energy of 2 g of substance is ___.(a) 18 x 10^13 J(b) 9 x 10^13 J(c) 6 x 10^11 J(d) 6 x 10^8 J
›Reveal solutionSolution
Mass-energy equivalence, E = m c^2, converts any mass entirely into energy; even a small mass corresponds to an enormous energy because c^2 is so large.
m = 2 g = 2 x 10^-3 kg, c = 3 x 10^8 m/s.
…
- GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQQ.According to Einstein's mass-energy equivalent relation, the energy equivalent of 1mg of substance is ___. (Speed of light in vacuum C = 3 x 10^8 m/s)(a) 9 x 10^13 J(b) 9 x 10^10 J(c) 9 x 10^-13 J(d) 9 x 10^-10 J
›Reveal solutionSolution
Mass-energy equivalence: E = mc², where m is the rest mass and c is the speed of light in vacuum.
Given m = 1 mg = 1 × 10⁻⁶ kg, c = 3 × 10⁸ m/s.
…
- GUJCET 2023Set 091 markMCQQ.In proton-proton cycle in Sun the energy released when an electron & its antiparticle combines is ______. (A) 1.021×10−13 J (B) 0.672×10−13 J (C) 1.126×10−13 J (D) 1.632×10−13 J
›Reveal solutionSolution
[!TLDR] 2mec2=1.022 MeV =1.63×10−13 J.
Concept
In electron–positron annihilation the entire rest mass of both particles converts to energy (usually two gamma photons). Each has rest energy mec2=0.511 MeV, so the total released is 2×0.511=1.022 MeV.
Solution
Convert to joules: …
- GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQQ.According to mass energy equivalence relation, 9 x 10^13 J of energy can be converted into ___ maximum mass. [Speed of light c = 3 x 10^8 m/s](a) 81 g(b) 9 g(c) 3 g(d) 1 g
›Reveal solutionSolution
Mass-energy equivalence m = E/c^2 gives 9x10^13/(9x10^16) = 10^-3 kg = 1 g.
Einstein's relation: E = m c^2, so m = E/c^2.
…
- GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQQ.One of the fusion reaction in Sun is given by (2,1)H + (1,1)H -> (3,2)He + gamma + ___. Fill in the blank with correct option.(a) 1.02 MeV(b) 5.49 MeV(c) 12.86 MeV(d) 0.42 MeV
›Reveal solutionSolution
The deuterium-proton fusion d + p -> He-3 + gamma releases 5.49 MeV; this is a known step of the solar proton-proton cycle.
The reaction (2,1)H + (1,1)H -> (3,2)He + gamma is the second step of the proton-proton fusion chain that powers the Sun.
…
- GUJCET 2022Set 171 markMCQQ.Given the following atomic masses: 92238U=238.05079 u, 24He=4.00260 u, 90234Th=234.04363 u. Calculate the energy released during the alpha decay of 92238U. (1u=931.5 MeV/c2) (A) 4.25 MeV (B) 6.23 MeV (C) 5.75 MeV (D) 3.25 MeV
›Reveal solutionSolution
Energy released =Δm×931.5 MeV.
Steps.
- Δm=238.05079−(234.04363+4.00260)=0.00456 u. …
- GUJCET 2020Set 071 markMCQQ.Calculate the energy equivalent of 1g of substance (A) 6×1011 J (B) 9×1013 J (C) 4×1012 J (D) 7×1012 J
›Reveal solutionSolution
Mass–energy equivalence: E=mc2 with m=1 g =10−3 kg.
Concept: E=mc2 where m=1g=10−3kg and c=3×108m/s: …
- GUJCET 2015Set C1 markMCQQ.The energy released by the fission of one uranium atom is 200 MeV. The number of fission per second required to produce 6.4 W power is _____. (A) 2×1011 (B) 1011 (C) 1010 (D) 2×1010
›Reveal solutionSolution
[!TLDR] n=P/Efission=2×1011 fissions per second.
Concept
Power is energy released per unit time. If each fission releases energy E, then producing power P requires n=P/E fissions per second.
Solution …
- GUJCET 2014Set A1 markMCQQ.The binding energy per nuclean of 8O16 is 7.97 MeV and that of 8O17 is 7.75 MeV. The energy required to remove one neutron from 8O17 is __________ MeV. (A) 3.52 (B) 3.62 (C) 4.23 (D) 7.86
›Reveal solutionSolution
[!TLDR]
Energy to remove a neutron = BE(O-17) - BE(O-16) = 131.75−127.52=4.23 MeV. Answer: (C).
Concept
The binding energy of a nucleus is (BE per nucleon) x (number of nucleons). Removing one neutron from 8O17 leaves 8O16; the energy needed equals the difference between the total binding energies of the two nuclei (NCERT/CBSE nuclei chapter).
Solution
Total binding energy of 8O17 (17 nucleons):
BE17=7.75×17=131.75 MeV …
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