CUET UG Physics Booster Test 2- Nuclear Reactions and Radioactivity -

🚀

CUET UG Physics Booster Test 2- Nuclear Reactions and Radioactivity

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Physics Booster Test 2- Nuclear Reactions and Radioactivity

1 / 20

QUESTION 1 OF 20

In the sequence of Becquerel's accident, the initial compound used and the radiation behaviour observed were respectively:

Answer Review

1 In the sequence of Becquerel's accident, the initial compound used and the radiation behaviour observed were respectively:

A Uranium-potassium sulphate, Penetration of opaque barriers

B Pure uranium, Reflection off silver

C Helium gas, Phosphorescence

D Gold foil, Alpha scattering

💡 Short Explanation

�� Becquerel studied uranium compounds. �� Radiation penetrated black paper. �� The discovery was accidental.

✅ Correct Answer Explanation

Becquerel used uranium-potassium sulphate while studying phosphorescent compounds. The emitted radiation penetrated opaque barriers such as black paper and exposed photographic plates. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Reflection from silver was not observed.
�� Option C → Helium gas was not used in the experiment.
�� Option D → Gold foil scattering relates to Rutherford's experiment.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Identify the correct historical experiment and observation.

Final Logic:

Becquerel's uranium compound emitted penetrating radiation.

🧠 Memory Trick

Uranium + Black Plate

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Discovery of Radioactivity

2 Photographic blackening statements in Becquerel's experiment:
1. The package was separated from the plate by a piece of silver.
2. The compound had to be constantly illuminated by visible light during the plate's exposure.
3. The emission could penetrate black paper.
4. The observation proved radioactivity is a nuclear phenomenon.

A (1), (3)

B (1), (2), (3)

C (2), (4)

D (1), (3), (4)

💡 Short Explanation

�� Silver and black paper were present. �� Radiation penetrated the covering. �� The nuclear nature was not established by this experiment alone.

✅ Correct Answer Explanation

Statement 1 is correct because a silver piece separated the package from the plate. Statement 2 is incorrect because continuous illumination was not required. Statement 3 is correct because the emitted radiation penetrated black paper. Statement 4 is incorrect because Becquerel's observation alone did not establish radioactivity as a nuclear phenomenon. Therefore, statements 1 and 3 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes statement 2, which is incorrect.
�� Option C → Includes statements 2 and 4, both incorrect.
�� Option D → Includes statement 4, which is incorrect.

🎯 Strategy

Used

Elimination

Application:

Verify each statement using Becquerel's experimental observations.

Final Logic:

Only statements 1 and 3 directly describe the experiment.

🧠 Memory Trick

Silver + Black Paper

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Discovery of Radioactivity

3 If hypothetical elements emit purely negatively charged particles representing a count of x and y electrons from two separate radioactive samples, the total charge emitted mathematically matches:

A (y - x)e

B (x + y)e

C (x - y)e

D -(x + y)e

💡 Short Explanation

�� Each electron has charge -e. �� Total electrons = x + y. �� Charges add algebraically.

✅ Correct Answer Explanation

The total number of emitted electrons is (x + y). Since each electron carries a charge of -e, the combined charge is: -(x + y)e Therefore, option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect algebraic expression.
�� Option B → Sign is incorrect.
�� Option C → Incorrect total electron count.

🎯 Strategy

Used

Substitution

Application:

Multiply total electrons by charge per electron.

Final Logic:

Charge = Number of electrons × (-e).

🧠 Memory Trick

Electron = Negative

📚 Class XI – NCERT Physics Part I – Chapter: Electric Charges and Fields – Topic: Quantisation of Charge

4 Correct statements about beta decay:
1. Positrons are particles with the same mass as electrons.
2. Positrons possess a charge exactly opposite to that of an electron.
3. Beta decay involves the emission of a helium nucleus.
4. It occurs to help radioactive fragments achieve stable end products.

A (1), (2), (4)

B (1), (3), (4)

C (2), (3)

D (1), (2), (3), (4)

💡 Short Explanation

�� Positrons are anti-electrons. �� Mass is the same as electron mass. �� Beta decay promotes stability.

✅ Correct Answer Explanation

Statement 1 is correct because positrons have the same mass as electrons. Statement 2 is correct because positrons carry a positive charge. Statement 3 is incorrect because helium nuclei are emitted in alpha decay. Statement 4 is correct because beta decay helps unstable nuclei move toward stability. Therefore, statements 1, 2 and 4 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes statement 3, which is incorrect.
�� Option C → Includes statement 3, which is incorrect.
�� Option D → Includes statement 3, which is incorrect.

🎯 Strategy

Used

Elimination

Application:

Distinguish beta decay from alpha decay.

Final Logic:

Beta decay emits electrons or positrons, not helium nuclei.

🧠 Memory Trick

β = e⁻ or e⁺

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Beta Decay

5 Incorrect statement about gamma decay

A It consists of high energy photons.

B The energies are typically in the range of hundreds of keV or more.

C It causes the atomic number of the nucleus to decrease by 2.

D It is an electromagnetic radiation.

💡 Short Explanation

�� Gamma rays are photons. �� Gamma emission changes energy state only. �� Atomic number remains unchanged.

✅ Correct Answer Explanation

Gamma decay emits electromagnetic radiation without changing the atomic number or mass number of the nucleus. A decrease of atomic number by 2 occurs in alpha decay, not gamma decay. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Correct description of gamma rays.
�� Option B → Correct energy range.
�� Option D → Gamma rays are electromagnetic radiation.

🎯 Strategy

Used

Odd One Out

Application:

Identify the statement that incorrectly attributes alpha-decay properties to gamma decay.

Final Logic:

Gamma emission does not alter atomic number.

🧠 Memory Trick

γ = Energy Only

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Gamma Decay

6 What is the energy of a 200 keV gamma photon expressed in Joules? (Given 1 eV = 1.6 × 10⁻¹⁹ J)

A 3.2 × 10⁻¹⁴ J

B 3.2 × 10⁻¹⁹ J

C 1.6 × 10⁻¹³ J

D 3.2 × 10⁻¹⁶ J

💡 Short Explanation

�� Convert keV into eV. �� Multiply by 1.6 × 10⁻¹⁹ J. �� Use unit conversion carefully.

✅ Correct Answer Explanation

200 keV = 200 × 10³ eV = 2 × 10⁵ eV Energy: = (2 × 10⁵)(1.6 × 10⁻¹⁹) = 3.2 × 10⁻¹⁴ J Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Equivalent to only 2 eV.
�� Option C → Five times larger than the correct value.
�� Option D → Corresponds to 2 keV.

🎯 Strategy

Used

Dimensional/Unit Analysis

Application:

Convert keV → eV → Joules.

Final Logic:

200 keV = 3.2 × 10⁻¹⁴ J.

🧠 Memory Trick

200 × 1.6 = 320

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Gamma-Ray Energy Units

7 In the nuclear reaction where Uranium-235 is bombarded with a neutron,

A the total mass of the products is strictly identical to the reactants down to the exact atomic mass unit

B the nucleus transmutes to an intermediate mass fragment without releasing any additional neutrons

C it breaks into two intermediate mass nuclear fragments, and the mass defect converts into kinetic energy and heat

D the binding energy per nucleon of the fragments is less than that of the original nucleus

💡 Short Explanation

�� U-235 undergoes fission. �� Mass defect produces energy. �� Intermediate-mass fragments are formed.

✅ Correct Answer Explanation

After absorbing a neutron, U-235 forms U-236 and splits into two medium-mass fragments. The mass defect is converted into kinetic energy, radiation and heat. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Exact mass equality does not hold because of mass defect.
�� Option B → Additional neutrons are emitted.
�� Option D → Fragments generally have higher binding energy per nucleon.

🎯 Strategy

Used

Elimination

Application:

Apply the basic mechanism of nuclear fission.

Final Logic:

Mass defect → Energy release.

🧠 Memory Trick

U-236 → Split + Energy

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Nuclear Fission

8 The total gain in binding energy during the fission of a heavy nucleus (A = 240) into two fragments (A = 120)

A is about 8.5 MeV per nucleon, yielding roughly 2000 MeV in total

B is about 0.9 MeV per nucleon, yielding roughly 216 MeV in total

C is negative because energy is required to break the nucleus apart

D depends entirely on the chemical combustion of uranium

💡 Short Explanation

�� Gain per nucleon ≈ 0.9 MeV. �� Total nucleons = 240. �� Energy released ≈ 216 MeV.

✅ Correct Answer Explanation

Gain in binding energy per nucleon: = 8.5 − 7.6 = 0.9 MeV Total energy released: = 240 × 0.9 = 216 MeV Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Uses total binding energy rather than gain.
�� Option C → Fission is exothermic.
�� Option D → Energy is nuclear, not chemical.

🎯 Strategy

Used

Substitution

Application:

Multiply gain per nucleon by total nucleons.

Final Logic:

0.9 × 240 = 216 MeV.

🧠 Memory Trick

0.9 × 240 = 216

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Binding Energy in Fission

9 Match the aspects of intermediate mass fragments (List I) with their specific details (List II) from U-235 fission.

List I	List II
1. Disintegration energy	a. Increases to about 8.5 MeV
2. Fragment stability	b. Ba-144 and Kr-89
3. Fragment elements (example)	c. Highly radioactive initially
4. Binding energy per nucleon	d. Appears first as kinetic energy

A (1)-d, (2)-c, (3)-b, (4)-a

B (1)-a, (2)-b, (3)-c, (4)-d

C (1)-c, (2)-d, (3)-a, (4)-b

D (1)-d, (2)-a, (3)-b, (4)-c

💡 Short Explanation

�� Fission energy first appears as kinetic energy. �� Fragments are radioactive. �� Binding energy per nucleon increases.

✅ Correct Answer Explanation

1 → d : Disintegration energy initially appears as kinetic energy. 2 → c : Fission fragments are highly radioactive. 3 → b : Ba-144 and Kr-89 are example fragments. 4 → a : Binding energy per nucleon rises to about 8.5 MeV. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Disintegration energy and binding energy are interchanged.
�� Option C → Multiple mismatches occur.
�� Option D → Fragment stability and binding energy are incorrectly matched.

🎯 Strategy

Used

Option Grouping

Application:

Match fragment properties with their corresponding descriptions.

Final Logic:

Energy → Kinetic, Fragments → Radioactive, Examples → Ba/Kr, E/A → 8.5 MeV.

🧠 Memory Trick

Energy → Motion First

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Fission Fragments

10 In a hypothetical calculation tracking negative charges emitted by beta decay in successive fragment stabilizations, if x electrons and y electrons are emitted from two fragment chains respectively, what is the mathematical expression for their combined emitted charge?

A -(x + y)e

B (x + y)e

C (x - y)e

D (y - x)e

💡 Short Explanation

�� Each electron carries charge -e. �� Charges add algebraically. �� Total electrons = x + y.

✅ Correct Answer Explanation

If x and y electrons are emitted, the total number of emitted electrons is (x + y). Since each electron has charge -e, the total emitted charge is: -(x + y)e Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Sign is incorrect.
�� Option C → Incorrect algebraic total.
�� Option D → Incorrect expression.

🎯 Strategy

Used

Substitution

Application:

Multiply total emitted electrons by charge per electron.

Final Logic:

Charge = (x + y)(-e).

🧠 Memory Trick

Electron = Minus

📚 Class XI – NCERT Physics Part I – Chapter: Electric Charges and Fields – Topic: Quantisation of Charge

11 Incorrect statement about the joining of light nuclei

A Light nuclei must have A ≤ 10 to be considered in this specific curve region.

B The binding energy per nucleon of the fused heavier nucleus is more than the lighter nuclei.

C The process consumes vast amounts of energy, making it an endothermic reaction.

D It is the exact process that powers the sun and other stars.

💡 Short Explanation

�� Fusion of light nuclei releases energy. �� Binding energy per nucleon increases. �� Fusion powers stars.

✅ Correct Answer Explanation

Fusion of light nuclei forms a more tightly bound nucleus with a higher binding energy per nucleon. The increase in binding energy corresponds to a release of energy, making fusion an exothermic process rather than an endothermic one. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Consistent with the light-nuclei region of the binding-energy curve.
�� Option B → Correct reason for energy release in fusion.
�� Option D → Fusion is the energy source of stars.

🎯 Strategy

Used

Odd One Out

Application:

Identify the statement that contradicts the energy-releasing nature of fusion.

Final Logic:

Fusion of light nuclei is exothermic, not endothermic.

🧠 Memory Trick

Fusion = Join + Energy

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Nuclear Fusion

12 Solar energy source statements:
1. The interior of the sun has a temperature of 1.5 × 10⁷ K.
2. The estimated temperature required for average energy particles to overcome the Coulomb barrier is 3 × 10⁹ K.
3. Fusion in the sun involves protons with energies much below the average energy.
4. The fuel is hydrogen in its core.

A (1), (2), (4)

B (1), (3), (4)

C (2), (3)

D (1), (2), (3), (4)

💡 Short Explanation

�� Solar core temperature is about 1.5 × 10⁷ K. �� Hydrogen is the primary fuel. �� Fusion occurs through quantum tunnelling.

✅ Correct Answer Explanation

Statement 1 is correct because the Sun's core temperature is approximately 1.5 × 10⁷ K. Statement 2 is correct as a classical estimate for overcoming the Coulomb barrier; however, it is not a statement describing the actual solar fusion process. Statement 3 is correct because solar fusion occurs at energies much lower than the classical barrier due to quantum tunnelling. Statement 4 is correct because hydrogen is the principal fuel in the solar core. Since the question concerns the solar energy source, the correct set is statements 1, 3 and 4. Therefore, the provided answer (A) is incorrect. The correct answer is B. (1), (3), (4).

❓ Why Other Options Are Incorrect

�� Option A → Includes statement 2 and omits statement 3.
�� Option C → Omits statements 1 and 4.
�� Option D → Includes statement 2, which does not describe the actual solar fusion mechanism.

🎯 Strategy

Used

Elimination

Application:

Separate classical Coulomb-barrier estimates from the actual solar-fusion process.

Final Logic:

Solar fusion occurs at 1.5 × 10⁷ K through tunnelling using hydrogen fuel.

🧠 Memory Trick

Sun = H + Tunnelling

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Thermonuclear Fusion in Stars

13 The Coulomb barrier height

A represents the strong nuclear force attraction

B is the repulsive barrier that two positive nuclei must overcome to fuse

C applies only to uncharged neutrons

D is approximately 1.02 MeV for two protons

💡 Short Explanation

�� Like charges repel. �� Fusion requires overcoming repulsion. �� This repulsion is the Coulomb barrier.

✅ Correct Answer Explanation

The Coulomb barrier arises from electrostatic repulsion between positively charged nuclei. For fusion to occur, nuclei must approach close enough for the strong nuclear force to dominate. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → The barrier is due to Coulomb repulsion, not strong-force attraction.
�� Option C → Neutrons are uncharged and do not face a Coulomb barrier.
�� Option D → The barrier for proton-proton interactions is not approximately 1.02 MeV.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Associate the barrier with electrostatic repulsion.

Final Logic:

Coulomb barrier = Repulsive electrostatic barrier.

🧠 Memory Trick

Coulomb = Charge Repulsion

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Coulomb Barrier

14 Calculate the kinetic energy K required to overcome a Coulomb barrier if T = 3 × 10⁹ K. (Use (3/2)kT = K, k = 1.38 × 10⁻²³ J/K, 1 eV = 1.6 × 10⁻¹⁹ J)

A ~ 400 keV

B ~ 26.7 MeV

C ~ 1.02 MeV

D ~ 5.49 MeV

💡 Short Explanation

�� Use K = (3/2)kT. �� Convert joules into eV. �� Express result in keV.

✅ Correct Answer Explanation

K = (3/2)(1.38 × 10⁻²³)(3 × 10⁹) = 6.21 × 10⁻¹⁴ J Converting to eV: K = (6.21 × 10⁻¹⁴)/(1.6 × 10⁻¹⁹) ≈ 3.88 × 10⁵ eV ≈ 388 keV ≈ 400 keV Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Much larger than calculated value.
�� Option C → Approximately three times larger than the result.
�� Option D → Far above the calculated energy.

🎯 Strategy

Used

Substitution

Application:

Directly substitute values into K = (3/2)kT.

Final Logic:

3 × 10⁹ K corresponds to about 400 keV.

🧠 Memory Trick

3 Billion K → 400 keV

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Thermonuclear Conditions

15 In the multi-step proton-proton cycle, the combination of e⁺ and e⁻ produces _____, while the combination of four hydrogen atoms ultimately releases _____ of energy.

A Neutrons, 200 MeV

B Gamma photons, 26.7 MeV

C Tritons, 12.86 MeV

D Alpha particles, 3.27 MeV

💡 Short Explanation

�� Positron-electron annihilation produces photons. �� Four protons ultimately form helium. �� Total energy released is about 26.7 MeV.

✅ Correct Answer Explanation

When a positron encounters an electron, annihilation occurs and gamma photons are emitted. The overall proton-proton chain converts four hydrogen nuclei into one helium nucleus, releasing approximately 26.7 MeV. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Electron-positron annihilation does not produce neutrons.
�� Option C → Tritons are not the annihilation products.
�� Option D → Alpha particles are formed later, not by e⁺e⁻ annihilation.

🎯 Strategy

Used

Memory-Based Recall

Application:

Recall the proton-proton chain and annihilation process.

Final Logic:

e⁺ + e⁻ → γ ; 4H → He + 26.7 MeV.

🧠 Memory Trick

4H → He + 26.7 MeV

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Proton-Proton Cycle

16 The depletion of hydrogen in the core and the subsequent start of helium burning respectively cause:

A Contraction of the core, Formation of carbon

B Expansion of the core, Formation of iron

C Cooling of the envelope, Fission of helium

D Stabilization of the star, Beta decay

💡 Short Explanation

�� Hydrogen depletion contracts the core. �� Core temperature rises. �� Helium fusion forms carbon.

✅ Correct Answer Explanation

When hydrogen is exhausted, the core contracts under gravity. The increasing temperature eventually initiates helium burning, producing carbon nuclei. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Iron is not formed directly from helium burning.
�� Option C → Helium undergoes fusion, not fission.
�� Option D → Beta decay is not the dominant process.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Follow the sequence of stellar evolution.

Final Logic:

Hydrogen exhaustion → Core contraction → Helium fusion → Carbon.

🧠 Memory Trick

H Ends → Core Shrinks → C Forms

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Stellar Evolution

17 Correct statements about plasma in controlled reactors:
1. Plasma is a mixture of positive ions and electrons.
2. It must be heated to temperatures in the range of 10⁸ K.
3. It must be confined using magnetic or alternative techniques because physical containers would melt.
4. It is used to generate steady power.

A (1), (2), (3), (4)

B (1), (3), (4)

C (2), (3)

D (1), (2), (4)

💡 Short Explanation

�� Plasma contains ions and electrons. �� Fusion requires extremely high temperatures. �� Magnetic confinement is necessary.

✅ Correct Answer Explanation

Statement 1 is correct because plasma consists of free ions and electrons. Statement 2 is correct because temperatures around 10⁸ K are required. Statement 3 is correct because no physical container can withstand such temperatures. Statement 4 is correct because controlled fusion aims to provide continuous power generation. Therefore, all four statements are correct.

❓ Why Other Options Are Incorrect

�� Option B → Omits statement 2.
�� Option C → Omits statements 1 and 4.
�� Option D → Omits statement 3.

🎯 Strategy

Used

Elimination

Application:

Verify each statement using fusion-reactor principles.

Final Logic:

All four statements accurately describe fusion plasma.

🧠 Memory Trick

Plasma = Ions + Electrons

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Controlled Thermonuclear Fusion

18 Match the energy process (List I) with its location/application (List II)

List I	List II
1. Uncontrolled nuclear fission	a. Stars (e.g., Sun)
2. Natural thermonuclear fusion	b. Nuclear reactors (current electricity)
3. Controlled nuclear fission	c. Atom bomb
4. Controlled thermonuclear fusion	d. Future unlimited power devices

A (1)-c, (2)-a, (3)-b, (4)-d

B (1)-a, (2)-c, (3)-d, (4)-b

C (1)-c, (2)-b, (3)-a, (4)-d

D (1)-b, (2)-d, (3)-c, (4)-a

💡 Short Explanation

�� Atom bombs use uncontrolled fission. �� Stars use natural fusion. �� Reactors use controlled fission.

✅ Correct Answer Explanation

1 → c : Atom bombs involve uncontrolled fission. 2 → a : Stars are powered by natural fusion. 3 → b : Nuclear power plants use controlled fission. 4 → d : Controlled fusion is expected to provide future large-scale power. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Applications are mismatched.
�� Option C → Fusion and fission locations are interchanged.
�� Option D → Multiple incorrect pairings.

🎯 Strategy

Used

Option Grouping

Application:

Associate each energy process with its real-world application.

Final Logic:

Bomb → Fission, Sun → Fusion, Reactor → Controlled Fission.

🧠 Memory Trick

Bomb-Sun-Reactor-Future

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Nuclear Energy Applications

19 Incorrect statement regarding energy scales

A Energy released in chemical reactions is of the order of electron volts (eV).

B Nuclear sources produce a million times less energy than chemical sources for the same quantity.

C Energy released in nuclear reactions is of the order of MeV.

D Both chemical and nuclear reactions involve mass-energy interconversions, though chemical mass defects are negligible.

💡 Short Explanation

�� Nuclear reactions release far more energy. �� Chemical energies are in eV. �� Nuclear energies are in MeV.

✅ Correct Answer Explanation

Nuclear reactions release approximately a million times more energy than chemical reactions for the same quantity of matter. Therefore, the statement claiming they produce a million times less energy is incorrect. Hence, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Correct energy scale for chemistry.
�� Option C → Correct energy scale for nuclear processes.
�� Option D → Correct description of mass-energy conversion.

🎯 Strategy

Used

Extreme Word Filter

Application:

Look for the statement reversing the known energy comparison.

Final Logic:

Nuclear energy ≫ Chemical energy.

🧠 Memory Trick

MeV > eV

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Nuclear vs Chemical Energy

20 When analyzing the output of 1 kg of uranium compared to 1 kg of coal,

A uranium relies on chemical binding energy to release 10¹⁴ J

B the energy generation from uranium fission is roughly 10⁷ times greater due to the magnitude of the nuclear mass defect

C coal provides 10¹⁴ J via exothermic combustion

D the mass defects involved in the coal reaction are identical to those in uranium fission

💡 Short Explanation

�� Uranium releases about 10¹⁴ J/kg. �� Coal releases about 10⁷ J/kg. �� Nuclear mass defects are much larger.

✅ Correct Answer Explanation

The energy from uranium fission is approximately: 10¹⁴ J / 10⁷ J = 10⁷ times greater than that obtained from burning coal. This difference arises from nuclear mass defects, which are vastly larger than chemical mass defects. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Uranium energy originates from nuclear binding energy, not chemical energy.
�� Option C → Coal releases about 10⁷ J, not 10¹⁴ J.
�� Option D → Nuclear and chemical mass defects are vastly different.

🎯 Strategy

Used

Substitution

Application:

Compare the given energy values directly.

Final Logic:

10¹⁴ ÷ 10⁷ = 10⁷.

🧠 Memory Trick

Uranium = 10⁷ × Coal

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Energy from Fission vs Chemical Reactions

CUET UG Physics Booster Test 2- Nuclear Reactions and Radioactivity

Test Complete!

Answer Review