Q.The atomic mass of $_8O^{16}$ is $16.0000$ amu. Calculate its binding energy per nucleon. Mass of electron $0.00055$ amu, mass of proton $1.007593$ amu and mass of neutron $1.008982$ amu and $1$ amu $= 931$ MeV.

Showing the 12 most recent of 15 on this concept.

1. CBSE 2026Set 55/1/11 markMCQ
   Q.For questions 13 to 16, two statements are given – one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) below: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Both Assertion (A) and Reason (R) are false. Assertion (A) : The mass of a nucleus is less than the sum of the masses of the constituent nucleons. Reason (R) : Energy is absorbed when the nucleons are bound together to form a nucleus.

   ›Reveal solutionSolution

   The key idea is mass defect: the mass of a stable nucleus is always less than the sum of its individual nucleon masses because the binding energy released during formation reduces the total mass. The Assertion is true, but the Reason is false — energy is released, not absorbed, when nucleons bind.

   This question tests your understanding of mass defect and binding energy — two of the most beautiful consequences of Einstein’s mass-energy equivalence, $E = mc^2$.

   When nucleons (protons and neutrons) come together to form a nucleus, they attract each other via the strong nuclear force. To pull them apart into isolated nucleons, you must supply energy — that energy is called the binding energy. Conversely, when they bind, that same amount of energy is released into the surroundings (usually as gamma rays).

   Here’s the crucial link: because energy is released, the system loses mass. The mass of the bound nucleus is less than the sum of the masses of its individual nucleons. This missing mass, multiplied by $c^2$, exactly equals the binding energy. That’s the mass defect.

   Now let’s examine each statement carefully.

   1. Assertion (A): “The mass of a nucleus is less than the sum of the masses of the constituent nucleons.”

      This is a well-established experimental fact for every stable nucleus. For example, a helium-4 nucleus has mass $4.0026\ \text{u}$, but two protons and two neutrons add up to $4.0330\ \text{u}$. The difference ($0.0304\ \text{u}$) is the mass defect. So Assertion (A) is true.

   2. Reason (R): “Energy is absorbed when the nucleons are bound together to form a nucleus.” …

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2. CBSE 2026Set ANNUAL1 mark
   Q.If the binding energy of oxygen nucleus 16/8 O is 128 MeV, then calculate the binding energy per nucleon.

   ›Reveal solutionSolution

   Binding energy per nucleon is simply the total binding energy divided by the number of nucleons (mass number A).

   For 16/8 O, mass number A = 16 (8 protons + 8 neutrons), and total binding energy = 128 MeV (given). …

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3. CBSE 2026Set ANNUAL1 markMCQ
   Q.Which of the following statements is true regarding the stability of a nucleus?

   (a) Binding energy alone determines nuclear stability.

   (b) Binding energy per nucleon is a better indicator of nuclear stability than total binding energy.

   (c) Neither binding energy nor binding energy per nucleon is related to nuclear stability.

   (d) Only the number of protons and neutrons determine nuclear stability.

   ›Reveal solutionSolution

   Total binding energy grows almost monotonically with the number of nucleons and so cannot distinguish stability between nuclei of different sizes; binding energy per nucleon — which rises, peaks near iron ($A\approx56$), and then falls for heavier nuclei — is the quantity that correctly tracks relative nuclear stability.

   Why total binding energy is a poor stability measure

   Binding energy (BE) is the energy required to completely separate a nucleus into its individual protons and neutrons (equivalently, the energy released when the nucleus is assembled from free nucleons):

   $BE = \left[Zm_p + (A-Z)m_n - M_{\text{nucleus}}\right]c^2$

   As the mass number $A$ (number of nucleons) increases, there are simply more nucleon–nucleon bonds contributing to the total, so total $BE$ tends to increase with $A$ across the periodic table — a very heavy nucleus like uranium has a much larger total binding energy than a light nucleus like helium, purely because it has far more nucleons, not necessarily because each individual nucleon is more tightly (stably) bound.

   Why binding energy per nucleon is the right measure

   Binding energy per nucleon, $BE/A$, measures the average energy binding each individual nucleon into the nucleus — this is what actually reflects how stable/tightly bound the nucleus is, independent of how many nucleons it happens to have.

   When $BE/A$ is plotted against $A$:

   • It rises steeply for light nuclei, …

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4. CBSE 2026Set ANNUAL1 markMCQ
   Q.The mass of a $^{7}_{3}\text{Li}$ nucleus is 0.042 u less than the sum of the masses of all its nucleons. The average binding energy per nucleon of $^{7}_{3}\text{Li}$ nucleus is nearly :

   (a) 23 MeV

   (b) 46 MeV

   (c) 5.6 MeV

   (d) 3.9 MeV

   ›Reveal solutionSolution

   Converting the given mass defect to energy and dividing by the 7 nucleons of $^7_3\text{Li}$ gives a binding energy per nucleon of about 5.6 MeV.

   Working

   Total binding energy: $BE = \Delta m \times 931.5\ \text{MeV/u}$ (using $E=mc^2$ with mass in atomic mass units).

   Given $\Delta m = 0.042$ u:

   $BE = 0.042 \times 931.5 \approx 39.12\ \text{MeV}$

   …

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5. CBSE 2025Set 55/5/11 markMCQ
   Q.Assertion (A): During the formation of a nucleus, the mass defect produced is the source of the binding energy of the nucleus. Reason (R): For all nuclei, the value of binding energy per nucleon increases with mass number. (A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true, but R is not the correct explanation of A. (C) A is true, but R is false. (D) Both A and R are false.

   ›Reveal solutionSolution

   The mass defect does provide the binding energy (A is true), but binding energy per nucleon does not increase monotonically with mass number—it peaks around iron-56 and then decreases (R is false). The answer is (C).

   Understanding Mass Defect and Binding Energy

   When protons and neutrons come together to form a nucleus, something remarkable happens: the mass of the resulting nucleus is less than the sum of the masses of its constituent nucleons. This "missing" mass hasn't vanished—it has been converted into energy that holds the nucleus together.

   Einstein's mass-energy equivalence $E = mc^2$ tells us that mass and energy are interchangeable. The mass defect $\Delta m$ is precisely the source of the binding energy:

   $BE = \Delta m \cdot c^2$

   This binding energy is what you would need to supply to completely disassemble the nucleus back into separate protons and neutrons. The assertion (A) captures this fundamental principle correctly.

   The Binding Energy Per Nucleon Curve

   Now let's examine the reason (R), which claims that binding energy per nucleon increases with mass number for all nuclei. This is where we need to look at experimental data.

   The binding energy per nucleon, $\frac{BE}{A}$ (where $A$ is the mass number), does not increase monotonically. Instead, it follows a characteristic curve:

   1. Light nuclei (hydrogen, helium): relatively low binding energy per nucleon, around 1–7 MeV.

   2. Medium nuclei (iron-56, nickel-62): the curve reaches its maximum at approximately 8.8 MeV per nucleon. Iron-56 sits near the peak of nuclear stability.

   3. Heavy nuclei (uranium, plutonium): the binding energy per nucleon decreases to around 7.5 MeV.

   Important

   The binding energy per nucleon curve peaks around $A \approx 56$ (iron) and then decreases for heavier elements. This is why both fusion (combining light nuclei) and fission (splitting heavy nuclei) release energy—both processes move toward the more stable middle region.

   Region	Mass Number Range	BE/nucleon	Trend
   Light	$A < 20$	1–7 MeV	Increasing
   Medium	$20 < A < 100$	7.5–8.8 MeV	Peak around Fe-56
   Heavy	$A > 100$	7.5–8 MeV	Decreasing

   The decrease in binding energy per nucleon for heavy nuclei occurs because: …

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6. CBSE 2025Set X11 markMCQ
   Q.Binding energy per nucleon of a nucleus is a measure of its

   (a) radius

   (b) mass

   (c) volume

   (d) stability

   ›Reveal solutionSolution

   (d) stability The binding energy per nucleon ($BE/A$) tells how tightly each nucleon is bound. A higher $BE/A$ means more energy is needed to remove a nucleon, so the nucleus is more stable. It is …

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7. CBSE 2024Set ANNUAL1 mark
   Q.Which nucleus has maximum average binding energy per nucleon?

   ›Reveal solutionSolution

   The binding-energy-per-nucleon curve peaks around mass number A ≈ 56, at iron — the most tightly bound (most stable) nucleus.

   The average binding energy per nucleon, BE/A, is plotted against mass number A to give the famous 'binding energy curve'. Its shape is:

   • It rises steeply for light nuclei (He, Li, Be, ...).
   • It reaches a broad maximum for nuclei with A roughly between 50 and 80, peaking at about 8.8 MeV per nucleon near A ≈ 56, which corresponds to iron (and its close neighbours like nickel).
   • It then slowly decreases for heavier nuclei, dropping to about 7.6 MeV/nucleon for uranium (A ≈ 238). …

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8. CBSE 2024Set ANNUAL1 mark
   Q.Which element in the periodic table shows maximum binding energy per nucleon ?

   ›Reveal solutionSolution

   Binding energy per nucleon peaks around mass number $A\approx 56$, at iron — this is why iron is exceptionally stable and is the endpoint of energy-releasing fusion in stars.

   The binding energy per nucleon ($BE/A$) of a nucleus measures how tightly its nucleons are bound — the higher this value, the more stable the nucleus (more energy would be needed to pull it apart, nucleon by nucleon).

   When $BE/A$ is plotted against mass number $A$ for all known nuclei, the curve rises steeply for light nuclei, reaches a broad maximum of about $8.7$–$8.8\ \text{MeV}$ per nucleon in the region $A\approx 50$–$60$, and then slowly decreases for heavier nuclei. The peak of this curve is at iron-56 ($_{26}\text{Fe}^{56}$, closely followed by nickel-62 which is marginally higher, but iron is the standard textbook answer), which is why: …

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9. CBSE 2024Set ANNUAL1 markMCQ
   Q.The generation of energy in the sun is mainly due to :

   (a) Fission of heavy nuclei

   (b) Fission of light nuclei

   (c) Fusion of heavy nuclei

   (d) Fusion of light nuclei

   ›Reveal solutionSolution

   The Sun's energy comes from nuclear fusion of light nuclei (hydrogen into helium).

   Inside the Sun's core, at extremely high temperature and pressure, light hydrogen nuclei (protons) fuse together in a chain of reactions (the proton-proton cycle) to form helium nuclei. The mass of the helium formed is slightly less than the mass of the hydrogen nuclei …

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10. CBSE 2023Set 55/3/11 markMCQ
    Q.Two statements are given – one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (a), (b),

    (c) and

    (d) below.

    (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

    (b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).

    (c) Assertion (A) is true, but Reason (R) is false.

    (d) Assertion (A) is false and Reason (R) is also false. Assertion (A) : The nucleus $^{7}_{3}X$ is more stable than the nucleus $^{4}_{3}Y$. Reason (R) : $^{7}_{3}X$ contains more number of protons.

    ›Reveal solutionSolution

    The assertion is true because $^{7}_{3}\text{X}$ has a higher binding energy per nucleon than $^{4}_{3}\text{Y}$ (both are isotopes of lithium, and $^{7}\text{Li}$ is more stable than $^{4}\text{Li}$). The reason is false — more protons do not guarantee greater stability; in fact, $^{4}_{3}\text{Y}$ has the same number of protons as $^{7}_{3}\text{X}$. So the correct option is (c).

    The key here is to understand what "more stable" means for a nucleus. Stability of a nucleus is measured by its binding energy per nucleon — the energy required to remove one nucleon from the nucleus. A higher binding energy per nucleon means the nucleus is more tightly bound and therefore more stable.

    Now look at the two nuclei: $^{7}_{3}\text{X}$ and $^{4}_{3}\text{Y}$. Both have the same atomic number $Z = 3$, so they are isotopes of the same element — lithium. The first has mass number $A = 7$, the second $A = 4$.

    The reason claims that $^{7}_{3}\text{X}$ is more stable because it "contains more number of protons." But both have exactly 3 protons — the number of protons is identical. So the reason is factually wrong. Even if it were about more nucleons, that alone doesn't guarantee stability; you need to look at the binding energy curve.

    Let’s walk through the reasoning step by step.

    1. Identify the nuclei.

       $^{7}_{3}\text{X}$ has 3 protons and 4 neutrons. $^{4}_{3}\text{Y}$ has 3 protons and 1 neutron. Both are lithium isotopes: $^{7}\text{Li}$ and $^{4}\text{Li}$.

    2. Check the stability of $^{4}\text{Li}$.

       $^{4}\text{Li}$ is extremely unstable — it has a half-life of about $10^{-22}$ seconds and decays almost instantly. It lies far from the valley of stability because the neutron-to-proton ratio is too low (1 neutron for 3 protons). In contrast, $^{7}\text{Li}$ is stable — it is one of the two stable isotopes of lithium (the other being $^{6}\text{Li}$).

    3. Binding energy per nucleon comparison. …

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11. CBSE 2023Set 55/4/11 markMCQ
    Q.The difference in mass of a $_7X$ nucleus and total mass of its constituent nucleons is $21.00\ \text{u}$. The binding energy per nucleon for this nucleus is equal to the energy equivalent of : (A) $3\ \text{u}$ (B) $3.5\ \text{u}$ (C) $7\ \text{u}$ (D) $21\ \text{u}$

    ›Reveal solutionSolution

    The mass defect of $21.00\ \text{u}$ for a nucleus with $7$ nucleons corresponds to a total binding energy. The binding energy per nucleon is found by dividing the mass defect by the number of nucleons, which gives $\boxed{3\ \text{u}}$ as its mass equivalent.

    Concept and Intuition

    At the heart of nuclear physics lies the concept that the mass of a nucleus is less than the sum of the masses of its individual constituent protons and neutrons (collectively called nucleons). This difference in mass is known as the mass defect ($\Delta m$).

    This "missing" mass is not actually lost; instead, it has been converted into energy according to Einstein's famous mass-energy equivalence principle, $E = mc^2$. This energy is called the binding energy ($E_{BE}$) of the nucleus. It represents the energy required to break the nucleus apart into its individual nucleons, or conversely, the energy released when the nucleons combine to form the nucleus.

    The binding energy per nucleon is a crucial quantity for understanding nuclear stability. It is simply the total binding energy divided by the total number of nucleons ($A$) in the nucleus. A higher binding energy per nucleon generally indicates a more stable nucleus.

    In this problem, we are given the mass defect directly. We need to find the binding energy per nucleon and express it as an "energy equivalent of X u". This means we are looking for a mass value $X$ such that $X \cdot c^2$ is equal to the binding energy per nucleon. Essentially, we need to calculate the mass defect per nucleon.

    Step-by-Step Solution

    1. Identify the given mass defect:

       The problem states that the difference in mass of the nucleus and the total mass of its constituent nucleons (which is the definition of mass defect) is $21.00\ \text{u}$.

       So, the mass defect $\Delta m = 21.00\ \text{u}$.

    2. Determine the number of nucleons ($A$):

       The nucleus is denoted as $_7X$. In standard nuclear notation, $_Z^AX$, $A$ represents the mass number (total number of nucleons) and $Z$ represents the atomic number (number of protons). If only one number is given as a subscript, it typically refers to $Z$. However, the question asks for "binding energy per nucleon", which requires knowing the total number of nucleons, $A$. Given the options, and the common practice in such problems, it is implied that the number $7$ refers to the mass number $A$. If $Z=7$ were intended, the mass number $A$ would be unknown, making the problem unsolvable. Therefore, we take the number of nucleons $A = 7$.

    3. Calculate the total binding energy ($E_{BE}$):

       The total binding energy is related to the mass defect by Einstein's mass-energy equivalence:

       $E_{BE} = \Delta m \cdot c^2$

       Substituting the given mass defect:

       $E_{BE} = 21.00\ \text{u} \cdot c^2$ …

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12. CBSE 2021Set TERM21 markMCQ
    Q.Assertion: If a heavy nucleus is split into two medium sized parts, each of the nuclei will have more binding energy per nucleon than the original nucleus. Reason: Joining two light nuclei together to give a single nucleus of medium size means more binding energy per nucleon than the two nuclei.

    (a) if both assertion and reason are true and reason is the correct explanation of the assertion

    (b) if both assertion and reason are true, but reason is not correct explanation of the assertion

    (c) if assertion is true, but reason is false

    (d) if both assertion and reason are false

    ›Reveal solutionSolution

    Both statements are true and both follow from the shape of the binding-energy-per-nucleon curve, but the reason describes fusion (light nuclei joining) while the assertion is about fission (a heavy nucleus splitting) — a different process, so it isn't a direct explanation of it.

    Assertion: True. The binding-energy-per-nucleon curve rises for light nuclei, peaks near mass number $A\approx56$ (iron region), then falls gradually for heavy nuclei. So when a heavy nucleus splits into two medium-mass fragments, each fragment (closer to the peak) has higher binding energy per nucleon than the original heavy nucleus — exactly what releases energy in nuclear fission.

    …

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