Nuclei
What to remember
- The nucleus has radius R = R₀A^(1/3) with R₀ = 1.2 fm, so its volume is proportional to A and its density is the same (about 2.3 × 10¹⁷ kg m⁻³) for all nuclei.
- Mass defect Δm = [Zm_p + (A − Z)m_n] − M(nucleus); binding energy = Δm × 931.5 MeV when Δm is in u.
- Binding energy per nucleon rises to a peak of about 8.8 MeV near A = 56 (iron region). Heavy nuclei release energy by fission; very light nuclei release energy by fusion.
- Nuclear force is strong, short-range (a few fm), charge independent, saturating, and attractive at about 1 fm but repulsive at a smaller distance.
Composition of the nucleus and atomic masses
An atom has a central nucleus and electrons around it. The nucleus contains protons and neutrons, which are together called nucleons.
- Proton: charge +e, mass m_p = 1.00727 u (about 1.673 × 10⁻²⁷ kg).
- Neutron: no charge, mass m_n = 1.00867 u (about 1.675 × 10⁻²⁷ kg).
- Atomic number Z = number of protons. Mass number A = number of protons + number of neutrons. Neutron number N = A − Z.
- A nuclide is written as ᴬ_Z X. For example, U-235 has Z = 92, A = 235 and N = 143.
Atomic mass unit (u): one u is 1/12 of the mass of one carbon-12 atom. 1 u = 1.66 × 10⁻²⁷ kg. By the mass-energy relation, 1 u is equivalent to 931.5 MeV (written as 931.5 MeV/c²).
| Term | Same | Different | Example |
|---|---|---|---|
| Isotopes | Z (proton number) | A and N | ¹H, ²H, ³H |
| Isobars | A (mass number) | Z and N | ³H and ³He |
| Isotones | N (neutron number) | Z and A | ³H (N = 2) and ⁴He (N = 2) |
Isotopes of an element have the same chemical properties because they have the same number of electrons. Most elements are mixtures of isotopes, so the atomic mass of an element is the weighted average of its isotopic masses.
Size and density of the nucleus
Experiments show that the nuclear radius depends on the mass number:
R = R₀A^(1/3), with R₀ = 1.2 × 10⁻¹⁵ m = 1.2 fm (1 fm = 10⁻¹⁵ m).
Consequences:
- Volume of the nucleus = (4/3)πR³ = (4/3)πR₀³A. So the volume is proportional to A.
- Each nucleon takes up about the same volume. Hence the density does not depend on A.
- Nuclear density ρ = (mass) / (volume) = A × (1.66 × 10⁻²⁷ kg) / [(4/3)π(1.2 × 10⁻¹⁵ m)³A] ≈ 2.3 × 10¹⁷ kg m⁻³.
- This is about 10¹⁴ times the density of water. All nuclei have nearly the same density.
- The ratio of the radii of two nuclei is R₁/R₂ = (A₁/A₂)^(1/3).
| Nucleus | A | Radius R = 1.2 × A^(1/3) fm |
|---|---|---|
| Helium-4 | 4 | about 1.9 fm |
| Carbon-12 | 12 | about 2.7 fm |
| Aluminium-27 | 27 | 3.6 fm |
| Uranium-238 | 238 | about 7.4 fm |
Mass-energy equivalence, mass defect and binding energy
Einstein showed that mass and energy are equivalent: E = mc². If a mass m disappears, an energy mc² appears. With c = 3 × 10⁸ m s⁻¹, 1 kg is equivalent to 9 × 10¹⁶ J.
Mass defect. The mass of a stable nucleus is always less than the sum of the masses of its separate nucleons. The difference is the mass defect:
Δm = [Z m_p + (A − Z) m_n] − M
where M is the mass of the nucleus. (When atomic masses are used, the proton mass is replaced by the mass of a hydrogen atom, m_H = 1.00783 u, so the electron masses cancel.)
Binding energy. The binding energy E_b is the energy needed to separate the nucleus into free nucleons. It equals the energy equivalent of the mass defect:
E_b = Δm c² = Δm (in u) × 931.5 MeV
Binding energy per nucleon = E_b / A. It measures how tightly a nucleon is held. A higher value means a more stable nucleus.
| Nucleus | Binding energy per nucleon (about) |
|---|---|
| Deuterium ²H | 1.1 MeV |
| Helium-4 | 7.1 MeV |
| Oxygen-16 | 8.0 MeV |
| Iron-56 | 8.8 MeV |
| Uranium-238 | 7.6 MeV |
The binding energy per nucleon curve, in words.
- For very light nuclei the value is small (about 1.1 MeV for ²H) and rises steeply with A.
- Helium-4 is a peak among the light nuclei: it is unusually stable (7.1 MeV per nucleon).
- Between A = 30 and A = 170 the curve is nearly flat, at about 8 MeV per nucleon. This is the range of the most stable nuclei.
- The maximum is about 8.8 MeV near A = 56 (iron region).
- Beyond the peak the curve falls slowly. At A = 238 it is about 7.6 MeV.
Consequences of the curve.
- Fission: a heavy nucleus (A about 240) splits into two middle-mass nuclei (A about 120). The products have a higher binding energy per nucleon, about 8.5 MeV against 7.6 MeV. The difference is released as energy.
- Fusion: two very light nuclei join to form a heavier nucleus. The product has a higher binding energy per nucleon. The difference is released as energy.
- In both processes the nuclei move towards the peak, and the total binding energy increases. That increase appears as released energy.
Nuclear force and its properties
The protons in a nucleus repel each other by the electric force, yet the nucleus is stable. So there must be a stronger attractive force between nucleons. This is the nuclear force.
| Property | Statement |
|---|---|
| Strength | Much stronger than the electric force and the gravitational force at nuclear distances |
| Range | Short: a few fm. It becomes negligible beyond about 2.5 fm to 3 fm |
| Nature | Attractive at distances of about 1 fm, repulsive if the nucleons come closer than about 0.8 fm |
| Charge | Charge independent: the p–p, n–n and n–p forces are nearly equal |
| Saturation | A nucleon interacts only with its near neighbours, so the binding energy per nucleon is nearly constant |
| Dependence | Not a central, inverse-square force; it also depends on spin |
Because of the short range, the binding energy per nucleon becomes nearly constant for medium and heavy nuclei. Because the electric repulsion has a long range, large nuclei become less tightly bound per nucleon, and the curve falls after A = 56.
Nuclear energy: fission
Nuclear fission is the splitting of a heavy nucleus into two lighter nuclei of comparable mass, with the release of energy and neutrons.
A typical reaction with U-235:
n + ²³⁵U → ²³⁶U* → ¹⁴¹Ba + ⁹²Kr + 3 n + energy
- A slow (thermal) neutron is absorbed by U-235 and forms an unstable compound nucleus U-236.
- The compound nucleus splits into two fragments and releases two or three neutrons.
- The energy released is about 200 MeV per fission of one U-235 nucleus.
- Most of this energy appears as the kinetic energy of the fragments, and it ends as heat.
Chain reaction. The neutrons released in one fission can cause fission in other nuclei, and these release further neutrons. The number of fissions grows rapidly if, on average, more than one released neutron causes another fission. This is a chain reaction.
- Uncontrolled chain reaction: the energy is released in a very short time, as in a nuclear bomb.
- Controlled chain reaction: exactly one neutron per fission is allowed to cause a further fission, so the reaction goes on at a steady rate. This is used in a nuclear reactor.
Nuclear reactor in brief.
| Part | Function |
|---|---|
| Fuel | Fissionable material, such as enriched uranium (U-235) |
| Moderator | Slows fast neutrons to thermal speeds; heavy water, graphite or ordinary water |
| Control rods | Absorb neutrons to control the rate of reaction; cadmium or boron rods |
| Coolant | Carries the heat away to produce steam; water or liquid sodium |
| Shielding | Thick concrete walls absorb the dangerous radiation |
Natural uranium contains only about 0.7% U-235, and slow neutrons are needed to cause fission of U-235. So a moderator is needed.
Nuclear energy: fusion
Nuclear fusion is the joining of two light nuclei to form a heavier nucleus, with the release of energy.
Examples:
- ¹H + ¹H → ²H + e⁺ + ν + 0.42 MeV
- ²H + ²H → ³He + n + 3.27 MeV
- ²H + ²H → ³H + ¹H + 4.03 MeV
- ²H + ³H → ⁴He + n + 17.6 MeV
- All nuclei are positively charged, so they repel. To fuse, the nuclei must come within the range of the nuclear force. This needs very high temperature, of the order of 10⁷ K to 10⁸ K, and high pressure. So fusion is called a thermonuclear reaction.
- Energy per unit mass: fusion releases more energy per nucleon than fission. Fission gives about 0.9 MeV per nucleon of fuel, and fusion about 6.7 MeV per nucleon of fuel in the conversion of hydrogen to helium.
Energy in stars. The Sun and other stars are powered by fusion in their hot cores. The net result of the proton–proton cycle is that four hydrogen nuclei (protons) fuse to form one helium-4 nucleus:
4 ¹H → ⁴He + 2 e⁺ + 2 ν + 26.7 MeV
The energy is about 26 MeV for each helium nucleus formed. The enormous gravitational pull in the star keeps the hot gas together at the required temperature.
| Feature | Fission | Fusion |
|---|---|---|
| Process | Heavy nucleus splits | Light nuclei join |
| Energy per reaction | About 200 MeV | About 3 to 25 MeV |
| Energy per unit mass | Less | More |
| Condition | Slow neutrons | Very high temperature (10⁷ K or more) |
| Control | Possible in a reactor | Not yet sustained on a large scale |
| Example | U-235 reactor | Sun, stars |
Worked numericals
Example 1. Radius and density. Find the radius of a nucleus of mass number 125, and compare it with the radius of a nucleus of mass number 8.
R(125) = 1.2 × 125^(1/3) = 1.2 × 5 = 6.0 fm. R(8) = 1.2 × 2 = 2.4 fm. The ratio of radii is 5 : 2. Both nuclei have the same density.
Example 2. Binding energy of helium-4. Given m_p = 1.0073 u, m_n = 1.0087 u and the mass of the helium nucleus 4.0015 u, find the binding energy and the binding energy per nucleon.
Mass of the separate nucleons = 2(1.0073) + 2(1.0087) = 2.0146 + 2.0174 = 4.0320 u.
Δm = 4.0320 − 4.0015 = 0.0305 u.
E_b = 0.0305 × 931.5 = 28.4 MeV (about).
Binding energy per nucleon = 28.4 / 4 = 7.1 MeV.
Example 3. Energy released in fission. A U-235 nucleus (binding energy per nucleon 7.6 MeV) splits into two fragments, each of mass number about 117 and binding energy per nucleon 8.5 MeV. Estimate the energy released.
Total binding energy before = 235 × 7.6 = 1786 MeV. After = 234 × 8.5 = 1989 MeV (taking 234 nucleons in the fragments and ignoring the released neutrons). Energy released is about 203 MeV, close to the standard 200 MeV. A quicker estimate: 235 × (8.5 − 7.6) = 211 MeV.
Example 4. Energy from mass. If 1 g of matter is completely converted to energy: E = mc² = 10⁻³ × (3 × 10⁸)² = 9 × 10¹³ J.
Example 5. Fusion energy. For the reaction ²H + ²H → ³He + n, take the masses 2.0141 u, 3.0160 u and 1.0087 u.
Mass before = 2 × 2.0141 = 4.0282 u. Mass after = 3.0160 + 1.0087 = 4.0247 u. Δm = 0.0035 u. Q = 0.0035 × 931.5 = 3.26 MeV, which is about 3.27 MeV.
Formula summary
Constants: 1 u = 1.66 × 10⁻²⁷ kg = 931.5 MeV/c²; c = 3 × 10⁸ m s⁻¹; 1 eV = 1.6 × 10⁻¹⁹ J; 1 MeV = 1.6 × 10⁻¹³ J; 1 fm = 10⁻¹⁵ m; N_A = 6.02 × 10²³ mol⁻¹; R₀ = 1.2 fm.
| Quantity | Formula | Symbols and units |
|---|---|---|
| Neutron number | N = A − Z | A mass number, Z atomic number |
| Nuclear radius | R = R₀A^(1/3) | R in m or fm, R₀ = 1.2 fm |
| Nuclear density | ρ = 3m_u / (4πR₀³) | about 2.3 × 10¹⁷ kg m⁻³ |
| Mass-energy relation | E = mc² | E in J, m in kg |
| Mass defect | Δm = Z m_p + (A − Z) m_n − M | in u or kg |
| Binding energy | E_b = Δm × 931.5 MeV | Δm in u |
| Binding energy per nucleon | E_b / A | MeV per nucleon |
| Energy released (Q value) | Q = (mass of reactants − mass of products) × 931.5 MeV | masses in u |
NEET traps
- 1. Isotopes have the same Z, isobars the same A, isotones the same N. Students swap them.
- 2. The nuclear density does not depend on A. Only the radius depends on A^(1/3).
- 3. The radius depends on the cube root of A, not on A. A nucleus with 8 times the mass number has only twice the radius.
- 4. The mass of a nucleus is less than the sum of its nucleons. If you add the masses and subtract the nuclear mass, the result must be positive.
- 5. 1 u is 931.5 MeV, not 931.5 J. Use MeV when Δm is in u.
- 6. Binding energy per nucleon is highest near A = 56, not for the heaviest nucleus and not for the lightest.
- 7. Iron is the most stable region, but helium-4 has a higher binding energy per nucleon than its neighbours; it is a local peak.
- 8. Fission and fusion both release energy, because the product nuclei are nearer to the peak. Fission of nuclei lighter than iron or fusion of nuclei heavier than iron would need energy.
- 9. The nuclear force is not the electric force. It acts between neutrons too, and it is charge independent.
- 10. The moderator slows neutrons; the control rods absorb them. They are not the same.
One-liners
- 1. A nucleon is a proton or a neutron.
- 2. Mass number A = Z + N.
- 3. Isotopes have the same Z, isobars the same A, isotones the same N.
- 4. 1 u = 1/12 of the mass of a carbon-12 atom = 1.66 × 10⁻²⁷ kg.
- 5. 1 u is equivalent to 931.5 MeV.
- 6. R = R₀A^(1/3), with R₀ = 1.2 fm.
- 7. The nuclear density is about 2.3 × 10¹⁷ kg m⁻³ and is the same for all nuclei.
- 8. Mass defect is the difference between the mass of the separate nucleons and the nuclear mass.
- 9. Binding energy is the energy needed to break a nucleus into its nucleons.
- 10. The binding energy per nucleon peaks at about 8.8 MeV near A = 56.
- 11. Nuclear force is short-ranged, strong, charge independent and saturating.
- 12. In fission a heavy nucleus splits and about 200 MeV is released for each U-235 nucleus.
- 13. In a controlled chain reaction exactly one neutron from each fission causes another fission.
- 14. A moderator slows neutrons; control rods (cadmium, boron) absorb neutrons.
- 15. Fusion needs a temperature of about 10⁷ K or more and powers the Sun.
Practice questions
The radius of a nucleus of mass number 64 is (take R₀ = 1.2 fm)
- 4.8 fm
- 76.8 fm
- 9.6 fm
- 3.6 fm
Answer
A. 4.8 fm
R = R₀A^(1/3) = 1.2 × 64^(1/3) = 1.2 × 4 = 4.8 fm. Check: 4.8/1.2 = 4 and 4³ = 64.
The ratio of the radius of a nucleus of mass number 216 to that of a nucleus of mass number 27 is
- 3 : 1
- 4 : 1
- 8 : 1
- 2 : 1
Answer
D. 2 : 1
R ∝ A^(1/3). (216/27)^(1/3) = 8^(1/3) = 2. Check: 6/3 = 2.
In nuclear physics, the energy equivalent of one atomic mass unit (1 u) is about
- 931.5 eV
- 931.5 MeV
- 93.15 MeV
- 9.315 MeV
Answer
B. 931.5 MeV
1 u = 1.66 × 10⁻²⁷ kg; E = mc² = 1.49 × 10⁻¹⁰ J = 931.5 MeV.
A helium nucleus has mass 4.0015 u. Take m_p = 1.0073 u and m_n = 1.0087 u and 1 u = 931.5 MeV. The binding energy of the nucleus is about
- 56.8 MeV
- 14.2 MeV
- 28.4 MeV
- 7.1 MeV
Answer
C. 28.4 MeV
Δm = 2(1.0073) + 2(1.0087) − 4.0015 = 0.0305 u. E_b = 0.0305 × 931.5 = 28.4 MeV (7.1 MeV per nucleon).
The total binding energy of an iron nucleus of mass number 56 is 492 MeV. Its binding energy per nucleon is about
- 88 MeV
- 8.8 MeV
- 17.6 MeV
- 4.4 MeV
Answer
B. 8.8 MeV
492 / 56 = 8.79 MeV. Check: 56 × 8.8 = 492.8.
If 1 g of matter is converted completely into energy, the energy released is (c = 3 × 10⁸ m s⁻¹)
- 3 × 10⁵ J
- 9 × 10¹⁰ J
- 9 × 10¹⁶ J
- 9 × 10¹³ J
Answer
D. 9 × 10¹³ J
m = 10⁻³ kg. E = mc² = 10⁻³ × 9 × 10¹⁶ = 9 × 10¹³ J.
The ratio of the density of a nucleus of mass number 64 to the density of a nucleus of mass number 8 is
- 1 : 1
- 2 : 1
- 1 : 2
- 8 : 1
Answer
A. 1 : 1
Mass ∝ A and volume ∝ R³ ∝ A, so the density is independent of A. The ratio is 1 : 1.
The number of neutrons in a nucleus of ²³⁵₉₂U is
- 92
- 327
- 143
- 235
Answer
C. 143
N = A − Z = 235 − 92 = 143.
Which pair of nuclides are isotones?
- ¹³₆C and ¹⁴₇N
- ¹²₆C and ¹⁴₆C
- ¹⁴₇N and ¹⁶₈O
- ¹⁴₆C and ¹⁴₇N
Answer
A. ¹³₆C and ¹⁴₇N
Isotones have equal neutron number. ¹³₆C has 13 − 6 = 7 and ¹⁴₇N has 14 − 7 = 7 neutrons. The second pair are isobars, the third are isotopes, and the fourth have 7 and 8 neutrons.
The binding energy per nucleon is greatest for nuclei of mass number about
- 4
- 56
- 120
- 238
Answer
B. 56
The curve peaks near A = 56 (iron region), at about 8.8 MeV per nucleon.
The energy released in the fission of one U-235 nucleus is about
- 20 MeV
- 200 eV
- 2000 MeV
- 200 MeV
Answer
D. 200 MeV
About 200 MeV is released per fission of U-235.
The range of the nuclear force is of the order of
- 10⁻¹⁰ m
- infinite, like gravitation
- a few femtometres (10⁻¹⁵ m)
- 10⁻²⁰ m
Answer
C. a few femtometres (10⁻¹⁵ m)
The nuclear force is short-ranged, a few fm; it is negligible beyond about 2.5 to 3 fm.
In a nuclear reactor the control rods are used to
- absorb neutrons and so control the rate of the chain reaction
- supply extra neutrons to the fuel
- slow down fast neutrons
- carry the heat to the steam generator
Answer
A. absorb neutrons and so control the rate of the chain reaction
Control rods (cadmium or boron) absorb neutrons. Slowing is done by the moderator and heat removal by the coolant.
The number of atoms in 1 kg of U-235 is taken as (1000/235) × 6.02 × 10²³. If every nucleus undergoes fission releasing 200 MeV (1 MeV = 1.6 × 10⁻¹³ J), the energy released is about
- 1.9 × 10¹³ J
- 8.2 × 10¹⁰ J
- 8.2 × 10¹⁶ J
- 8.2 × 10¹³ J
Answer
D. 8.2 × 10¹³ J
N = 2.56 × 10²⁴. E = 2.56 × 10²⁴ × 200 × 1.6 × 10⁻¹³ = 8.2 × 10¹³ J. Check: 8.2 × 10¹³ J is 5.1 × 10²⁶ MeV, and 5.1 × 10²⁶ / 2.56 × 10²⁴ = 200 MeV.
A reactor produces 200 MW of power from fission, each fission giving 200 MeV (1 MeV = 1.6 × 10⁻¹³ J). The number of fissions per second is
- 1.25 × 10¹⁹
- 6.25 × 10²⁴
- 6.25 × 10¹⁸
- 6.25 × 10¹⁵
Answer
C. 6.25 × 10¹⁸
Energy per fission = 200 × 1.6 × 10⁻¹³ = 3.2 × 10⁻¹¹ J. Rate = 2 × 10⁸ / 3.2 × 10⁻¹¹ = 6.25 × 10¹⁸ s⁻¹.
In the reaction ²H + ²H → ³He + n, the masses are 2.0141 u, 3.0160 u and 1.0087 u. Using 1 u = 931.5 MeV, the energy released is about
- 33 MeV
- 3.3 MeV
- 6.5 MeV
- 0.0035 MeV
Answer
B. 3.3 MeV
Δm = 4.0282 − 4.0247 = 0.0035 u. Q = 0.0035 × 931.5 = 3.26 MeV.
Given below are two statements: Statement I: All nuclei have nearly the same density. Statement II: The radius of a nucleus is proportional to the cube root of its mass number, so its volume is proportional to its mass number.
- Both Statement I and Statement II are correct
- Both Statement I and Statement II are incorrect
- Statement I is correct but Statement II is incorrect
- Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
Volume ∝ A and mass ∝ A, so the density is constant. Both statements are correct.
Given below are two statements: Statement I: Isotopes have the same mass number. Statement II: Isobars have the same atomic number.
- Both Statement I and Statement II are correct
- Both Statement I and Statement II are incorrect
- Statement I is correct but Statement II is incorrect
- Statement I is incorrect but Statement II is correct
Answer
B. Both Statement I and Statement II are incorrect
Isotopes have the same Z but different A; isobars have the same A but different Z. Both are incorrect.
Given below are two statements: Statement I: The mass of a stable nucleus is greater than the sum of the masses of its free nucleons. Statement II: The binding energy of a nucleus equals the energy equivalent of its mass defect.
- Both Statement I and Statement II are correct
- Both Statement I and Statement II are incorrect
- Statement I is correct but Statement II is incorrect
- Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
The nuclear mass is less than the sum of the nucleon masses, so Statement I is incorrect. E_b = Δm c², so Statement II is correct.
Given below are two statements: Statement I: The nuclear force is charge independent. Statement II: The nuclear force is a long-range force like the gravitational force.
- Both Statement I and Statement II are correct
- Both Statement I and Statement II are incorrect
- Statement I is correct but Statement II is incorrect
- Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
The p–p, n–n and n–p forces are nearly equal, so Statement I is correct. The nuclear force has a short range of a few fm, so Statement II is incorrect.
Which of the following statements are correct? 1. Fusion of light nuclei needs a very high temperature. 2. Control rods absorb neutrons in a reactor. 3. A moderator speeds up the neutrons in a reactor.
- 1 and 2 only
- 2 and 3 only
- 1 and 3 only
- 1, 2 and 3
Answer
A. 1 and 2 only
Fusion needs about 10⁷ K or more and control rods absorb neutrons, so 1 and 2 are correct. A moderator slows neutrons, so 3 is wrong.
Which of the following statements about the binding energy per nucleon curve are correct? 1. Its maximum is about 8.8 MeV near A = 56. 2. It is nearly flat at about 8 MeV for A between 30 and 170. 3. It keeps rising steadily up to the heaviest nuclei.
- 1 only
- 2 and 3 only
- 1 and 3 only
- 1 and 2 only
Answer
D. 1 and 2 only
Statements 1 and 2 describe the curve. After the peak the curve falls slowly (7.6 MeV at A = 238), so 3 is wrong.
Given below are two statements: Statement I: Fusion of nuclei takes place at a lower temperature than fission. Statement II: In fusion the positively charged nuclei must overcome their mutual electric repulsion.
- Both Statement I and Statement II are correct
- Both Statement I and Statement II are incorrect
- Statement I is correct but Statement II is incorrect
- Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
Fusion needs a very high temperature, about 10⁷ K or more, so Statement I is incorrect. Statement II is correct.
Given below are two statements: Statement I: The radius of a nucleus of mass number 27 is 3.6 fm. Statement II: The radius of a nucleus of mass number 64 is 4.8 fm. (Take R₀ = 1.2 fm.)
- Both Statement I and Statement II are correct
- Both Statement I and Statement II are incorrect
- Statement I is correct but Statement II is incorrect
- Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
R = 1.2 × 3 = 3.6 fm for A = 27 and 1.2 × 4 = 4.8 fm for A = 64. Both are correct.
Given below are two statements: Statement I: The energy equivalent of 1 u is about 931.5 MeV. Statement II: A neutron is lighter than a proton.
- Both Statement I and Statement II are correct
- Both Statement I and Statement II are incorrect
- Statement I is correct but Statement II is incorrect
- Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
m_n = 1.00867 u is greater than m_p = 1.00727 u, so Statement II is incorrect.
Given below are two statements: Statement I: About 200 MeV is released when one U-235 nucleus undergoes fission. Statement II: In every fission of U-235 the two fragments have exactly equal masses.
- Both Statement I and Statement II are correct
- Both Statement I and Statement II are incorrect
- Statement I is correct but Statement II is incorrect
- Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
The energy is about 200 MeV. The fragments are not always equal; one example is ¹⁴¹Ba and ⁹²Kr.
Assertion (A): The density of nuclear matter is the same for all nuclei. Reason (R): The volume of a nucleus is proportional to its mass number.
- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
Mass ∝ A and volume ∝ A, so the density is constant.
Assertion (A): A moderator is used in a nuclear reactor. Reason (R): A moderator absorbs neutrons so that the chain reaction stops.
- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- A is false but R is true
Answer
C. A is true but R is false
A moderator slows fast neutrons to thermal speed; absorbing neutrons is the job of control rods. R is false.
Assertion (A): Energy is released when a heavy nucleus undergoes fission. Reason (R): A neutron has a slightly greater mass than a proton.
- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- A is false but R is true
Answer
B. Both A and R are true but R is not the correct explanation of A
Both are true, but fission releases energy because the binding energy per nucleon increases. R is not the explanation.
Assertion (A): The binding energy per nucleon of U-238 is higher than that of Fe-56. Reason (R): The binding energy per nucleon is maximum near A = 56.
- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- A is false but R is true
Answer
D. A is false but R is true
U-238 has about 7.6 MeV per nucleon and Fe-56 about 8.8 MeV. A is false and R is true.
Assertion (A): Fusion reactions occur in the core of the Sun. Reason (R): The temperature in the core is of the order of 10⁷ K, which lets nuclei overcome their electric repulsion.
- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
The high temperature gives nuclei enough kinetic energy to come within the range of the nuclear force.
Match List I with List II. List I: a. ¹H and ²H b. ³H and ³He c. ³H and ⁴He d. ¹H and ⁴He. List II: i. Isotopes ii. Isobars iii. Isotones iv. None of these relations
- a-ii, b-i, c-iii, d-iv
- a-i, b-ii, c-iii, d-iv
- a-i, b-iii, c-ii, d-iv
- a-iv, b-ii, c-i, d-iii
Answer
B. a-i, b-ii, c-iii, d-iv
¹H and ²H: same Z. ³H and ³He: same A. ³H and ⁴He: N = 2 for both. ¹H (N = 0) and ⁴He (N = 2) have different Z, A and N.
Match List I with List II. List I: a. R₀ in R = R₀A^(1/3) b. Nuclear density c. Energy equivalent of 1 u d. Energy per fission of U-235. List II: i. 1.2 fm ii. 2.3 × 10¹⁷ kg m⁻³ iii. 931.5 MeV iv. About 200 MeV
- a-i, b-ii, c-iv, d-iii
- a-iii, b-ii, c-i, d-iv
- a-i, b-ii, c-iii, d-iv
- a-ii, b-i, c-iii, d-iv
Answer
C. a-i, b-ii, c-iii, d-iv
Standard values: R₀ = 1.2 fm, ρ ≈ 2.3 × 10¹⁷ kg m⁻³, 1 u = 931.5 MeV, about 200 MeV per fission.
Match List I with List II. List I (reactor part): a. Moderator b. Control rods c. Coolant d. Fuel. List II (function): i. Absorb neutrons ii. Slow down neutrons iii. Remove heat iv. Undergoes fission
- a-ii, b-iii, c-i, d-iv
- a-ii, b-i, c-iii, d-iv
- a-iv, b-i, c-iii, d-ii
- a-i, b-ii, c-iii, d-iv
Answer
B. a-ii, b-i, c-iii, d-iv
Moderator slows neutrons, control rods absorb them, the coolant removes heat and the fuel (enriched uranium) undergoes fission.
Match List I with List II. List I (reaction): a. ²H + ³H → ⁴He + n b. 4 ¹H → ⁴He + 2 e⁺ + 2 ν c. ²H + ²H → ³He + n d. Fission of ²³⁵U by a slow neutron. List II (energy released): i. 3.27 MeV ii. 17.6 MeV iii. 26.7 MeV iv. About 200 MeV
- a-iii, b-ii, c-i, d-iv
- a-ii, b-i, c-iii, d-iv
- a-i, b-iii, c-ii, d-iv
- a-ii, b-iii, c-i, d-iv
Answer
D. a-ii, b-iii, c-i, d-iv
The values are 17.6 MeV (D–T), 26.7 MeV (proton–proton cycle), 3.27 MeV (D–D) and about 200 MeV (fission).
Match List I with List II. List I (nuclide): a. ⁴₂He b. ¹²₆C c. ¹⁶₈O d. ²³⁵₉₂U. List II (number of neutrons): i. 143 ii. 8 iii. 2 iv. 6
- a-iv, b-iii, c-ii, d-i
- a-ii, b-iv, c-iii, d-i
- a-iii, b-iv, c-i, d-ii
- a-iii, b-iv, c-ii, d-i
Answer
D. a-iii, b-iv, c-ii, d-i
N = A − Z: 4 − 2 = 2; 12 − 6 = 6; 16 − 8 = 8; 235 − 92 = 143.
Match List I with List II. List I (feature of nuclear force): a. Range b. Dependence on charge c. Saturation d. Separation less than about 0.8 fm. List II: i. A few fm ii. Independent of charge iii. A nucleon interacts only with its near neighbours iv. Repulsive
- a-i, b-ii, c-iv, d-iii
- a-i, b-ii, c-iii, d-iv
- a-ii, b-i, c-iii, d-iv
- a-iii, b-ii, c-i, d-iv
Answer
B. a-i, b-ii, c-iii, d-iv
The force is short-ranged, charge independent, saturating, and becomes repulsive at very small separations.
A U-235 nucleus has a binding energy of 7.6 MeV per nucleon. It splits into fragments in which all 235 nucleons have a binding energy of 8.5 MeV per nucleon (neglect the released neutrons). The energy released is about
- 1786 MeV
- 1989 MeV
- 211.5 MeV
- 21.2 MeV
Answer
C. 211.5 MeV
Gain per nucleon = 8.5 − 7.6 = 0.9 MeV. Energy = 235 × 0.9 = 211.5 MeV. Check: 235 × 8.5 − 235 × 7.6 = 1989 − 1786 = 203 MeV when 234 nucleons are used, close to 200 MeV.
Two deuterons (binding energy 1.1 MeV per nucleon) fuse to make a helium-4 nucleus (binding energy 7.1 MeV per nucleon). The energy released is
- 24.0 MeV
- 6.0 MeV
- 4.4 MeV
- 28.4 MeV
Answer
A. 24.0 MeV
Final binding energy = 4 × 7.1 = 28.4 MeV. Initial = 4 × 1.1 = 4.4 MeV. Released = 28.4 − 4.4 = 24.0 MeV. Check: 4 × 6.0 = 24.0 MeV.
The Sun radiates energy at the rate of 3.6 × 10²⁶ W. Taking c = 3 × 10⁸ m s⁻¹, the mass converted into energy every second is about
- 4 × 10⁹ kg
- 3.2 × 10⁴³ kg
- 1.2 × 10¹⁸ kg
- 4 × 10²⁵ kg
Answer
A. 4 × 10⁹ kg
m = E/c² = 3.6 × 10²⁶ / 9 × 10¹⁶ = 4 × 10⁹ kg per second. Check: 4 × 10⁹ × 9 × 10¹⁶ = 3.6 × 10²⁶ J.
Natural uranium contains only about 0.7% of U-235, and U-235 is fissioned most readily by slow neutrons. Which arrangement helps a chain reaction in natural uranium?
- Placing the uranium among cadmium rods so that the neutrons are absorbed
- Heating the uranium to 10⁷ K so that the nuclei fuse
- Placing the uranium in a moderator such as heavy water so that the neutrons are slowed down
- Surrounding the uranium with thick concrete only
Answer
C. Placing the uranium in a moderator such as heavy water so that the neutrons are slowed down
A moderator slows the fast fission neutrons to thermal speed. Cadmium absorbs neutrons and stops the reaction; concrete only shields.
A nucleus of mass number 240 splits into two nuclei of mass number about 120 each. Energy is released because
- the binding energy per nucleon does not change but the number of nucleons increases
- the binding energy per nucleon of the products is greater than that of the original nucleus
- the electric repulsion between the products becomes zero
- the binding energy per nucleon of the products is less than that of the original nucleus
Answer
B. the binding energy per nucleon of the products is greater than that of the original nucleus
Nuclei of A ≈ 120 are nearer the peak of the curve (about 8.5 MeV) than A = 240 (about 7.6 MeV). The gain in total binding energy is released.
The binding energy of a deuteron is 2.2 MeV. Using 1 u = 931.5 MeV, its mass defect is about
- 2.2 u
- 1.2 × 10⁻³ u
- 2.0 × 10³ u
- 2.4 × 10⁻³ u
Answer
D. 2.4 × 10⁻³ u
Δm = 2.2 / 931.5 = 2.4 × 10⁻³ u. Check: 2.4 × 10⁻³ × 931.5 = 2.2 MeV.
A power station of 1000 MW runs for one day (86400 s). Taking c = 3 × 10⁸ m s⁻¹, the mass equivalent of the energy produced is
- 9.6 × 10⁻¹ kg
- 9.6 × 10⁻⁷ kg
- 9.6 × 10⁻⁴ kg
- 3.6 × 10⁻⁵ kg
Answer
C. 9.6 × 10⁻⁴ kg
E = 10⁹ × 86400 = 8.64 × 10¹³ J. m = E/c² = 8.64 × 10¹³ / 9 × 10¹⁶ = 9.6 × 10⁻⁴ kg.