Nuclei – Mass-Energy Relation: NEET MCQs with Answers

15 free questions on Nuclei – Mass-Energy Relation for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

Syllabus covered: Mass-Energy Relation, Binding Energy, Nuclear Fission and Fusion: Mass-energy relation, mass defect; binding energy per nucleon and its variation with mass number, nuclear fission, and fusion

Practise all 156 Nuclei – Mass-Energy Relation questions free Take a free timed test

Question 1

Previous-year question
When a uranium isotope ₉₂U²³⁵ is bombarded with a neutron, it generates ₃₆Kr⁸⁹, three neutrons and [NEET 2020]
  1. ₄₀Zr⁹¹
  2. ₃₆Kr¹⁰¹
  3. ₃₆Kr¹⁰³
  4. ₅₆Ba¹⁴⁴
Show answer
Answer: (D) ₅₆Ba¹⁴⁴
Explanation: Mass number: 235 + 1 = 89 + 3 + A ⇒ A = 144. Charge: 92 = 36 + Z ⇒ Z = 56. The missing fragment is therefore ₅₆Ba¹⁴⁴.

Question 2

Previous-year question
The binding energy per nucleon of ₃Li⁷ and ₂He⁴ nuclei are 5.60 MeV and 7.06 MeV respectively. In the nuclear reaction ₃Li⁷ + ₁H¹ → 2 ₂He⁴ + Q, the value of the energy Q released is [AIPMT 2014]
  1. 19.6 MeV
  2. –2.4 MeV
  3. 8.4 MeV
  4. 17.3 MeV
Show answer
Answer: (D) 17.3 MeV
Explanation: Q = (binding energy of products) – (binding energy of reactants) = 8 × 7.06 – 7 × 5.60 = 56.48 – 39.2 = 17.28 ≈ 17.3 MeV.

Question 3

Ten free protons and ten free neutrons can combine in different ways. The binding energies per nucleon are 7.1 MeV for ⁴He, 7.7 MeV for ¹²C and 8.0 MeV for ¹⁶O. Which product set releases the most energy, and how much?
  1. One ¹⁶O and one ⁴He nucleus, 156.4 MeV
  2. One ¹²C and two ⁴He nuclei, 149.2 MeV
  3. Five ⁴He nuclei, 142.0 MeV
  4. One ¹²C and two ⁴He nuclei, 156.4 MeV
Show answer
Answer: (A) One ¹⁶O and one ⁴He nucleus, 156.4 MeV
Explanation: The energy released in forming nuclei from free nucleons equals the total binding energy of the products. Five ⁴He give 20 × 7.1 = 142.0 MeV. One ¹²C and two ⁴He give 12 × 7.7 + 8 × 7.1 = 92.4 + 56.8 = 149.2 MeV. One ¹⁶O and one ⁴He give 16 × 8.0 + 4 × 7.1 = 128.0 + 28.4 = 156.4 MeV, the largest value.

Question 4

Consider α-decay, β⁻ decay, fission of a heavy nucleus and fusion of two light nuclei, each of which releases energy. An emitted electron is counted with mass number 0 and charge number −1. Which statement is true of all four processes?
  1. Total rest mass and total mass number stay the same, while total charge number rises
  2. Total rest mass falls, total mass number falls and total charge number stays the same
  3. Total rest mass falls, while total mass number and total charge number stay the same
  4. Total rest mass stays the same, while total mass number falls
Show answer
Answer: (C) Total rest mass falls, while total mass number and total charge number stay the same
Explanation: The energy released in each process comes from a loss of rest mass, so the products have less total rest mass than the starting system. The nucleons are only rearranged or, in β⁻ decay, a neutron becomes a proton, so the total mass number is unchanged. Charge is conserved in every process, so the total charge number is also unchanged.

Question 5

Suppose the nuclear force were long-ranged, so that every nucleon attracted all the other nucleons of the nucleus equally strongly instead of only its nearest neighbours. Ignoring Coulomb repulsion, how would the binding energy per nucleon of a large nucleus depend on the mass number A?
  1. It would fall roughly in proportion to 1/A.
  2. It would rise roughly in proportion to A.
  3. It would stay nearly constant, as it does for real nuclei of medium mass.
  4. It would rise roughly in proportion to A^(1/3).
Show answer
Answer: (B) It would rise roughly in proportion to A.
Explanation: If every nucleon were bound to all the others, the number of interacting pairs would be A(A − 1)/2. The total binding energy would then grow nearly as A², so the binding energy per nucleon would grow nearly as A. In real nuclei the short range of the force lets a nucleon influence only nucleons close to it. This saturation is why the binding energy per nucleon stays nearly constant over a wide range of A.

Question 6

The proton-proton cycle consists of these steps: (i) ¹H + ¹H → ²H + e⁺ + ν + 0.42 MeV; (ii) e⁺ + e⁻ → γ + γ + 1.02 MeV; (iii) ²H + ¹H → ³He + γ + 5.49 MeV; (iv) ³He + ³He → ⁴He + ¹H + ¹H + 12.86 MeV. For each ⁴He nucleus finally formed, the number of neutrinos emitted and the total energy released are nearly
  1. 2 neutrinos and 26.7 MeV
  2. 4 neutrinos and 40.6 MeV
  3. 2 neutrinos and 19.8 MeV
  4. 1 neutrino and 19.8 MeV
Show answer
Answer: (A) 2 neutrinos and 26.7 MeV
Explanation: Step (iv) needs two ³He nuclei, so steps (i), (ii) and (iii) must each occur twice. Step (i) gives one neutrino each time, making 2 neutrinos. The energy is 2(0.42 + 1.02 + 5.49) + 12.86 = 26.72 MeV.

Question 7

The Coulomb barrier between two protons is about 400 keV. Setting the average thermal energy (3/2)kT equal to this barrier gives T of about 3 × 10⁹ K. The interior of the Sun is at only about 1.5 × 10⁷ K, yet protons fuse there. What explains this?
  1. Protons in the Sun's core first capture electrons and become neutrons, which feel no Coulomb barrier.
  2. Gravity in the Sun's core cancels the Coulomb repulsion between protons, so the barrier there is practically zero.
  3. Fusion in the Sun involves only those protons whose energies are much above the average energy at that temperature.
  4. The 400 keV for every collision is supplied by gamma rays arriving from the Sun's cooler outer layers.
Show answer
Answer: (C) Fusion in the Sun involves only those protons whose energies are much above the average energy at that temperature.
Explanation: The estimate of 3 × 10⁹ K is the temperature at which a proton of average energy could cross the barrier. Protons in a gas have a range of energies, and the average at 1.5 × 10⁷ K is only about 2 keV. The fusion in the Sun is therefore carried by the small fraction of protons with energies far above this average.
Doing well? Create a free account to take timed tests on this topic and see your weak areas.

Question 8

A controlled thermonuclear fusion device heats its nuclear fuel to about 10⁸ K to obtain steady power. In what state is the fuel at this temperature, and what is the main challenge?
  1. It is a dense liquid of neutral molecules, and the challenge is to keep it from freezing on the walls of the vessel.
  2. It is a plasma of positive ions and electrons, and the challenge is to absorb the extra neutrons with control rods to stop a runaway chain reaction.
  3. It is a plasma of positive ions and electrons, and the challenge is to confine it, since no container can stand such a temperature.
  4. It is a gas of neutral atoms, and the challenge is to slow down its fast neutrons with a moderator.
Show answer
Answer: (C) It is a plasma of positive ions and electrons, and the challenge is to confine it, since no container can stand such a temperature.
Explanation: At temperatures of about 10⁸ K the atoms are fully ionised, so the fuel is a plasma of positive ions and electrons. No material container can withstand this temperature, so confining the plasma is the central problem. Moderators and control rods belong to fission reactors, where a neutron chain reaction has to be managed.

Question 9

Fusion in the core of a star has converted most of the hydrogen there into helium, so hydrogen burning in the core dies down. What must happen next for helium nuclei in the core to fuse into carbon?
  1. Helium nuclei begin to fuse at once at the same core temperature, because the Coulomb barrier between them is lower than that between two protons.
  2. Neutrons released by the helium nuclei split them into smaller pieces, which recombine to form carbon without any change in core temperature.
  3. The core cools, the star collapses under its own gravity, and this raises the core temperature to about 10⁸ K, at which helium nuclei fuse.
  4. The star expands strongly at the end of hydrogen burning, and this expansion raises the core temperature enough to fuse helium.
Show answer
Answer: (C) The core cools, the star collapses under its own gravity, and this raises the core temperature to about 10⁸ K, at which helium nuclei fuse.
Explanation: When the core hydrogen is used up, the core starts to cool and can no longer hold the star up, so the star collapses under gravity. This collapse heats the core, and at about 10⁸ K helium nuclei fuse into carbon. A higher temperature is needed because helium nuclei carry more charge than protons, which makes the Coulomb barrier higher.

Question 10

When a uranium-235 nucleus undergoes neutron-induced fission, about 200 MeV is released. In what form does this energy first appear, and what is the usual nature of the two fragments?
  1. Almost entirely as gamma rays; the fragments are stable nuclei
  2. As rest mass of the extra neutrons; the fragments are stable nuclei
  3. As kinetic energy of the fragments and neutrons; the fragments are radioactive and emit beta particles in succession
  4. As heat in the uranium; the fragments are alpha emitters that return to uranium
Show answer
Answer: (C) As kinetic energy of the fragments and neutrons; the fragments are radioactive and emit beta particles in succession
Explanation: The energy released in fission first appears as kinetic energy of the fragments and the emitted neutrons. It becomes heat only later, when these particles are stopped by the surrounding matter. The fragments are radioactive and emit beta particles one after another until stable end products are reached.

Question 11

Assertion: Thermonuclear fusion in the core of a star cannot build up nuclei much more massive than those near the peak of the binding energy curve. Reason: Beyond that peak the binding energy per nucleon falls as mass number rises, so fusion there would not release energy.
  1. The Assertion is true, but the Reason is false
  2. Both Assertion and Reason are true, and the Reason correctly explains the Assertion
  3. Both Assertion and Reason are true, but the Reason does not explain the Assertion
  4. The Assertion is false, but the Reason is true
Show answer
Answer: (B) Both Assertion and Reason are true, and the Reason correctly explains the Assertion
Explanation: Binding energy per nucleon peaks near mass number 56. Fusing lighter nuclei raises the binding energy per nucleon and releases energy, which keeps the star's core burning. Past the peak, joining nuclei would lower the binding energy per nucleon, so energy would have to be supplied instead. Both statements are true and the second explains the first.

Question 12

In the proton-proton cycle in the Sun, four hydrogen atoms (each of mass 1.008 u) are finally converted into one helium atom, with a total energy release of 26.7 MeV. Roughly what percentage of the mass of hydrogen consumed is converted into energy? (1 u = 931.5 MeV/c²)
  1. 0.7%
  2. 0.07%
  3. 27%
  4. 7%
Show answer
Answer: (A) 0.7%
Explanation: The rest energy of four hydrogen atoms is 4 × 1.008 × 931.5 MeV ≈ 3756 MeV. The fraction converted is 26.7/3756 ≈ 0.0071. This is about 0.7% of the mass of hydrogen consumed.

Question 13

Suppose 1 mg of matter could be converted completely into electrical energy. For roughly how long could this energy run a 3 kW heater continuously? (Take c = 3 × 10⁸ m/s.)
  1. About 10 years
  2. About 3 days
  3. About 1 year
  4. About 1 month
Show answer
Answer: (C) About 1 year
Explanation: From E = mc², E = 10⁻⁶ kg × (3 × 10⁸ m/s)² = 9 × 10¹⁰ J. Dividing by the power gives t = 9 × 10¹⁰ / 3000 = 3 × 10⁷ s. A year is about 3.15 × 10⁷ s, so the heater would run for about one year.

Question 14

The nuclei ²H, ⁴He, ⁵⁶Fe and ²³⁸U are arranged by the average energy needed per nucleon to separate each nucleus completely into its nucleons. Which order runs from the least to the greatest?
  1. ²H, ⁴He, ⁵⁶Fe, ²³⁸U
  2. ²H, ⁴He, ²³⁸U, ⁵⁶Fe
  3. ²H, ²³⁸U, ⁴He, ⁵⁶Fe
  4. ²³⁸U, ⁵⁶Fe, ⁴He, ²H
Show answer
Answer: (B) ²H, ⁴He, ²³⁸U, ⁵⁶Fe
Explanation: The average energy per nucleon needed is the binding energy per nucleon. It is about 1.1 MeV for ²H, 7.1 MeV for ⁴He, 7.6 MeV for ²³⁸U and 8.8 MeV for ⁵⁶Fe. The curve peaks near mass number 56, and very heavy nuclei lie below the peak, so uranium is less tightly bound than iron.

Question 15

Three ⁴He nuclei combine to form one ¹²C nucleus. The binding energy per nucleon is 7.1 MeV for ⁴He and 7.7 MeV for ¹²C. How much energy is released in this process?
  1. 0.6 MeV
  2. 85.2 MeV
  3. 7.2 MeV
  4. 92.4 MeV
Show answer
Answer: (C) 7.2 MeV
Explanation: The total binding energy of the three ⁴He nuclei is 12 × 7.1 = 85.2 MeV. The binding energy of ¹²C is 12 × 7.7 = 92.4 MeV. The energy released is the increase in binding energy, 92.4 − 85.2 = 7.2 MeV.

Want more Nuclei – Mass-Energy Relation questions?

E-GYAN NIDHI has 156 questions on this topic and 63,000+ in total, all free: timed CBT tests, NEET marking and instant analysis.

Register free Free Sunday Mock Test

More topics in ATOMS AND NUCLEI

↑