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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🔒 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" …
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 16 on this concept.
- COMEDK 2026Set 2026-A1 markMCQQ.The atomic mass of an element 10X20 is 19.98170 amu. The binding energy per nucleon of that element is: [given mass of neutron = 1.00867amu and mass of proton = 1.00783 amu and 1amu = 931 MeV ] (A) 17.66MeV/ nucleon (B) 8.533MeV/ nucleon (C) 85.33MeV/ nucleon (D) 170.66MeV/ nucleon
›Reveal solutionSolution
The binding energy per nucleon is found by computing the mass defect (difference between the sum of individual nucleon masses and the actual nuclear mass), converting it to energy using E=Δm⋅931 MeV/amu, then dividing by the number of nucleons. The result is approximately 8.533 MeV/nucleon, which corresponds to option (B).
Concept & Intuition:
The nucleus is made of protons and neutrons. If you add up the masses of all these individual nucleons, you get a number larger than the actual mass of the nucleus. That missing mass — the mass defect — is converted into the energy that holds the nucleus together (binding energy). To compare how tightly bound different nuclei are, we divide by the number of nucleons to get the binding energy per nucleon. Here, the element is 10X20, meaning 10 protons and 10 neutrons (since mass number = 20).
Step-by-step solution:
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Identify the composition
The notation 10X20 tells us:
- Atomic number Z=10 → 10 protons
- Mass number A=20 → 20 nucleons total
- Number of neutrons N=A−Z=20−10=10
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Calculate the total mass of the individual nucleons
Mass of 10 protons: 10×1.00783 amu=10.0783 amu
Mass of 10 neutrons: 10×1.00867 amu=10.0867 amu
Total nucleon mass = 10.0783+10.0867=20.1650 amu
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Find the mass defect
Actual nuclear mass = 19.98170 amu
Mass defect Δm= (sum of nucleon masses) – (actual nuclear mass)
Δm=20.1650−19.98170=0.18330 amu
- Convert mass defect to binding energy Using 1 amu=931 MeV:
Binding energy=0.18330×931 MeV
Let’s compute:
0.18330×900=164.97
0.18330×31=5.6823
Sum = 164.97+5.6823=170.6523 MeV
(More precisely: 0.18330×931=170.6523 MeV)
- Compute binding energy per nucleon …
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- COMEDK 2026Set 2026-M1 markMCQQ.A nucleus of uranium -235 absorbs a slow neutron and undergoes nuclear fission according to the reaction: 92235U+01n→56141Ba+3692Kr+301n+Q If the average energy released per fission is 202 MeV , the energy released when 2.35 g of U235 undergoes complete fission is approximately; [Given 1eV=1.6×10−19 J, Avogadro number =6.02×1023 ] (A) 1.945×1010J (B) 19.45×1011J (C) 1.945×1011J (D) 19.45×1010 J
›Reveal solutionSolution
[!TLDR]
Count the nuclei in 2.35 g of U-235, multiply by 202 MeV and convert to joules to get ≈1.945×1011 J.
Concept
Nuclear fission energetics (CBSE Class-12 Nuclei): total energy released = (number of nuclei that fission) × (energy per fission). The number of nuclei comes from moles × Avogadro’s number.
Solution
Number of moles of U-235:
n=2352.35=0.01 mol.
Number of nuclei:
N=nNA=0.01×6.02×1023=6.02×1021.
Total energy in MeV:
E=N×202=6.02×1021×202=1.216×1024 MeV.
Convert to joules using 1 MeV=106×1.6×10−19=1.6×10−13 J: …
- COMEDK 2026Set 2026-M1 markMCQQ.Two deuterons are fused to form one alpha particle. If binding energy per nucleon of deuterium is 1.05 MeV and that of alpha particle is 7 MeV , what is the energy released in the formation of one alpha particle from the fusing of two deuterons? (A) 24.8 MeV (B) 23.8 MeV (C) 26.8 MeV (D) 28.3 MeV
›Reveal solutionSolution
The energy released is the difference between the total binding energy of the products and that of the reactants. For two deuterons fusing into one alpha particle, the released energy is 23.8 MeV, corresponding to option (B).
The key idea here is that binding energy is the energy required to break a nucleus into its individual nucleons. When lighter nuclei fuse into a heavier one, the difference in binding energy before and after the reaction is released as kinetic energy (or gamma rays). This is because the final nucleus is more tightly bound — its nucleons are in a lower energy state.
We are given binding energy per nucleon, so we must multiply by the number of nucleons to get the total binding energy of each nucleus.
-
Find the total binding energy of one deuteron.
Deuterium (²H) has 2 nucleons. Binding energy per nucleon = 1.05 MeV.
Total binding energy of one deuteron = 2×1.05=2.10 MeV.
-
Find the total binding energy of two deuterons (the reactants).
Since we start with two separate deuterons:
Total binding energy of reactants = 2×2.10=4.20 MeV.
-
Find the total binding energy of one alpha particle (the product).
An alpha particle (⁴He) has 4 nucleons. Binding energy per nucleon = 7 MeV.
Total binding energy of alpha particle = 4×7=28.0 MeV.
-
Calculate the energy released.
The energy released in fusion is the increase in binding energy:
Energy released=(Total BE of products)−(Total BE of reactants)
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- COMEDK 2025Set 2025-A1 markMCQQ.To get 300 MW electric power for half an hour, how much mass is to be completely converted into energy? (A) 6×10−2 kg (B) 3×10−6 kg (C) 6×10−3 kg (D) 6×10−6 kg
›Reveal solutionSolution
Energy delivered =Pt=5.4×1011 J; mass =E/c2=6×10−6 kg — option (D).
Energy required:
E=Pt=(300×106 W)×(1800 s)=5.4×1011 J,
where half an hour =1800 s.
Mass–energy equivalence (E=mc2, c=3×108 m/s): …
- COMEDK 2025Set 2025-A1 markMCQQ.Fusion reaction is more energetic than fission reaction because (A) Uncontrolled chain reaction is taking place In the fusion reaction. (B) Fusion reaction is taking place at very high temperature (C) The energy released per unit mass of the fuel in fusion reaction is larger than the energy released per unit mass of the fuel in fission reaction. (D) In the fusion reaction lighter nuclei combine to form a heavier nucleus
›Reveal solutionSolution
The key idea is that fusion releases more energy per unit mass of fuel than fission, making option (C) correct. The other options describe conditions or processes, not the fundamental reason for greater energy output.
The question asks why a fusion reaction is more energetic than a fission reaction. The answer lies in the physics of nuclear binding energy and the mass defect, not in the temperature or the type of chain reaction.
Concept and Intuition:
The energy released in any nuclear reaction comes from the conversion of a tiny amount of mass into energy, as described by Einstein’s famous equation E=mc2. The key measure is the binding energy per nucleon — the energy needed to hold a nucleus together. For light elements (like hydrogen isotopes), fusing them into a heavier nucleus (like helium) increases the binding energy per nucleon dramatically. For very heavy elements (like uranium), splitting them into medium-mass nuclei also increases binding energy per nucleon, but the gain per nucleon is smaller. This means fusion releases more energy per kilogram of fuel than fission does.
Now, let’s evaluate each option step by step.
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Option (A): "Uncontrolled chain reaction is taking place in the fusion reaction."
This is false. Fusion reactions (like in the sun or hydrogen bombs) are not chain reactions in the same sense as fission. A chain reaction involves neutrons causing subsequent fissions. Fusion requires extremely high temperatures and pressures to overcome electrostatic repulsion; it does not sustain itself via a chain mechanism. Even if it did, that wouldn’t explain why it’s more energetic — it would only describe how it proceeds.
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Option (B): "Fusion reaction is taking place at very high temperature."
This is true — fusion requires temperatures of millions of degrees to give nuclei enough kinetic energy to overcome repulsion. However, high temperature is a condition for fusion to occur, not the reason it releases more energy. The energy output per reaction is determined by nuclear forces, not by the temperature of the environment.
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Option (C): "The energy released per unit mass of the fuel in fusion reaction is larger than the energy released per unit mass of the fuel in fission reaction." …
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- COMEDK 2025Set 2025-A1 markMCQQ.The nucleus of oxygen atom contains 8 protons and 8 neutrons. What is the mass defect in amu? [Given Mass of proton =1.00727amu Mass of neutron =1.00866amu and the mass of oxygen nucleus =15.99053amu. ] (A) 0.12691 amu (B) 0.13692 amu (C) 0.13691 amu (D) 0.12961 amu
›Reveal solutionSolution
The mass defect is the difference between the sum of the individual masses of the protons and neutrons and the actual mass of the nucleus. For oxygen-16, this comes out to 0.13691 amu, which corresponds to option (C).
Concept and Intuition
The mass defect is a direct consequence of Einstein’s famous equation E=mc2. When protons and neutrons bind together to form a nucleus, some of their mass is converted into binding energy — the energy that holds the nucleus together. So the nucleus always weighs less than the sum of its individual parts. That missing mass is the mass defect. To find it, we simply subtract the actual nuclear mass from the total mass of the separate nucleons.
Step-by-Step Solution
-
Identify the composition of the oxygen nucleus
Oxygen has 8 protons and 8 neutrons (since its atomic number is 8 and mass number is 16). So we have:
- Number of protons = 8
- Number of neutrons = 8
-
Calculate the total mass of the separate nucleons
Mass of one proton = 1.00727amu
Mass of one neutron = 1.00866amu
Total mass of protons = 8×1.00727=8.05816amu
Total mass of neutrons = 8×1.00866=8.06928amu
Sum = 8.05816+8.06928=16.12744amu
-
Subtract the actual nuclear mass
Given mass of oxygen nucleus = 15.99053amu …
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- COMEDK 2025Set 2025-E1 markMCQQ.Radium having mass number 200 and binding energy per nucleon 5.6 MeV , splits into two fragments Cadmium of mass number 112 and Hassium of mass number 108. If the binding energy per nucleon for Cadmium and Hassium is approximately 8.0 MeV , then the energy Q released per fission will be: (A) 598 MeV (B) 176 MeV (C) 640 MeV (D) 475 MeV
›Reveal solutionSolution
Energy released equals the gain in total binding energy: Q=Bproducts−Bparent=(112+108)×8.0−200×5.6=1760−1120=640 MeV — option (C).
Concept
In fission, energy is released because the fragments are more tightly bound than the parent. The energy released equals the increase in the total binding energy of the system:
Q=Bfragments−Bparent.
Step-by-step solution
- Binding energy of the parent (Radium, A=200). Bparent=200×5.6=1120 MeV. …
- COMEDK 2025Set 2025-M1 markMCQQ.If the binding energy per nucleon in 3Li7 and 2He4 nuclei are respectively 5.60 MeV and 7.06 MeV , then energy of p in the reaction p+3Li7→22He4 is (A) 12.28 MeV (B) 13.28 MeV (C) 28.28 MeV (D) 17.28 MeV
›Reveal solutionSolution
The key idea is to use the difference in binding energies to find the energy released in the reaction, then apply conservation of energy to find the proton’s kinetic energy. The correct answer is 17.28 MeV.
Concept and Intuition
Binding energy is the energy needed to break a nucleus into its individual protons and neutrons. When a reaction rearranges nucleons into more tightly bound nuclei (higher binding energy per nucleon), the excess binding energy is released as kinetic energy. Here, a proton plus lithium-7 yields two alpha particles (helium-4). Since alpha particles have a higher binding energy per nucleon, the reaction is exothermic. The energy released (the Q-value) equals the difference in total binding energy before and after. That released energy plus the proton’s initial kinetic energy must equal the total kinetic energy of the two alpha particles. But the problem asks for the proton’s energy, assuming the alpha particles are produced at rest? Actually, the question likely means: what must the proton’s kinetic energy be so that the reaction can occur? In many such problems, the proton’s energy is the Q-value itself if the products are at rest, but here we need to check.
Let’s work it out step by step.
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Write the reaction and identify the nuclei
Reaction: p+37Li→224He
The proton is 11H. Lithium-7 has 3 protons and 4 neutrons. Helium-4 has 2 protons and 2 neutrons.
-
Find total binding energy before the reaction
- For 37Li: binding energy per nucleon = 5.60 MeV, so total binding energy = 7×5.60=39.20 MeV.
- For a free proton, binding energy is 0 (it’s a single nucleon). Total binding energy before = 39.20+0=39.20 MeV.
-
Find total binding energy after the reaction
- For each 24He: binding energy per nucleon = 7.06 MeV, so total per alpha = 4×7.06=28.24 MeV.
- Two alphas: total = 2×28.24=56.48 MeV.
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Calculate the energy released (Q-value)
The increase in binding energy is the energy released:
Q=56.48−39.20=17.28 MeV. …
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- COMEDK 2025Set 2025-M1 markMCQQ.In a nuclear fusion reaction, two nuclei, A and B fuse to produce a nucleus C, releasing an amount of energy ΔE in the process. If the mass defects of the three nuclei are ΔMA,ΔMB and ΔMC respectively, then which of the following relations is true? ( c is the speed of light). (A) ΔMA+ΔMB=ΔMC+c2ΔE (B) ΔMA−ΔMB=ΔMC+c2ΔE (C) ΔMA−ΔMB=ΔMC−c2ΔE (D) ΔMA+ΔMB=ΔMC−c2ΔE
›Reveal solutionSolution
Writing each nuclear mass as M=(nucleon masses)−ΔM and applying ΔE=(MA+MB−MC)c2, the nucleon terms cancel to give ΔMA+ΔMB=ΔMC−c2ΔE — option (D).
Concept. The mass defect of a nucleus is ΔM=(sum of free nucleon masses)−(actual nuclear mass), i.e. the mass equivalent of its binding energy. So the actual mass is M=∑(nucleons)−ΔM.
Step 1 — Nucleon conservation.
In A+B→C the total number of protons and neutrons is conserved, so the summed free-nucleon mass of A and B equals that of C. Call it Σ.
Step 2 — Express the actual masses.
MA+MB=Σ−(ΔMA+ΔMB),MC=Σ−ΔMC.
Step 3 — Energy released.
ΔE=(MA+MB−MC)c2.
Substituting and cancelling Σ: …
- COMEDK 2024Set 2024-A1 markMCQQ.A nucleus with mass number 190 initially at rest emits an alpha particle. If the Q value of the reaction is 4.5 MeV, the kinetic energy of the alpha particle is (A) 4 MeV (B) 3.2 MeV (C) 0.43 MeV (D) 4.4 MeV
›Reveal solutionSolution
In a nuclear decay where the parent nucleus is initially at rest, the Q‑value is shared between the alpha particle and the recoil daughter nucleus in inverse proportion to their masses. For mass numbers 190 (parent) and 4 (alpha), the alpha gets about 97% of the Q‑value, so its kinetic energy is roughly 4.4 MeV, corresponding to option (D).
Concept & Intuition
When a stationary nucleus emits an alpha particle, momentum must be conserved — the daughter nucleus recoils in the opposite direction with equal and opposite momentum. The Q‑value (the energy released) becomes the total kinetic energy of the two products. Because kinetic energy depends on both mass and velocity, the lighter alpha particle carries away most of the energy. The exact split follows from combining conservation of momentum and energy.
Step‑by‑step solution
- Set up the reaction Let the parent nucleus have mass number A=190. It emits an alpha particle (mass number 4) and becomes a daughter nucleus of mass number Ad=190−4=186. The Q‑value is the total kinetic energy released:
Q=Kα+Kd=4.5 MeV.
- Apply conservation of momentum Initially, the parent is at rest, so total momentum is zero. After decay:
mαvα=mdvd⇒vd=mdmαvα.
(Masses are proportional to mass numbers, so we can use mα=4u, md=186u, where u is the atomic mass unit.)
- Express kinetic energies in terms of vα
Kα=21mαvα2,Kd=21mdvd2=21md(mdmαvα)2=21mdmα2vα2.
- Find the ratio of kinetic energies
KαKd=21mαvα221mdmα2vα2=mdmα=1864=932.
So Kd=932Kα.
- Use the Q‑value equation
- COMEDK 2024Set 2024-E1 markMCQQ.The binding energy per nucleon for C12 is 7.68 MeV and that for C13 is 7.47 MeV. The energy required to remove a neutron from C13 is (A) 7.92×10−13 MeV (B) 4.95×10−13eV (C) 7.92×10−13 J (D) 7.92×10−19 J
›Reveal solutionSolution
The energy to remove a neutron from C13 is the difference between its total binding energy and that of C12, giving 4.95MeV, which converts to 7.92×10−13J. The correct option is (C).
Concept & Intuition
The binding energy per nucleon tells us how tightly each nucleon is held on average. To remove a neutron from C13, we must supply enough energy to overcome the binding of that neutron. This is the neutron separation energy. A neat way to find it: compare the total binding energy of C13 with that of C12. The difference is exactly the energy needed to pluck one neutron away, because the leftover nucleus is C12. No need to look up masses — the given data is enough.
Step-by-step reasoning
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Find total binding energies
For C12:
Binding energy per nucleon = 7.68MeV, and it has 12 nucleons.
Total binding energy B12=12×7.68=92.16MeV.
For C13:
Binding energy per nucleon = 7.47MeV, and it has 13 nucleons.
Total binding energy B13=13×7.47=97.11MeV.
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Interpret the removal process
Removing a neutron from C13 leaves C12. The energy required is the difference in total binding energy:
Sn=B13−B12=97.11−92.16=4.95MeV.
This makes sense: the extra neutron in C13 is less bound (lower per-nucleon average) than the average in C12, so the separation energy is less than the average binding energy.
- Convert to joules The answer choices are in joules or weird multiples, so convert: 1MeV=1.602×10−13J.
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- COMEDK 2024Set 2024-M1 markMCQQ.Find the binding energy of the tritium nucleus: [Given: mass of 1H3=3.01605 u; mp=1.00782 u; mn=1.00866 u.] (A) 8.5 MeV (B) 8.5 J (C) 0.00909 MeV (D) 0.00909 eV
›Reveal solutionSolution
The binding energy is the energy equivalent of the mass defect — the difference between the sum of the masses of the constituent nucleons and the actual nuclear mass. For tritium, this comes out to about 8.5 MeV, so the correct option is (A).
Concept & Intuition
The nucleus of tritium (13H) contains 1 proton and 2 neutrons. If you add up the masses of these three separate nucleons, you get a number larger than the measured mass of the tritium nucleus. That missing mass — the mass defect — has been converted into the energy that holds the nucleus together. By Einstein’s E=mc2, we convert that mass difference into energy units (MeV). The result is the binding energy.
Step-by-step solution
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Identify the nucleon composition
Tritium has atomic number Z=1 (one proton) and mass number A=3.
Number of neutrons = A−Z=3−1=2.
So the nucleus consists of 1 proton and 2 neutrons.
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Calculate the total mass of the separate nucleons
Given:
mp=1.00782u
mn=1.00866u
Total mass = 1×mp+2×mn
=1.00782+2×1.00866=1.00782+2.01732=3.02514u
- Find the mass defect Actual mass of tritium nucleus = 3.01605u (given). Mass defect Δm = (sum of nucleon masses) – (actual nuclear mass)
Δm=3.02514−3.01605=0.00909u
- Convert mass defect to energy The standard conversion: 1u=931.5MeV/c2. Binding energy Eb=Δm×931.5MeV
Eb=0.00909×931.5≈8.47MeV
Rounding gives 8.5 MeV. …
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