Nuclear Physics
SYLLABUS
Nucleus: Discovery of nucleus, Nuclear density; Mass number; Atomic number, Atomic mass; Isotopes, Einstein’s mass-energy relation, Mass Defect, packing fraction, BE per nucleon, Creation and annihilation, Nuclear fission and fusion, energy released
What is This Chapter About?
At the heart of every atom lies a nucleus — a tiny, incredibly dense region containing almost all of the atom’s mass, packed into a volume roughly 100,000 times smaller than the atom itself. Inside this nucleus, protons and neutrons are bound together by forces far stronger than any other force in nature. Understanding the nucleus — how it is structured, how stable it is, and how enormous amounts of energy can be released by rearranging its contents — is the domain of nuclear physics.
Chapter 24 of NEB Grade 11 Physics introduces students to the fundamental concepts of nuclear structure and nuclear energy. Carrying 6 teaching hours, it covers the discovery of the nucleus, nuclear density, isotopes, Einstein’s mass-energy relation, mass defect, binding energy, and the mechanisms of nuclear fission and fusion. These ideas connect directly to nuclear power generation, nuclear medicine, and the physics of stars — some of the most consequential applications of physics in human history.
Key Concepts You Will Learn
Discovery of the Nucleus — Rutherford’s Experiment
Before 1909, the dominant model of atomic structure was Thomson’s “plum pudding model” — electrons embedded in a diffuse positive charge spread throughout the atom. This was overturned by Geiger and Marsden’s famous alpha particle scattering experiment, directed by Ernest Rutherford. When alpha particles were fired at a thin gold foil:
Most passed straight through with no deflection — consistent with mostly empty space.
Some were deflected at small angles — consistent with small deflections from a distributed charge.
A tiny fraction — about 1 in 8000 — bounced back at angles greater than 90° — completely inconsistent with the plum pudding model.
Rutherford concluded that the atom consists of a tiny, dense, positively charged nucleus containing almost all the atomic mass, surrounded by electrons at relatively large distances. The large-angle scattering occurred when alpha particles came close to or collided with this dense nucleus. This nuclear model of the atom — essentially correct — replaced the plum pudding model and launched modern nuclear physics.
Nuclear Density, Mass Number, and Atomic Number
The atomic number (Z) is the number of protons in the nucleus — it determines which element the atom is. The mass number (A) is the total number of protons plus neutrons (collectively called nucleons) in the nucleus. The number of neutrons is N = A − Z.
A nucleus of element X is denoted [latex]_Z^A X[/latex]. The radius of a nucleus is approximately: [latex]r = r_0 A^{1/3}[/latex], where [latex]r_0 \approx 1.2 \times 10^{-15}[/latex] m = 1.2 fm. The volume of the nucleus is therefore proportional to A.
The nuclear density is the ratio of nuclear mass to nuclear volume. Since mass ∝ A and volume ∝ A, nuclear density is approximately constant for all nuclei:
[latex]\rho_{nuclear} \approx 2.3 \times 10^{17} \text{ kg/m}^3[/latex]
This is an almost incomprehensibly large density — a teaspoon of nuclear material would weigh about 2 billion tonnes. The fact that nuclear density is essentially the same for all nuclei suggests that nucleons are in close contact and the nuclear force is short-range — each nucleon interacts only with its immediate neighbours, not with all others.
Atomic Mass and Isotopes
The atomic mass unit (u) is defined as exactly [latex]\frac{1}{12}[/latex] of the mass of a carbon-12 atom: 1 u = 1.66054 × 10⁻²⁷ kg. The mass of a proton is approximately 1.007276 u and the mass of a neutron is approximately 1.008665 u.
Isotopes are atoms of the same element (same Z) but with different numbers of neutrons (different A). They have identical chemical properties (because chemistry depends only on electron configuration, which depends on Z) but different nuclear properties (different mass, different stability). For example, hydrogen has three isotopes: protium (¹H, no neutrons), deuterium (²H, one neutron), and tritium (³H, two neutrons). Carbon has several isotopes, including stable ¹²C and ¹³C and radioactive ¹⁴C (used in carbon dating).
Einstein’s Mass-Energy Relation
One of the most famous equations in science, derived by Einstein in 1905 as a consequence of special relativity, states that mass and energy are equivalent and interconvertible:
[latex]E = mc^2[/latex]
where c = 3 × 10⁸ m/s is the speed of light. The conversion factor c² is enormous — a tiny mass corresponds to an enormous energy. The annihilation of just 1 kg of matter would release [latex]9 \times 10^{16}[/latex] J — equivalent to the energy output of a large power station running for several years.
In nuclear physics, energy is often expressed in MeV (mega-electron volts), where 1 MeV = 10⁶ eV = 1.6 × 10⁻¹³ J. The mass-energy equivalence is: 1 u ≡ 931.5 MeV. This conversion is used constantly in nuclear calculations.
Mass Defect and Binding Energy
A remarkable fact about nuclei: the mass of a nucleus is always slightly less than the sum of the masses of its constituent protons and neutrons. This difference is called the mass defect ([latex]\Delta m[/latex]):
[latex]\Delta m = Z m_p + (A-Z)m_n – M_{nucleus}[/latex]
where [latex]m_p[/latex] and [latex]m_n[/latex] are the masses of a free proton and neutron respectively and [latex]M_{nucleus}[/latex] is the actual nuclear mass.
By Einstein’s relation, this mass defect corresponds to an energy called the binding energy (BE) — the energy that was released when the nucleus was assembled from its constituent nucleons, or equivalently, the energy that must be supplied to completely separate the nucleus into individual protons and neutrons:
[latex]BE = \Delta m \cdot c^2[/latex]
The binding energy per nucleon (BE/A) is the average binding energy per nucleon and is a measure of nuclear stability — higher BE/A means more tightly bound and more stable. A plot of BE/A versus mass number A reveals the famous binding energy curve:
BE/A rises steeply for light nuclei, reaches a peak of about 8.7 MeV/nucleon around iron (Fe-56) — the most stable nucleus — and then gradually decreases for heavier nuclei. This curve explains both fission and fusion as energy-releasing processes.
Creation and Annihilation
Pair creation occurs when a high-energy photon (gamma ray) spontaneously converts into an electron-positron pair: [latex]\gamma \rightarrow e^- + e^+[/latex]. The minimum photon energy needed is twice the rest mass energy of an electron: [latex]E_{min} = 2m_e c^2 = 1.022[/latex] MeV. This process demonstrates mass being created from energy.
Pair annihilation is the reverse: when an electron and positron meet, they annihilate and produce two gamma ray photons: [latex]e^- + e^+ \rightarrow 2\gamma[/latex]. Each photon carries energy [latex]m_e c^2 = 0.511[/latex] MeV. This process demonstrates mass converting entirely into energy. Pair annihilation is the operating principle of PET (Positron Emission Tomography) scanners used in medical imaging.
Nuclear Fission
Nuclear fission is the splitting of a heavy nucleus into two lighter nuclei (fission fragments) plus several neutrons and a large release of energy. The classic example is the fission of uranium-235 when struck by a slow neutron:
[latex]^{235}{92}U + ^1_0n \rightarrow ^{141}{56}Ba + ^{92}_{36}Kr + 3^1_0n + \text{energy} (\approx 200 \text{ MeV})[/latex]
The energy released per fission event (about 200 MeV) is enormous compared to chemical reactions (a few eV per reaction) — nuclear fission releases about 50 million times more energy per reaction than burning coal.
The released neutrons can trigger further fissions in neighbouring uranium nuclei — if enough uranium is present (exceeding the critical mass), this becomes a self-sustaining chain reaction. In a nuclear reactor, the chain reaction is controlled (usually by neutron-absorbing control rods) to release energy at a steady rate for power generation. In an atomic bomb, the chain reaction is uncontrolled and explosive.
The energy release in fission is explained by the binding energy curve: the fission fragments (around A ≈ 90–140) have higher BE/A than uranium (A = 235), so the products are more tightly bound — the difference in binding energy is released as kinetic energy and radiation.
Nuclear Fusion
Nuclear fusion is the combining of light nuclei to form a heavier nucleus, releasing energy. The classic example is the fusion of deuterium and tritium (isotopes of hydrogen):
[latex]^2_1H + ^3_1H \rightarrow ^4_2He + ^1_0n + 17.6 \text{ MeV}[/latex]
Fusion releases even more energy per unit mass than fission and produces far less radioactive waste. The Sun and all stars are powered by fusion reactions — in the Sun’s core, hydrogen nuclei fuse into helium through the proton-proton chain, releasing the energy that makes life on Earth possible.
Fusion requires enormously high temperatures (100 million °C) to overcome the electrostatic repulsion between positively charged nuclei and bring them close enough for the strong nuclear force to take over — this is why fusion is called a thermonuclear reaction. Achieving and sustaining these temperatures for useful energy production on Earth is the central challenge of nuclear fusion research — if successful, it would provide virtually unlimited clean energy from hydrogen, the most abundant element in the universe.
Why This Chapter Matters for NEB Exams
Nuclear Physics is examined thoroughly in NEB Grade 11 Physics. Mass defect and binding energy calculations — finding [latex]\Delta m[/latex], BE, and BE/A for given nuclei — are standard numerical problems. The binding energy per nucleon curve — drawing it, identifying the peak at iron, and using it to explain why fission and fusion both release energy — is a classic examination question. Einstein’s mass-energy relation and energy calculations using 1 u = 931.5 MeV appear in every paper. The distinction between fission and fusion, their energy releases, and their applications are standard short-answer questions. Rutherford’s scattering experiment and the nuclear model of the atom are frequent conceptual questions.
Important Topics to Focus On
- Rutherford’s alpha scattering experiment — setup, observations, and conclusions
- Atomic number Z, mass number A, and nuclear notation [latex]_Z^A X[/latex]
- Nuclear radius: [latex]r = r_0 A^{1/3}[/latex]
- Nuclear density — calculation and significance of its constancy
- Isotopes — definition, examples, and properties
- Atomic mass unit (u) and its energy equivalent: 1 u = 931.5 MeV
- Einstein’s mass-energy relation: [latex]E = mc^2[/latex]
- Mass defect: [latex]\Delta m = Zm_p + (A-Z)m_n – M_{nucleus}[/latex]
- Binding energy: [latex]BE = \Delta m \cdot c^2[/latex]
- Binding energy per nucleon — the BE/A curve and why iron is most stable
- Creation and annihilation of matter-antimatter pairs
- Nuclear fission — mechanism, chain reaction, energy released per fission
- Nuclear fusion — mechanism, conditions needed, energy released, solar fusion
- Why both fission and fusion release energy (explained by BE curve)
- Numerical problems on mass defect, binding energy, and energy release
Connection to Other Chapters
Nuclear Physics connects to Grade 11 Chapter 5 (Work, Energy and Power) through Einstein’s mass-energy relation — the most profound extension of the concept of energy. Grade 11 Chapter 19 (Electric Charges) provides the Coulomb repulsion that must be overcome in fusion. In Grade 12, Chapter 24 (Radioactivity and Nuclear Reaction) directly extends this chapter — radioactive decay is the spontaneous transformation of unstable nuclei, while nuclear reactions include both induced fission and natural decay. Chapter 20 (Electrons) connects through pair creation and annihilation — positrons are the antimatter counterparts of electrons. Chapter 25 (Recent Trends in Physics) connects through particle physics — quarks and gluons are the constituents of protons and neutrons, and nuclear forces are understood as residual strong forces between quarks.
Discovery of nucleus:
In 1911, Rutherford and his teams discovered the atomic nucleus from α scattering experiment.
Constituents of nucleus:
All atomic nuclei, are made up of protons and neutrons except H-atom. Proton has +ve charge while the electrons are negatively charged. The neutrons are electrically neutral. The collective forms of protons and neutrons in an atom are called neucleons.
An atom is denoted by [latex]_Z X^A[/latex], where, Z is the atomic number and A is the mass no. The no. of neutron is given as: N = A-Z.
General properties of nucleus:
1. Size:
According to Rutherford, radius of atomic nucleus is of the order of [latex]10^{-14}[/latex] m to [latex]10^{-15}[/latex] m, while an atom has radius of about [latex]10^{-10}[/latex] m. Thus, nucleus is about 10000 times smaller than atom.
2. Charge:
A nucleus consists of protons and neutrons. A proton has positive charge, while a neutron is chargeless. So, the nuclei are positively charged and has charge of Ze.
3. Mass:
Since, a nucleus consists of protons and neutrons, the mass of nucleus is equal to sum of masses of nucleons.
∴ mass of nucleus = Zmp + Nmn, where, Z = atomic no., N = neutron no., mp = mass of proton and mn = mass of neutrons.
4. Nuclear density:
The mass per unit volume of a nucleus is called nuclear density. i.e.
Nuclear density, [latex]\rho = \frac{Nuclear , Mass}{nuclear , volume} = \frac{Am_N}{\frac{4}{3}\pi R^3} = \frac{Am_N}{\frac{4}{3}\pi (r_o A^{\frac{1}{3}})^3} = \frac{3m_N A}{4\pi r_o^3 A} = \frac{3m_N}{4\pi r_o^3}[/latex]
Where, mn = mass of nucleon
R = [latex]r_oA^{1/3}[/latex], empirical formula for nuclear radius, A = mass no. and [latex]r_o = 1.2 \times 10^{-15}[/latex] m.
So, [latex]\rho = 2.3 \times 10^{17} Kg/m^3[/latex]. Since, nuclear density, is independent of mass no. (A) it is almost constant for all nuclei.
5. Nuclear Spin and magnetic moment:
Protons and neutrons are in continuous motion in the orbits. So, this orbital motion produces the nucleons with angular momentum. Besides this, nucleons have internal angular momentum known as spins. So, they possess magnetic moment associated with their orbital motion and spin.
Atomic mass unit:
The mass is usually measured in unified atomic mass unit (u). The atomic mass unit is defined as the [latex]1/12^{th}[/latex] of the mass of carbon atoms ([latex]_6C^{12}[/latex]).
We know, the number of molecules in 1 mole of carbon is [latex]6.02 \times 10^{23}[/latex] and mass of 1 mole of carbon is 12 gram.
So, we can write, the mass of [latex]6.023 \times 10^{23}[/latex] C-atoms = 12 gram.
The mass of 1 C-atom = [latex]\frac{12}{6.023 \times 10^{23}}[/latex] gm.
By the definition of atomic mass unit (U)
1 U = [latex]\frac{1}{12}[/latex] x mass of 1 carbon atom.
= [latex]\frac{1}{12} \times \frac{12}{6.02 \times 10^{26}}[/latex] Kg.
∴ 1 U = [latex]1.66 \times 10^{-27}[/latex] kg.
Relation between the atomic mass unit (U) and MeV:
From mass energy relation, we have, E = [latex]mc^2[/latex]
So, the energy related to the mass 1U is:
E = ([latex]1.66 \times 10^{-27}) \times (3 \times 10^8)^2[/latex] J
= [latex]1.66 \times 10^{-27}\times 9 \times 10^{16}[/latex] J
∴ E = [latex]\frac{9 \times 1.66 \times 10^{-27+16}}{1.6 \times 10^{-13}}[/latex] MeV [∵ 1MeV = [latex]1.6 \times 10^{-13}[/latex] J]
= 931 MeV
∴ 1U = 931 MeV.
Einstein’s Mass Energy Relation:
Earlier to Einstein, mass and energy was considered two separate entities and hence, there were two distinct laws for their conservation. But, in 1905, Einstein from his theory of relativity showed that mass and energy are interconvertible to each other. So, only a single mass energy conservation law is considered.
The change in energy ∆E is related to the change of mass ∆m, by the equation, [latex]\Delta E = \Delta m c^2[/latex] where, c = velocity of light. i.e. c = [latex]3 \times 10^8[/latex] m/s.
For the change of mass 1 kg, the equivalent energy is: E= 1 x ([latex]3 \times 10^8)^2 = 9\times 10^{16}[/latex] J
In nuclear physics, energy is usually expressed in electron volt. We have, the electron volt is the energy gained by a charge equal to that of an electron in moving through a p.d. of one volt.
1 eV = [latex]1.66 \times 10^{-19}\ C \times 1V = 1.66 \times 10^{-19}[/latex] J
For the larger unit, 1 MeV = [latex]10^6[/latex] eV
∴ 1 MeV = [latex]1.66 \times 10^{-13}[/latex] J.
Types of nuclei:
i. Isotopes:
The atoms of an element which have the same atomic number but different mass no. are called isotopes. In isotopes, the no. of protons inside the nucleus or the no. of electrons outside the nucleus is the same but the no. of neutrons inside the nucleus is not same. The chemical properties of isotopes are same. For example, [latex]_8O^{16}, _8O^{17}, _8O^{18}\ and\ _6C^{14}, _6C^{12}[/latex] are the isotopes of oxygen and carbon respectively.
ii. Isobars:
The atoms which have same mass no. but different atomic no. are called isobars. Isobars have same no. of nucleons (neutrons and protons) in their nuclei and hence they have same atomic weight. The chemical properties of isobars are different. For example: [latex]_18Ar^{40}\ and\ _20Ca^{40}[/latex].
iii. Isotones:
Isotones are the nuclei, which have same neutron no. For example, i) =[latex]_2He^4\ \&\ _1H^3[/latex] (N = 2), ii) [latex]_1H^2\ and\ _2He^3[/latex](N = 1).
Binding energy and Mass Defect:
Binding energy:
It has been found that the rest mass of the nucleus of a stable atom is always less than the sum of the masses of its constituent particles i.e. protons and neutrons. The difference between the sum of the masses of nucleons and the rest mass of the nucleus is called mass defect. The energy equivalent of this mass defect is called binding energy of the nucleus.
Hence, binding energy may be measured as the work required to be done to separate the nucleus on infinite distance apart, so that they no longer interact with each other.
If ∆m is the mass defect, then according the mass energy relation:
BE = [latex]∆mc^2[/latex]
If M is the rest mass of a nucleus, [latex]_ZX^A[/latex], [latex]m_n\ and\ m_p[/latex] are the masses of neutron and proton, then
B.E. = [Zmp + (A-Z)mn – M][latex]c^2[/latex]
In the terms of amu,
BE = [Zmp + (A-Z)mn – M] x 931.5 MeV
BE per nucleons curve:
Binding energy per nucleons is the ratio of total B.E. to the mass no. i.e.
[latex]\overline{B.E.} = \frac{Total , B.E.}{A}[/latex]
It is also known as average B.E. It determines the measure of stability of the nucleus. Greater the value of B.E. per nucleons, greater the stability. The B.E. per nucleon with the different mass no is shown in the graph. From the graph, it is clear that the curve first rises sharply and then gradually until it reaches the maximum value of 8.79 MeV corresponding to iron. Then it drops slowly to 7.6 MeV for Uranium. It is clear that the intermediate nuclei are more stable than the lighter or lower nuclei.
Nuclear Reaction:
i. Nuclear fission:
The splitting up of a heavy nucleus into two nuclei of nearly equal or comparable masses with liberation of energy is called nuclear fission.
Bohr Wheeler theory of nuclear fission:
Nuclear fission was first observed by Fermi and his co-workers and was explained by Neil Bohr and J.A. Wheeler. They explained the nuclear fission on the basis of liquid drop model of nucleus. When the neutron strikes [latex]_{92}U^{235}[/latex] nucleus (which behaves like a liquid drop and owing to surface tension it is perfectly spherical in shape), some energy called excitation energy is imparted to the nucleus. The energy due to surface tension tries to keep the shape of the nucleus spherical but the excitation energy tries to deform it. As a result, the oscillations are set up inside the compound nucleus, and they come to a dumbbell shape. Then, Coulomb’s repulsive force between protons separate the two dumbbells into two spherical shape nuclei.
In this way, nuclear fission occurs. It can be explained by the following nuclear reaction:
[latex]_{92}U^{235} + _0n^1 → _{92}U^{236} → _{56}Ba^{141} + _{36}Kr^{92} + 3_0n^1 + Q[/latex]
In the above reaction [latex]_{92}U^{235}[/latex] is bombarded with neutron thus forming a compound nucleus [latex]_{92}U^{236}[/latex] which disintegrates into two nuclei of [latex]_{56}Ba^{141}\ and\ _{36}Kr^{92}[/latex] with release of three neutrons and a lot of thermal energy Q.
Energy released in Fission:
The initial mass of nucleus is greater than the sum of masses of the products and the difference in mass (∆m) is converted into energy according to Einstein’s mass-energy relation, i.e. ∆E = [latex]∆mc^2[/latex]. We have,
Mass of neutron, [latex]_0n^1[/latex] = 1.008665 u.
Mass of [latex]_{92}U^{235}[/latex] = 235.045733 u.
∴ Total initial mass = 236.054398 u.
Mass of [latex]_{56}Ba^{141}[/latex] = 140.91770 u.
Mass of [latex]_{36}Kr^{92}[/latex] = 91.8854 u.
Mass of 3[latex]_0n^1[/latex] = 3.0260 u.
∴ Total final mass = 235.829100 u.
∴ Mass difference (∆m) = 236.054398 – 235.829100
= 0.2253 u.
∴ Energy released per fission = 0.2253 x 931 MeV
≈ 200 MeV
Which is very large amount of energy.
Nuclear fusion:
When two or more than two light nuclei fuse together to form heavy nucleus with the liberation of energy, the process is called nuclear fusion.
E.g., [latex]_1H^2 + _1H^2 → _2He^4[/latex] + 24 MeV
The above nuclear reaction is energetically possible only if the mass of [latex]_2He^4[/latex] nucleus is less than sum of masses of the two deuteron nuclei. The difference in the initial mass and the final mass of the product nuclei is liberated as the energy of fusion reaction.
Energy released in Fusion:
The energy released in this reaction can be calculated as:
Mass of deuteron = 2 x 2.014102 u = 4.028204 u.
Mass of [latex]_2He^4[/latex] = 4.002604 u.
∴ Decrease in mass = 0.025600 u.
∴ Energy released = 0.0256 x 931 MeV ≈ 24 MeV.
Thus, the energy released in fusion = 24 MeV.
Chain reaction:
A nuclear reaction initiated is capable of triggering the nuclear reaction and the process goes on until all the atoms of the target material are bombarded. This is known as chain reaction.
When a neutron bombards an atom of Uranium in a piece of uranium, three neutrons are produced along with the release of energy of about 200 MeV. These three neutrons can further bombard other three Uranium atoms producing 9 neutrons and the process goes on until all the uranium atoms are bombarded.
Types of chain reaction:
i. Controlled chain reaction:
The no. of neutrons and energy is controlled to a desired level in controlled chain reaction. This principle is used in nuclear reaction.
ii. Uncontrolled chain reaction:
The neutrons are allowed to multiply independently in this reaction. Hence, the energy is released all the time. This is the principle behind atom bomb.
Multiplication factor:
It is defined as the ratio of secondary neutrons produced to the initial number of neutrons. It is denoted by K. i.e.
K = [latex]\frac{\text{No. of neutrons in any generations}}{\text{No. of neutrons in previous generation}}[/latex]
For K = 1, the reaction is steady or critical.
For K > 1, the reaction is building up or super-critical.
For K < 1, the reaction is dying down or sub-critical.
Packing fraction:
The packing fraction of the isotope is defined as the ratio of its mass defect to its mass number. i.e.
Packing fraction = [latex]\frac{\text{mass defect}}{\text{mass number}}[/latex]
Creation and Annihilation:
A subatomic particle having same mass as another particle but opposite value of charge is called antiparticle. For eg. the antiparticle of electron is positron (<sub>+1</sub>e<sup>o</sup>), which has same mass as that of electron but has a positive charge equal in magnitude to that of electrons.
The antiparticle of proton is antiproton. Particles form matter but antiparticles form antimatter. When an energetic [latex]\gamma[/latex] – ray photon falls on a heavy substance, it is absorbed by the nucleus of the substance and its energy gives rise to the production of an electron and a positron. This phenomenon, in which energy is converted into mass is called creation or pair production and is represented by the equation:
[latex]h\nu[/latex] = [latex]_{-1}e^o + _{+1}e^0[/latex]
([latex]\gamma[/latex] – photon) (electron) (positron)
When a particle and antiparticle meet each other, the destruction of particle and antiparticle occur with the emission of energy. This is called annihilation of matter. For example, when electron meets positron, there is emission of energy and [latex]\gamma[/latex] – ray. i.e.
[latex]e^- + e^+ → \gamma[/latex] + 1.02 MeV
MCQs
The mass defect for the nucleus of helium is 0.0303 amu. What is the binding energy per nucleon for helium in MeV?
Show answer
What is the expected source of energy in the sun and other starts?
Show answer
What is the rest mass energy of an electron?
Show answer
The mean density of the nuclei is proportional to
Show answer
For an isobaric family members of nuclei, which of the following statement is true
Show answer
For a nucleus, the required binding energy is provided by
Show answer
Energy liberated per nucleon in nuclear fusion of hydrogen to helium is
Show answer
Controlled fusion reaction is used in
Show answer
Which one of the following have minimum nuclear radius?
Show answer
Fractional atomic weight exists due to the pressure of
Show answer
The B.E. per nucleon is constant for many nucleus. It shows that the nuclear force are
Show answer
The neutron was discovered by
Show answer
In the first observed nuclear reaction, [latex]{}_7N^{14} was bombarded with [latex]\alpha[/latex] – particles. The reaction could be represented as [latex]{}_7N^{14} + {}_2He^4 \rightarrow X + {}_1H^{1}[/latex]. The element X in this reaction is
Show answer
The binding energy of a nucleus is a measure of its:
Show answer
What were the fissionable materials used in the atomic bomb dropped in Nagasaki and Hiroshima?
Show answer
The size of nucleus is of the order of
Show answer
Theoretical questions
Neutron is considered the most effective bombarding particle in a nuclear reaction. Why?
A nucleus consists of positively charged protons and electrically neutral neutrons in a small volume. How can this be possible as the like charges repel each other?
All the nuclei have nearly the same density. Justify.
Why is neutron considered the most effective bombarding particle in a nuclear reaction?
According to properties of charges, like charges repel each other. Then, how do the protons in a nucleus stay together?
Why is the mass of a nucleus slightly less than the mass of constituent nucleons?
Define atomic mass unit (amu). Hence convert the mass of a neutron, (1840 me), into amu where me is the mass of ‘an electron’.
Diameter of Al27 nucleus is DAl. How can one express the diameter of Cu64 in terms of DAl? Explain.
By what factor must the mass number of a nucleus increase to double its volume? Explain.
What is the significance of binding energy per nucleon?
Define atomic mass unit and convert it into MeV.
Does a nucleus contain electrons? Explain.
‘Nuclear charge determines the chemical elements.’ Justify.
What does the energy balance (Q-value) of a nuclear reaction signify? Explain.
Why are neutrons used to initiate fission reaction?
Why does a mountain of uranium not explode as a bomb?
Point out the difference between’ nuclear fission and fusion.
Explain the significance of Einstein’s mass energy equivalence relation.
Define mass defect and packing fraction of a nucleus.
Distinguish between isotopes and isobar?
Explain binding energy in terms of packing fraction.
Define binding energy and binding energy per nucleon. How does binding energy per nucleon vary with mass number? What is its significance?
Differentiate between nuclear fission and fusion. Explain the production of energy in the Sun.
Discuss fission and fusion with an example of each. In which reaction is the energy released greater?
Define mass defect and binding energy of a nucleus. Draw a graph showing the variation of binding energy per nucleon and atomic number of the elements. Also interpret the graph.
Write down the representative nuclear fission and fusion reactions. Explain, how the energy release in the case of four protons fused into doubly ionized helium can be estimated?
Discuss four important properties of nuclei.
What are meant by mass defect and binding energy per nucleon? Draw a graph showing the relation between binding energy per nucleon and atomic number. Explain its significances.
Write down the schemes for nuclear fusion and nuclear fission. How can the release energy in any of these reactions estimated? What do you mean by Q – value of a nuclear reaction?
What is nuclear fission? How energy is released in nuclear fission reaction?
What is nuclear fission? Give an example of nuclear reaction.
State and explain Einstein’s mass energy relation with example.
Numerical problems
Calculate the binding energy per nucleon of 26Fe56. Atomic mass of 26Fe56 is 55.9349 u and that of 1H1 is 1.00783u. Mass of 0n1 = 1.00867u and 1u = 931 MeV. Ans: [latex]1.41\times 10^{-12}[/latex] J
Calculate the binding energy per nucleon of calcium nucleus (20Ca40).
Given: mass of 20Ca40 = 39.962589 u
Mass of neutron, mn = 1.008665u
Mass of proton, mp = 1.007825u
1u = 931 MeV.
A city requires 107 watts of electrical power on the average. If this is to be supplied by a nuclear reactor of efficiency 20%. Using 92U235 as the fuel source, calculate the amount of fuel required per day. (Energy released per fission 92U235 = 200 MeV). Ans: 0.0527 kg
A nucleus of 92U238 disintegrates according to
92U238 → 90Th234 + 2He4
Calculate:
(i) the total energy released in the disintegration process.
(ii) the K.E. of the [latex]\alpha[/latex] – particle, the nucleus at rest before disintegration. (Mass of 92U238 = [latex]3.859\times 10^{-25}[/latex] kg, mass of 90Th234 = [latex]3.787\times 10^{-25}[/latex] kg, 2He4 = [latex]6.648\times 10^{-27}[/latex] kg). Ans: 4.236 MeV, 4.16 MeV
The mass of 17Cl35 is 34.9800 amu. Calculate its binding energy and binding energy per nucleon. Mass of one proton = 1.007825 amu and mass of one neutron = 1.008665 amu. Ans: 287.66 MeV, 8.21 MeV/nucleon
The energy liberated in the fission of single uranium – 235 atom is [latex]3.2\times 10^{-11}[/latex] J. Calculate the power production corresponding to the fission of 1 gm of uranium per day. Assume Avogadro constant as [latex]6\times 10^{23}\ mol^{-1}[/latex]. Ans: [latex]9.46\times 10^5[/latex] W
What will be the amount of energy released in the fusion of three alpha particles into a C12 nucleus if mass of He4 and C12 nuclei are respectively 4.00263 amu and 12 amu. Ans: 7.34 MeV
The mass of the nucleus of the isotope Lithium 3Li7 is 7.014351 u. Find its binding energy and binding energy per nucleon. (Given mass of proton = 1.00727 u, mass of neutron = 1.008665 u). Ans: 39.3 MeV, 5.6 MeV/nucleon
28Ni62 may be described as the most strongly bound nucleus because it has the highest B.E. per nucleon. Its neutral atomic mass is 61.928349 amu. Find its mass defect, its total binding energy and binding energy per nucleon.
Given, mass of neutron = 1.008665 amu
Mass of proton = 1.007825 amu
1 amu = 931.5 MeV. Ans: [latex]\Delta m[/latex] = 0.583561 amu, B.E. = 543.58 MeV, B.E./Nucleon = 8.76 MeV per nucleon
The energy released per fission of one U235 atom is 200 MeV. Calculate the energy released in Kwh, when one gram of uranium undergoes fission. Ans: 2.278 x 104 kwh
Calculate the binding energy per nucleon for a helium nucleus. Given that mass of helium nucleus = 4.001509 amu, mass of proton = 1.007277 amu and mass of neutron = 1.008666 amu. Ans: 7.07 MeV per nucleon
The most common isotope of uranium 92U238, has atomic mass 238.050783 u. Calculate the a) mass defect, b) binding energy, c) Binding energy per nucleon. (Mass of proton = 1.007825 u, mass of neutron = 1.008665 u). Ans: 1.934207 amu, 1800.75 MeV, 7.57 MeV/Nucleon
Estimate the binding energy per nucleon of 3Li7. Mass of 3Li7, a proton & a neutron are respectively 7.01435 amu, 1.00728 amu and 1.00867 amu. Ans: 5.6 MeV/Nucleon
Assuming that about 200 MeV energy is released per fission of 92U235 nuclei. What would be the mass of U235 consumed per day in the fission of power 1 MW approximately. Ans: 1.05 gm
A nucleus of uranium – 238 disintegrates according to the reaction 92U238 → 90Th234 + 2He4. Calculate: i) the total energy released in the disintegration ii) the K.E. of the alpha particle, the nucleus being at rest before disintegration. (Mass of 92U238 = 238.12492 u, mass of 90Th234 = 234.1165 u, mass of 2He4 = 4.00387 u, 1u = 931 MeV). Ans: 4.236 MeV, 4.16 MeV
The energy liberated in the fission of a single uranium – 235 atom is [latex]3.2\times 10^{-11}[/latex] J. Calculate the power production corresponding to the fission of 1 kg of uranium per day. (Avogadro constant = [latex]6.0\times 10^{23}\ mole^{-1}[/latex]). Ans: [latex]9.47\times 10^8 [/latex] W
Calculate the Q – value of the reaction & mention the type of reaction (endo – thermic or exothermic). 2He4 = 4.00377 amu, 8O17 = 17.00450 amu
7N14 = 14.00783 amu, 1H1 = 1.00814 amu. Ans: – 0.96824 MeV
Calculate the speed of particle of the mass of it is equal to 5 times its rest value. Ans: [latex]2.93\times 10^8[/latex] m/s
Calculate the binding energy per nucleon of 26Fe56. Atomic mass of 26Fe56 is 55.9349 u and that of 1H1 is 1.00783u. Mass of 0n1 = 1.00867u and 1u = 931 MeV.
Solution
Given,
Atomic mass of 26Fe56 (M) = 55.9349 u.
Here, Z = 26 & A = 56
Mass of 1H1 (mp) = 1.00783 u.
Mass of 0n1 (mn) = 1.00867 u
1u = 931 MeV
We know that,
B.E. = [Zmp + (A – Z) mn – M]. 931 MeV
= [latex]
[26\times 1.00783 + (56 – 26)\times 1.00867 – 55.9349]\times 931 MeV
[/latex]
= 492.294 MeV
Now, B.E. per nucleon
[latex]
(\bar{B.E.}) = \frac{B.E.}{A} = \frac{492.294}{56}
[/latex]
= 8.791 Mev/nucleon.
Calculate the binding energy per nucleon of calcium nucleus (20Ca40).
Given: mass of 20Ca40 = 39.962589 u
Mass of neutron, mn = 1.008665u
Mass of proton, mp = 1.007825u
1u = 931 MeV.
Solution
Given,
Atomic mass of 20Ca40 (M) = 39.962589 u.
Here, Z = 20 & A = 40
Mass of 1H1 (mp) = 1.007825 u.
Mass of 0n1 (mn) = 1.008665 u
1u = 931 MeV
We know that, B.E. = [Zmp + (A – Z) mn – M]. 931 MeV
= [latex][20\times 1.007825 + (40 – 20)\times 1.008665 – 39.962589]\times 931 MeV [/latex]
= 341.873 MeV
Now, B.E. per nucleon [latex](\bar{B.E.}) = \frac{B.E.}{A} = \frac{341.873}{40}[/latex] = 8.547 Mev/nucleon.
A city requires 107 watts of electrical power on the average. If this is to be supplied by a nuclear reactor of efficiency 20%. Using 92U235 as the fuel source, calculate the amount of fuel required per day. (Energy released per fission 92U235 = 200 MeV).
Solution
Given,
Electrical power required by city (Pout) = 107 Watt
Efficiency [latex](\eta)[/latex] = 20%
Amount of fuel required per day (m) = ?
Energy released per fission (E) = 200 MeV = 200 x 1.6 x 10-13
= 3.2 x 10-11 J
We have, efficiency
[latex]
(\eta) = \frac{P_{out}}{P_{in}}\times 100%
[/latex]
Or, 20 % = [latex]\frac{P_{out}}{P_{in}}\times 100%[/latex]
Or, 0.2 = [latex]\frac{10^7}{P_{in}}[/latex]
Or, Pin = [latex]\frac{10^7}{0.2} = 5\times 10^7[/latex] Watt
Now, Time (t) = 1 day = 86400 Sec
So, Power = [latex]\frac{Total\ Energy\ (E_T)}{Time\ taken\ (t)}[/latex]
Or, ET = Pin . t = 5 x 107 x 86400
= 4.320 x 1012 J
Now, No. of atom released per fission (N) = [latex]\frac{E_T}{E} = \frac{4.320\times 10^{12}}{3.2\times 10^{-11}}[/latex]
= [latex]1.350\times 10^{23}[/latex]
Using Avogadro’s Hypothesis,
∵ 6.023 x 1023 atom of U – 235 is contained in 235 g of it.
∴ 1 atom of U – 235 is contained in [latex]\frac{235}{6.023\times 10^{23}}[/latex] g of it.
∴ 1.35 x 1023 atom of U – 235 is contained in [latex]\frac{235}{6.023\times 10^{23}}\times 1.35\times 10^{23}[/latex] g of it.
= 0.0527 Kg of it.
A nucleus of 92U238 disintegrates according to
92U238 → 90Th234 + 2He4
Calculate:
(i) the total energy released in the disintegration process.
(ii) the K.E. of the [latex]\alpha[/latex] – particles, the nucleus at rest before disintegration. (Mass of 92U238 = [latex]3.859\times 10^{-25}[/latex] kg, mass of 90Th234 = [latex]3.787\times 10^{-25}[/latex] kg, mass of 2He4 = [latex]6.648\times 10^{-27}[/latex] kg).
Solution
Given,
92U235 disintegrates according to 92U238 → 90Th234 + 2He4
We have,
Mass of 92U238 = 3.859 x 10-25 Kg
Mass of 90Th234 = 3.787 x 10-25 Kg
Mass of 2He4 = 6.648 x 10-27 Kg
Total energy released in the disintegration process (E)
= [latex][\sum_{reactant} – \sum_{product}]\times c^2[/latex]
= [3.859 x 10-25 – (3.787 x 10-25 + 6.648 x 10-27)] x c2
= 4.968 x 10-11 J
= 310.5 MeV
Now, K.E. of [latex]\alpha[/latex] – particles = [latex]\frac{M_{Th}}{M_{Th} + M_{He}}\times E[/latex]
= [latex]\frac{3.787\times 10^{-25}}{(3.787\times 10^{-25}) + (6.648\times 10^{-27})}\times 310.5[/latex]
= 305.143 MeV.
The mass of 17Cl35 is 34.9800 amu. Calculate its binding energy and binding energy per nucleon. Mass of one proton = 1.007825 amu and mass of one neutron = 1.008665 amu.
Solution
Given,
Atomic mass of 17Cl35 (M) = 34.9800 u
Here, Z = 17 & A = 35
Mass of proton (mp) = 1.007825 u.
Mass of neutron (mn) = 1.008665 u
1u = 931 MeV
We know that, B.E. = [Zmp + (A – Z) mn – M]. 931 MeV
= [latex][17\times 1.007825 + (35 – 17)\times 1.008665 – 34.98]\times 931 MeV [/latex]
= 287.674 MeV
Now, Binding Energy per nucleon [latex](\bar{B.E.}) = \frac{B.E.}{A} = \frac{287.674}{35}[/latex] = 8.219 Mev/nucleon.
The energy liberated in the fission of single uranium – 235 atom is [latex]3.2\times 10^{-11} J[/latex]. Calculate the power production corresponding to the fission of 1 gm of uranium per day. Assume Avogadro constant as [latex]6\times 10^{23} mol^{-1}[/latex].
Solution
Given,
Energy liberated in the fission of single uranium atom (E) = 3.2 x 10-11 J
Power production (P) =?
Mass of uranium (m) = 1 gm = 1 x 10-3 kg
Time (t) = 1 day = 86400 s
We know that,
According to Avogadro’s hypothesis,
[latex]\because[/latex] 235 gm of U – 235 contains [latex]6.023\times 10^{23}[/latex] atoms.
[latex]\therefore[/latex] 1 gm of U – 235 contains [latex]\frac{6.023\times 10^{23}\ atoms}{235}[/latex].
Or, N = 2.563 x 1021 atoms.
Total energy liberated (T.E.) = N.E
= (2.563 x 1021) x (3.2 x 10-11)
= 8.202 x 1010 J
Power production (P) = [latex]\frac{T.E.}{t}[/latex]
= [latex]\frac{8.202\times 10^{10}}{86400}[/latex]
= 9.493 x 105 Watt.
What will be the amount of energy released in the fusion of three alpha particles into a C12 nucleus if mass of He4 and C12 nuclei are respectively 4.00263 amu and 12 amu.
Solution
Given,
Amount of energy released (E) =?
Mass of He4 = 4.00263 u
Mass of C12 = 12 u
We have, the reaction involved in the fusion is written as:
32He4 → 6C12 + Q
So, Q – energy = [latex][\sum_{reactant} – \sum_{product}]\times 931 MeV[/latex]
= [(3 x 4.00263) – 12] x 931 MeV
= 7.346 MeV.
The mass of the nucleus of the isotope Lithium 3Li7 is 7.014351 u. Find its binding energy and binding energy per nucleon. (Given mass of proton = 1.00727 u, mass of neutron = 1.008665 u).
Solution
Given,
Atomic mass of 3Li7 (M) = 7.014351 u
Here, Z = 3 & A = 7
Mass of proton (mp) = 1.00727 u
Mass of neutron (mn) = 1.008665 u
1u = 931 MeV
We know that, B.E. = [Zmp + (A – Z) mn – M]. 931 MeV
= [latex][3\times 1.00727 + (7 – 3)\times 1.008665 – 7.014351]\times 931 MeV[/latex]
= 39.213 MeV
Now, Binding Energy per nucleon [latex](\bar{B.E.}) = \frac{B.E.}{A} = \frac{39.213}{7}[/latex] = 5.602 Mev/nucleon.
28Ni62 may be described as the most strongly bound nucleus because it has the highest B.E. per nucleon. Its neutral atomic mass is 61.928349 amu. Find its mass defect, its total binding energy and binding energy per nucleon.
Given, mass of neutron = 1.008665 amu
Mass of proton = 1.007825 amu
1 amu = 931.5 MeV.
Solution
Given,
Atomic mass of 28Ni62 (M) = 61.928349 u
Here, Z = 28 & A = 62
Mass of proton (mp) = 1.007825 u.
Mass of neutron (mn) = 1.008665 u
1u = 931.5 MeV
We know that,
Mass defect [latex](\Delta m)[/latex] = [Zmp + (A – Z) mn – M]
= [latex][28\times 1.007825 + (62 – 28)\times 1.008665 – 61.928349][/latex]
= 0.585361 amu
B.E. = [latex][\Delta m][/latex]. 931 MeV
= [latex][0.585361]\times 931.5\ MeV[/latex]
= 545.264 MeV
Now, Binding Energy per nucleon [latex](\bar{B.E.}) = \frac{B.E.}{A} = \frac{545.264}{62}[/latex] = 8.795 Mev/nucleon.
The energy released per fission of one U235 atom is 200 MeV. Calculate the energy released in Kwh, when one gram of uranium undergoes fission.
Solution
Given,
Energy liberated per fission (E) = 200 MeV
= 200 x (1.6 x 10-13) J
= 3.2 x 10-11 J
Energy released in Kwh =?
Mass of uranium (m) = 1 gm = 1 x 10-3 kg
Time (t) = 1 hour = 3600 s
We know that,
According to Avogadro’s hypothesis,
∵ 235 gm of U – 235 contains 6.023 × 1023 atoms.
∴ 1 gm of U – 235 contains [latex]\dfrac{6.023 \times 10^{23}}{235}[/latex] atoms.
N = 2.563 x 1021 atoms.
Total energy liberated (T.E.) = N.E
[latex]= (2.563 \times 10^{21}) \times (3.2 \times 10^{-11})[/latex]
[latex]= 8.202 \times 10^{10}\ J[/latex]
Power production [latex](P) = \dfrac{T.E.}{t}[/latex]
[latex]= \dfrac{8.202 \times 10^{10}}{3600}[/latex]
[latex]= 2.278 \times 10^7\ Watt[/latex]
[latex]= \dfrac{2.278 \times 10^7}{1000}\ KW[/latex]
[latex]= 2.278 \times 10^4\ KW.[/latex]
Calculate the binding energy per nucleon for a helium nucleus. Given that mass of helium nucleus = 4.001509 amu, mass of proton = 1.007277 amu and mass of neutron = 1.008666 amu.
Solution
Given,
Atomic mass of [latex]{}_2He^4[/latex] (M) = 4.001509 u
Here, Z = 2 & A = 4
Mass of proton [latex](m_p)[/latex] = 1.007277 u
Mass of neutron [latex](m_n)[/latex] = 1.008666 u
1u = 931 MeV
We know that, [latex]B.E. = [Zm_p + (A-Z)m_n – M] \times 931\ MeV[/latex]
[latex]= [2 \times 1.007277 + (4-2) \times 1.008666 – 4.001509] \times 931\ MeV[/latex]
[latex]= 28.281\ MeV[/latex]
Now, Binding Energy per nucleon [latex]\bar{B.E.} = \dfrac{B.E.}{A} = \dfrac{28.281}{4}[/latex]
[latex]= 7.07\ MeV/nucleon.[/latex]
The most common isotope of uranium 92U238, has atomic mass 238.050783 u. Calculate the a) mass defect, b) binding energy, c) Binding energy per nucleon. (Mass of proton = 1.007825 u, mass of neutron = 1.008665 u).
Solution
Given,
Atomic mass of [latex]{}_{92}U^{238}[/latex] (M) = 238.050783 u
Here, Z = 92 & A = 238
Mass of proton [latex](m_p)[/latex] = 1.007825 u
Mass of neutron [latex](m_n)[/latex] = 1.008665 u
1u = 931 MeV
We know that,
Mass defect [latex](\Delta m) = [Zm_p + (A-Z)m_n – M][/latex]
[latex]= [92 \times 1.007825 + (238-92) \times 1.008665 – 238.050783][/latex]
[latex]= 1.934207\ amu[/latex]
[latex]B.E. = [\Delta m] \times 931\ MeV[/latex]
[latex]= 1.934207 \times 931\ MeV[/latex]
[latex]= 1800.747\ MeV[/latex]
Now, Binding Energy per nucleon [latex](\bar{B.E.}) = \dfrac{B.E.}{A} = \dfrac{1800.747}{238}[/latex]
[latex]= 7.566\ MeV/nucleon.[/latex]
Estimate the binding energy per nucleon of 3Li7. Mass of 3Li7, a proton & a neutron are respectively 7.01435 amu, 1.00728 amu and 1.00867 amu.
Solution
Given,
Atomic mass of [latex]{}_3Li^7[/latex] (M) = 7.01435 u
Here, Z = 3 & A = 7
Mass of proton [latex](m_p)[/latex] = 1.00728 u
Mass of neutron [latex](m_n)[/latex] = 1.00867 u
1u = 931 MeV
We know that, [latex]B.E. = [Zm_p + (A-Z)m_n – M] \times 931\ MeV[/latex]
[latex]= [3 \times 1.00728 + (7-3) \times 1.00867 – 7.01435] \times 931\ MeV[/latex]
[latex]= 39.26\ MeV[/latex]
Now, Binding Energy per nucleon [latex](\bar{B.E.}) = \dfrac{B.E.}{A} = \dfrac{39.26}{7}[/latex]
[latex]= 5.609\ MeV/nucleon.[/latex]
Assuming that about 200 MeV energy is released per fission of [latex]{}_{92}U^{235}[/latex] nuclei. What would be the mass of [latex]U^{235}[/latex] consumed per day in the fission of power 1 MW approximately.
Solution
Given,
Energy liberated per fission (E) = 200 MeV
= [latex]200 \times (1.6 \times 10^{-13})[/latex]
= [latex]3.2 \times 10^{-11}[/latex] J
Mass of [latex]U^{235}[/latex] (m) =?
Time (t) = 1 day = 86400 s
Power (P) = 1 MW
= [latex]1 \times 10^6[/latex] W
We know that,
Power production [latex](P) = \dfrac{T.E.}{t}[/latex]
[latex]T.E. = P.t[/latex]
[latex]= 10^6 \times 86400[/latex]
[latex]= 8.640 \times 10^{10}\ J[/latex]
Total energy liberated (T.E.) = N.E
[latex]N = \dfrac{T.E.}{E}[/latex]
[latex]= \dfrac{8.64 \times 10^{10}}{3.2 \times 10^{-11}}[/latex]
[latex]= 2.7 \times 10^{21}\ J[/latex]
According to Avogadro’s hypothesis,
∵ [latex]6.023 × 10^{23}[/latex] atoms of U – 235 is contained in 235 gm of it.
∴ [latex]2.7 × 10^{21}[/latex] atoms of U – 235 is contained in [latex]\dfrac{235}{6.023 \times 10^{23}} \times (2.7 \times 10^{21})[/latex]
m = 1.053 gm.
A nucleus of uranium – 238 disintegrates according to the reaction [latex]{}_{92}U^{238} → {}_{90}Th^{234} + {}_2He^4[/latex]. Calculate: i) the total energy released in the disintegration ii) the K.E. of the alpha particle, the nucleus being at rest before disintegration. (Mass of [latex]{}_{92}U^{238} = 238.12492\ u[/latex], mass of [latex]{}_{90}Th^{234} = 234.1165\ u[/latex], mass of [latex]{}_2He^4[/latex] = 4.00387 u, 1u = 931 MeV).
Solution
Given,
[latex]{}_{92}U^{235}[/latex] disintegrates according to [latex]{}_{92}U^{238} → {}_{90}Th^{234} + {}_2He^4[/latex]
We have,
Mass of [latex]{}_{92}U^{238}[/latex] = 238.12492 u
Mass of [latex]{}_{90}Th^{234}[/latex] = 234.1165 u
Mass of [latex]{}_2He^4[/latex] = 4.00384 u
Total energy released in the disintegration process (E)
[latex]= [\sum reactant – \sum product] \times 931\ MeV[/latex]
[latex]= [238.12492 – (234.1165 + 4.00384)] \times 931\ MeV[/latex]
[latex]= 4.264\ MeV[/latex]
Now, K.E. of [latex]\alpha[/latex] – particles [latex]= \dfrac{M_{Th}}{M_{Th} + M_{He}} \times E[/latex]
[latex]= \dfrac{234.1165}{(234.1165+4.00384)} \times 4.264[/latex]
[latex]= 4.192\ MeV.[/latex]
The energy liberated in the fission of a single uranium – 235 atom is [latex]3.2 × 10^{−11}[/latex] J. Calculate the power production corresponding to the fission of 1 kg of uranium per day. (Avogadro constant = [latex]6.0 × 10^{23} mole^{−1}[/latex]).
Solution
Given,
Energy liberated per fission (E) = [latex]3.2 \times 10^{-11}[/latex] J
Power production (P) =?
Mass of uranium (m) = 1 kg = 1000 gm
Time (t) = 1 day = 86400 s
Avogadro’s constant [latex](N_A) = 6.0 \times 10^{23}\ mole^{-1}[/latex]
We know that,
According to Avogadro’s hypothesis,
∵ 235 gm of U – 235 contains [latex]6.0 \times 10^{23}[/latex] atoms.
∴ 1000 gm of U – 235 contains [latex]\dfrac{6.0 \times 10^{23}}{235} \times 1000[/latex] atoms
N = [latex]2.553 \times 10^{24}[/latex] atoms.
Total energy liberated (T.E.) = N.E
[latex]= (2.553 \times 10^{24}) \times (3.2 \times 10^{-11})[/latex]
[latex]= 8.17 \times 10^{13}\ J[/latex]
Power production [latex](P) = \dfrac{T.E.}{t}[/latex]
[latex]= \dfrac{8.17 \times 10^{13}}{86400}[/latex]
[latex]= 9.456 \times 10^8\ Watt.[/latex]
Calculate the Q – value of the reaction & mention the type of reaction (endo – thermic or exothermic). [latex]{}_2He^4[/latex] = 4.00377 amu, [latex]{}_8O^{17}[/latex] = 17.00450 amu, [latex]{}_7N^{14}[/latex] = 14.00783 amu, [latex]{}_1H^1[/latex] = 1.00814 amu.
Solution
Given,
Q – value of the reaction (Q) =?
Mass of [latex]{}_2He^4[/latex] = 4.00377 u
Mass of [latex]{}_8O^{17}[/latex] = 17.00450 u
Mass of [latex]{}_7N^{14}[/latex] = 14.00783 u
Mass of [latex]{}_1H^1[/latex] = 1.00814 u
We have, the reaction involved is written as:
[latex]{}_2He^4 + {}_7N^{14} → {}_8O^{17} + {}_1H^1[/latex] + Q
So, Q – value [latex]= [\sum reactant – \sum product] \times 931\ MeV[/latex]
[latex]= [(4.00377+14.00783) – (17.00450+1.00814)] \times 931\ MeV[/latex]
[latex]= -0.96824\ MeV[/latex]
Since, Q – value of the reaction is negative, the reaction is endothermic.
Calculate the speed of particle of the mass of it is equal to 5 times its rest value.
Solution
Given,
Speed of particle (v) =?
Mass of particle (m) = 5 times rest mass [latex](m_o)[/latex] i.e.
m = [latex]5m_o[/latex]
We know that,
From relativistic relation of mass,
[latex]m = \dfrac{m_o}{\sqrt{1-\dfrac{v^2}{c^2}}}[/latex]
[latex]5m_o = \dfrac{m_o}{\sqrt{1-\dfrac{v^2}{c^2}}}[/latex]
[latex]\sqrt{1-\dfrac{v^2}{c^2}} = \dfrac{1}{5}[/latex]
Squaring both sides,
[latex]1 – \dfrac{v^2}{c^2} = \dfrac{1}{25}[/latex]
[latex]\dfrac{v^2}{c^2} = 1 – \dfrac{1}{25}[/latex]
[latex]v^2 = \dfrac{24}{25} \times (3 \times 10^8)^2[/latex]
[latex]v^2 = 8.64 \times 10^{16}[/latex]
[latex]v = \sqrt{8.64 \times 10^{16}}[/latex]
[latex]= 2.939 \times 10^8\ ms^{-1}.[/latex]