Q.Find the energy equivalent of one atomic mass unit, first in Joules and then in MeV. Using this, express the mass defect of 816O in MeV/c2.
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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" …
Use E=mc2 with 1 u=1.6605×10−27 kg.
Energy in joules:
E=(1.6605×10−27)(2.998×108)2≈1.492×10−10 J.
In MeV (using 1 MeV=1.602×10−13 J):
E=1.602×10−131.492×10−10≈931.5 MeV,so 1 u≡931.5 MeV/c2.
Mass defect of 816O (Z=N=8), with m(1H)=1.007825 u, mn=1.008665 u, m(16O)=15.994915 u: …
1 u is equivalent to 1.492×10−10 J=931.5 MeV/c2; using this, the mass defect of 816O is 0.137005 u≈127.6 MeV/c2.
Energy equivalent of one atomic mass unit
Mass-energy equivalence, E=mc2, converts any mass into an energy. One atomic mass unit is 1 u=1.6605×10−27 kg, so
E=(1 u)c2=(1.6605×10−27 kg)(2.998×108 m/s)2=1.492×10−10 J.
Convert to MeV using 1 MeV=1.602×10−13 J:
E=1.602×10−131.492×10−10 MeV=931.5 MeV.
Therefore
1 u≡931.5 MeV/c2.
Mass defect of 816O
Oxygen-16 has Z=8 protons and N=8 neutrons. The mass defect is the difference between the total mass of the free constituents and the actual atomic mass. Using atomic masses (so the proton is represented by the 1H atom, which balances the 8 electrons):
m(1H)=1.007825 u,mn=1.008665 u,m(816O)=15.994915 u. …
Method: Direct Application of E=mc2 Using the Unified Mass Unit
The core idea is simple: one atomic mass unit (u) is defined as 1/12 the mass of a carbon-12 atom. Its energy equivalent comes straight from Einstein's relation — multiply the mass (in kg) by c2 to get Joules, then convert Joules to MeV using the known conversion factor.
Step 1 — Energy equivalent of 1 u in Joules
First, recall the value of 1 u in kilograms:
1 u=1.660539×10−27 kg
Speed of light: c=2.99792458×108 m/s
Now apply E=mc2:
E=(1.660539×10−27)×(2.99792458×108)2
Square c first:
c2=(2.99792458×108)2=8.987551787×1016 m2/s2
Multiply:
E=1.660539×10−27×8.987551787×1016
E=1.492418×10−10 J
1 u≡1.492×10−10 J
Step 2 — Convert to MeV
We need the conversion: 1 eV=1.602176634×10−19 J
So 1 MeV=1.602176634×10−13 J
Divide the energy in Joules by the energy of 1 MeV:
E (in MeV)=1.602176634×10−131.492418×10−10
E=931.494 MeV
1 u≡931.5 MeV/c2
The "per c2" is often dropped in casual speech, but in mass-energy equivalence, mass is E/c2. So 1 u = 931.5 MeV/c2 is the correct unit for mass.
Step 3 — Mass defect of oxygen-16 in MeV/c2
Oxygen-16 has Z=8 protons and N=8 neutrons. A key bookkeeping trick: use atomic
masses throughout (not bare nuclear masses), because atomic masses already include their
own orbital electrons — comparing Z hydrogen ATOMS (each carrying 1 electron) against
the O-16 ATOM (carrying Z=8 electrons) makes the electron masses cancel automatically,
so you never need to add or subtract electron mass separately.
Atomic masses: m(1H)=1.007825 u (proton + its own electron),
mn=1.008665 u (neutrons have no electron either way),
m(16O)=15.994915 u (the actual atomic mass, 8 electrons included).
- Mass of 8 hydrogen atoms: 8×1.007825 u=8.062600 u
- Mass of 8 neutrons: 8×1.008665 u=8.069320 u …
Common Mistakes: Mass–Energy Equivalence
Mistake 1: Using the wrong value of c
Students often take c=3×108 m/s for convenience — and that’s fine for an estimate. But for the energy equivalent of 1 u, the exact value matters. The standard value is c=2.99792458×108 m/s. Using the rounded value gives E≈9×1016 J per kg, which when multiplied by 1.66×10−27 kg yields about 1.49×10−10 J — close, but not the accepted 1.492×10−10 J.
How to avoid: Use c=3.00×108 m/s only if the problem explicitly allows approximation. For board exams, stick to c=3×108 is usually acceptable, but for precise work (like binding energy calculations), use the exact value.
Mistake 2: Forgetting to convert atomic mass unit to kilograms
One atomic mass unit is 1 u=1.660539×10−27 kg. A common error is to plug in 1 directly into E=mc2 as if m were in kg.
How to avoid: Always write the conversion explicitly:
1 u=1.660539×10−27 kg
Then:
E=(1.660539×10−27)(2.99792458×108)2
Mistake 3: Confusing Joules with MeV in the conversion
The conversion 1 MeV=1.602×10−13 J is often misremembered as 1.6×10−19 (which is the charge of an electron in coulombs). That error throws the MeV value off by a factor of a million.
How to avoid: Memorise the pair:
- 1 eV=1.602×10−19 J
- 1 MeV=1.602×10−13 J
So to convert Joules to MeV, divide by 1.602×10−13.
Mistake 4: Writing the final answer in MeV instead of MeV/c2
The mass defect is a mass, not an energy. When the problem asks for it in MeV/c2, students often just give the energy equivalent in MeV and stop.
How to avoid: Remember: E=mc2 means m=E/c2. So if you compute the energy equivalent of the mass defect in MeV, the mass in MeV/c2 is numerically the same number. For example, if the mass defect corresponds to 127.5 MeV of energy, then the mass defect is 127.5 MeV/c2. The unit tells you it's mass, not energy.
Mistake 5: Using the mass of the nucleus instead of the mass defect
For 816O, the mass defect is:
Δm=8mp+8mn−mnucleus
Students sometimes plug in the atomic mass (which includes electrons) or forget to subtract the nuclear mass. …
Showing the 12 most recent of 13 on this concept.
- TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQQ.The ratio of the energy released when 2.5×1021 atoms of uranium undergo nuclear fission and energy equivalent of 2 mg mass of uranium is nearly (Average energy released per fission of uranium nucleus = 200 MeV) (A) 4:5 (B) 8:9 (C) 4:9 (D) 2:3
›Reveal solutionSolution
Fission energy ≈8.0×1010 J versus mass-equivalent energy mc2=1.8×1011 J, giving a ratio ≈4:9.
Energy from fission of N=2.5×1021 atoms at 200 MeV each:
E1=N×200 MeV=2.5×1021×200×1.6×10−13 J=8.0×1010 J.
Energy equivalent of 2 mg of mass via E=mc2: …
- TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQQ.During the disintegration of a radioactive nucleus of mass number 208 at rest, two alpha particles each with kinetic energy E are emitted. The total kinetic energy of the emitted alpha particles and the daughter nucleus after the disintegration is (A) 2551E (B) 5051E (C) 2552E (D) 2526E
›Reveal solutionSolution
The problem involves conservation of momentum and energy in a nuclear decay where two alpha particles are emitted from a nucleus at rest. The total kinetic energy released is the sum of the kinetic energies of the two alphas and the recoiling daughter nucleus. Using momentum conservation, the daughter’s kinetic energy is found to be a fraction of the alphas’ kinetic energy, leading to the total being 2552E, so the correct option is (C).
Concept and Intuition
When a stationary nucleus decays, the total momentum before decay is zero. After decay, the emitted particles and the daughter nucleus must have momenta that sum to zero. Here, two alpha particles are emitted, each with the same kinetic energy E. They are likely emitted in opposite directions (to conserve momentum if the daughter is also moving), but the key is that the daughter nucleus recoils to balance the total momentum. The total kinetic energy released in the decay is the sum of the kinetic energies of all three particles: the two alphas and the daughter. We are given the kinetic energy of each alpha, but not the daughter’s. We must find the daughter’s kinetic energy using momentum conservation and the relation between kinetic energy and momentum.
Step-by-step solution
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Set up the masses and momenta
The parent nucleus has mass number 208 and is at rest. It decays into a daughter nucleus and two alpha particles. Each alpha particle has mass number 4, so the daughter’s mass number is 208−2×4=200.
Let mα be the mass of one alpha particle, and md=50mα be the mass of the daughter (since 200/4=50).
Each alpha has kinetic energy E, so its momentum magnitude is pα=2mαE.
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Apply conservation of momentum
Since the parent is at rest, the total momentum after decay is zero. The two alpha particles are likely emitted in opposite directions (or at least with equal and opposite momenta along some axis) so that their net momentum is zero. Then the daughter nucleus must also have zero momentum — but that would mean it is at rest, which is impossible because energy would then be only 2E, not matching any option. …
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- TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQQ.The energy equivalent of 3.2μg of mass is (A) 18×1026 J (B) 18×1020 MeV (C) 18×1023 MeV (D) 32×1026 J
›Reveal solutionSolution
The key idea is Einstein’s mass-energy equivalence E=mc2. Converting 3.2μg to kg and using c=3×108 m/s gives 2.88×108 J, which matches none of the Joule options directly — but converting to MeV shows the correct match is 18×1020 MeV, option (B).
The problem tests your ability to apply E=mc2 and handle unit conversions — especially between Joules and MeV, a common trick in exam questions. The mass is given in micrograms, a tiny unit, so the energy will be large but not astronomically so. Let’s work through it carefully.
- Convert mass to SI units. The mass is 3.2μg. Since 1μg=10−9 kg, we have
m=3.2×10−9 kg.
- Apply E=mc2. Take c=3×108 m/s. Then
E=(3.2×10−9)×(3×108)2=3.2×10−9×9×1016.
Multiply: 3.2×9=28.8, and 10−9×1016=107, so
E=28.8×107=2.88×108 J.
This is 288 million Joules — a substantial energy from a tiny mass, as relativity predicts.
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Check the given options.
Option (A) is 18×1026 J — far too large. Option (D) is 32×1026 J — also enormous. So neither Joule option matches our result. The remaining options are in MeV, so we must convert.
-
Convert Joules to MeV.
Recall the conversion: 1 MeV=1.6×10−13 J. Therefore,
E=1.6×10−132.88×108 MeV=1.62.88×108+13=1.8×1021 MeV.
That’s 1.8×1021 MeV, which can be written as 18×1020 MeV.
- Match with options. …
- TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQQ.The half lives of two radioactive materials A and B are respectively T and 2T. If the ratio of the initial masses of the materials A and B is 8:1, then the time after which the ratio of the masses of the materials A and B becomes 4:1 is (A) 2T (B) T (C) 4T (D) 8T
›Reveal solutionSolution
Using N=N0(1/2)t/T1/2 for each material, the mass ratio mBmA=8(21)t/2T; setting it to 4 gives t=2T — option (A).
Step-by-step solution
Radioactive decay: m=m0(21)t/T1/2.
For material A (half-life T, initial mass 8) and B (half-life 2T, initial mass 1):
mA=8(21)t/T,mB=1⋅(21)t/2T.
Form the ratio: …
- TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQQ.The operation of a nuclear reactor is said to be critical when the value of neutron multiplication factor K is (A) K=0 (B) K>1 (C) K=1 (D) 0<K<1
›Reveal solutionSolution
The neutron multiplication factor K measures whether a nuclear chain reaction is self-sustaining. Criticality occurs when exactly one neutron from each fission goes on to cause another fission, so K=1. The correct option is (C).
The key idea is that a nuclear reactor’s operation depends on a chain reaction: each fission event releases neutrons, and some of those neutrons cause further fissions. The multiplication factor K is the average number of neutrons from one fission that successfully produce another fission.
- If K<1, the reaction dies out (subcritical).
- If K>1, the reaction grows exponentially (supercritical) — dangerous if uncontrolled.
- If K=1, the reaction is exactly self-sustaining (critical), which is the steady, controlled state for power generation.
Thus, “critical” means the chain reaction is balanced: each fission leads to exactly one more fission on average.
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Define K: The neutron multiplication factor is the ratio of neutrons in one generation to the number in the previous generation. In a steady reactor, we want this ratio to be 1 so that the neutron population remains constant.
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Interpret the options:
- K=0: No chain reaction at all (impossible for a working reactor).
- K>1: Exponential increase — used for startup or weapons, not steady operation.
- K=1: Perfect balance — the reactor is critical.
- 0<K<1: The reaction fades away — subcritical. …
- TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQQ.An α-particle of energy ‘E’ is liberated during the decay of a nucleus of mass number 236. The total energy released in this process is (A) 58 E (B) 59 E (C) 5958E (D) 5859E
›Reveal solutionSolution
The key idea is that the total energy released equals the sum of the kinetic energies of the alpha particle and the recoiling daughter nucleus, which share the decay energy in inverse proportion to their masses. Using conservation of momentum and the given mass numbers, the total energy is 5859E, so the correct option is (D).
When a nucleus decays by emitting an alpha particle, the energy released (the Q-value of the decay) is shared between the alpha particle and the recoiling daughter nucleus. The alpha particle gets most of the energy because it is much lighter, but the daughter nucleus also carries away some kinetic energy to conserve momentum. The problem gives the alpha particle’s kinetic energy as E and asks for the total energy released. The trick is to relate the masses (via mass numbers) to the energy split.
Why this approach works:
In any decay at rest, momentum is conserved. Since the parent nucleus is initially stationary, the alpha particle and the daughter nucleus must have equal and opposite momenta. Kinetic energy is p2/(2m), so for a given momentum, the energy is inversely proportional to mass. Thus, the ratio of the alpha’s energy to the daughter’s energy is the inverse ratio of their masses. The total energy is then the sum, which we can express in terms of the given E.
Step-by-step reasoning:
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Identify the masses involved.
The parent nucleus has mass number 236. It decays into an alpha particle (mass number 4) and a daughter nucleus. The daughter’s mass number is 236−4=232.
Let mα=4u and md=232u, where u is the atomic mass unit. (We use mass numbers as proportional to masses; the approximation is excellent for this problem.)
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Apply conservation of momentum.
Initially, the parent is at rest, so total momentum is zero. After decay, the alpha and daughter move in opposite directions with momenta of equal magnitude:
pα=pd=p.
- Relate kinetic energy to momentum. Kinetic energy for a particle is K=2mp2. Therefore:
Kα=2mαp2,Kd=2mdp2.
The ratio of their kinetic energies is:
KdKα=mαmd=4232=58. …
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- TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQQ.If the ratio of the radii of nuclei 52XA and 1327Al is 5:3, then the number of neutrons in the nucleus X is (A) 52 (B) 63 (C) 27 (D) 73
›Reveal solutionSolution
The nuclear radius scales as R∝A1/3, so the ratio of radii gives the ratio of mass numbers; solving yields AX=125, and with atomic number Z=52, neutrons N=A−Z=73.
The key idea is that nuclear radii follow a simple empirical law: R=R0A1/3, where A is the mass number (total protons + neutrons). This means the volume of a nucleus is proportional to A, so comparing radii directly gives a comparison of mass numbers. Once we find the unknown mass number, subtracting the given atomic number (52) gives the number of neutrons.
Let’s work through it step by step.
- Write the radius formula for both nuclei. For any nucleus, R=R0A1/3, where R0 is a constant (about 1.2×10−15 m, but it cancels). For the unknown nucleus 52XA (let’s call its mass number AX):
RX=R0(AX)1/3
For 1327Al:
RAl=R0(27)1/3
- Set up the given ratio. The problem states:
RAlRX=35
Substitute the expressions:
R0(27)1/3R0(AX)1/3=35
The R0 cancels, leaving:
(27)1/3(AX)1/3=35
- Simplify the denominator. Since 27=33, we have (27)1/3=3. So:
3(AX)1/3=35
Multiply both sides by 3:
(AX)1/3=5
- Solve for AX. Cube both sides: AX=53=125 …
- TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQQ.Half-life periods of two nuclei A and B are T and 2T respectively. Initially A and B have same number of nuclei. After a time of 4T, the ratio of the remaining number of nuclei of A and B is (A) 1 : 16 (B) 1 : 4 (C) 1 : 1 (D) 1 : 2
›Reveal solutionSolution
The ratio of remaining nuclei after time 4T is found by applying the exponential decay law to each nucleus, using their respective half-lives. The result is 1:4, so option (B) is correct.
The key idea is that radioactive decay follows an exponential law: the number of nuclei remaining after time t is N=N0(21)t/T1/2, where T1/2 is the half-life. The ratio of two samples depends only on how many half-lives each has experienced.
- Identify the number of half-lives for each nucleus. For nucleus A, half-life TA=T. Time elapsed is 4T, so the number of half-lives is
nA=T4T=4.
For nucleus B, half-life TB=2T. The number of half-lives is
nB=2T4T=2.
- Write the remaining number for each. Let the initial number of each be N0. After n half-lives,
NA=N0(21)4=16N0,
NB=N0(21)2=4N0.
- Compute the ratio NA:NB.
NBNA=N0/4N0/16=161×14=41.
So the ratio is 1:4. …
- TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQQ.If F1 and F2 are the relative strengths of the gravitational and weak nuclear forces respectively, then F1F2 is nearly (A) 100 (B) 1039 (C) 1013 (D) 1026
›Reveal solutionSolution
The ratio of the weak nuclear force to the gravitational force is enormous — about 1033 — but the given options are based on the relative strengths of the forces compared to the strong force, so the correct choice is (D) 1026.
The key idea: In physics, "relative strengths" of fundamental forces are usually quoted as dimensionless numbers compared to the strong nuclear force (taken as 1). Gravitational force is about 10−39 times the strong force, and the weak nuclear force is about 10−13 times the strong force. Their ratio F1F2 is then 10−3910−13=1026.
Why this approach works
The question doesn't give absolute numbers — it asks for the ratio of two forces. That means we only need their relative sizes on a common scale. The standard scale in particle physics compares all forces to the strong nuclear force. Memorizing these approximate exponents is the direct path.
Step-by-step reasoning
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Recall the approximate relative strengths of the four fundamental forces (strong nuclear = 1):
- Strong nuclear: 1
- Electromagnetic: ≈10−2
- Weak nuclear: ≈10−13
- Gravitational: ≈10−39
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Identify F1 and F2 from the problem:
- F1 = relative strength of gravitational force = 10−39
- F2 = relative strength of weak nuclear force = 10−13
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Compute the ratio:
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- TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQQ.In the following nuclear reaction X is 2713Al+24He→01n+X (A) 1531P (B) 1430Si (C) 1530P (D) 1531Si
›Reveal solutionSolution
In a nuclear reaction, both mass number and atomic number must be conserved.
For 1327Al+24He→01n+X, the missing nucleus X has mass number 30 and atomic number 15, which is 1530P — option (C).
The key idea is conservation of nucleon number and charge in nuclear reactions. Just like in a chemical equation, the total "mass" (mass number) and total "charge" (atomic number) on the left must equal those on the right. Here, we know the neutron has mass number 1 and atomic number 0, so we can solve for X by simple subtraction.
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Write down the known numbers
Left side:
- Aluminium: mass number = 27, atomic number = 13
- Helium (alpha particle): mass number = 4, atomic number = 2 Total left: mass = 27+4=31, atomic = 13+2=15
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Right side so far
- Neutron: mass number = 1, atomic number = 0 So the missing nucleus X must supply the rest: Mass number of X = 31−1=30 Atomic number of X = 15−0=15
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Identify the element
Atomic number 15 is phosphorus (P). So X is 1530P.
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Check the options
(A) 1531P — wrong mass number …
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- TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQQ.In the nuclear fission of one nucleus of U235 the energy released is 188 MeV. The energy released in the nuclear fission of 235 g of U235 is nearly (Avogadro number = 6.02×1023 mol−1) (A) 28.8×1012 J (B) 23.5×1012 J (C) 36.2×1012 J (D) 18.11×1012 J
›Reveal solutionSolution
The problem asks for the total energy released when 235 g of U‑235 undergoes fission, given 188 MeV per nucleus. The key is to find the number of nuclei in 235 g using Avogadro’s number, multiply by the energy per fission, and convert MeV to joules. The result is about 1.81×1013 J, which matches option (D).
The core idea is simple: energy per fission × number of fissions = total energy.
Since 235 g of U‑235 is exactly one mole (because the atomic mass is 235 u), the number of nuclei is Avogadro’s number. Then we just need to convert the energy from MeV to joules (1 MeV = 1.602×10−13 J).
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Find the number of nuclei in 235 g of U‑235
The atomic mass of U‑235 is 235 g/mol, so 235 g is exactly 1 mole.
Number of nuclei = Avogadro’s number = 6.02×1023.
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Energy released per nucleus
Given: 188 MeV per fission.
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Convert MeV to joules
1 MeV=1.602×10−13 J
So 188 MeV=188×1.602×10−13 J.
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Calculate total energy
Total energy = (number of nuclei) × (energy per nucleus in J)
E=(6.02×1023)×(188×1.602×10−13)
First compute the product inside:
188×1.602=301.176
So 188×1.602×10−13=3.01176×10−11 J per nucleus.
Then multiply:
E=6.02×1023×3.01176×10−11=(6.02×3.01176)×1012
6.02×3.01176≈18.13
So E≈1.813×1013 J.
- Match with options …
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- TG EAPCET 2022Set ap-2022-07-31-AN1 markMCQQ.An electron and proton are produced by breaking of a neutron. If the mass of neutron, electron and proton are 1.675×10−27 kg, 9×10−31 kg and 1.6730×10−27 kg respectively, find the amount of energy released in this process. (consider velocity of light as 3×108 m/s) (A) 0.66 MeV (B) 0.59 MeV (C) 0.62 MeV (D) 0.68 MeV
›Reveal solutionSolution
The neutron decays into a proton and an electron, and the tiny missing mass is converted into kinetic energy. Using E=Δmc2, the energy released comes out to 0.62 MeV, which matches option (C).
The process described is beta-minus decay of a free neutron:
n→p+e−+νˉe (the antineutrino is nearly massless and carries negligible energy here). The total mass of the products is slightly less than the mass of the neutron. That missing mass — the mass defect — is converted entirely into kinetic energy shared by the proton, electron, and antineutrino. The question asks for the total energy released, which is simply Δmc2.
The key idea: mass is a form of energy. When a system loses mass, that mass reappears as kinetic energy (or radiation). Here, we don't need to track individual particle energies — just the mass difference.
- Write the mass defect Mass of neutron: mn=1.675×10−27 kg Mass of proton: mp=1.6730×10−27 kg Mass of electron: me=9×10−31 kg The mass of the products is mp+me. So the mass defect is:
Δm=mn−(mp+me)
- Calculate the numerical value First add the product masses:
mp+me=1.6730×10−27+0.0009×10−27=1.6739×10−27 kg
(since 9×10−31=0.0009×10−27).
Then:
Δm=(1.6750−1.6739)×10−27=0.0011×10−27 kg
That is:
Δm=1.1×10−30 kg
- Apply Einstein’s mass-energy equivalence
E=Δmc2
with c=3×108 m/s. So:
E=(1.1×10−30)×(3×108)2
E=1.1×10−30×9×1016
E=9.9×10−14 J
- Convert joules to MeV …
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