Q.A given coin has a mass of $3.0\ \text{g}$. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of $^{63}_{29}\text{Cu}$ atoms (of mass $62.92960\ \text{u}$).

Showing the 12 most recent of 16 on this concept.

1. COMEDK 2026Set 2026-A1 markMCQ
   Q.The atomic mass of an element $10 X^{20}$ is 19.98170 amu. The binding energy per nucleon of that element is: [given mass of neutron = 1.00867amu and mass of proton = 1.00783 amu and 1amu = 931 MeV ] (A) $17.66 \mathrm{MeV} /$ nucleon (B) $8.533 MeV /$ nucleon (C) $85.33 \mathrm{MeV} /$ nucleon (D) $170.66 \mathrm{MeV} /$ nucleon

   ›Reveal solutionSolution

   The binding energy per nucleon is found by computing the mass defect (difference between the sum of individual nucleon masses and the actual nuclear mass), converting it to energy using $E = \Delta m \cdot 931\ \text{MeV/amu}$, then dividing by the number of nucleons. The result is approximately $8.533\ \text{MeV/nucleon}$, which corresponds to option (B).

   Concept & Intuition:

   The nucleus is made of protons and neutrons. If you add up the masses of all these individual nucleons, you get a number larger than the actual mass of the nucleus. That missing mass — the mass defect — is converted into the energy that holds the nucleus together (binding energy). To compare how tightly bound different nuclei are, we divide by the number of nucleons to get the binding energy per nucleon. Here, the element is $_{10}X^{20}$, meaning 10 protons and 10 neutrons (since mass number = 20).

   Step-by-step solution:

   1. Identify the composition

      The notation $_{10}X^{20}$ tells us:

      • Atomic number $Z = 10$ → 10 protons
      • Mass number $A = 20$ → 20 nucleons total
      • Number of neutrons $N = A - Z = 20 - 10 = 10$

   2. Calculate the total mass of the individual nucleons

      Mass of 10 protons: $10 \times 1.00783\ \text{amu} = 10.0783\ \text{amu}$

      Mass of 10 neutrons: $10 \times 1.00867\ \text{amu} = 10.0867\ \text{amu}$

      Total nucleon mass = $10.0783 + 10.0867 = 20.1650\ \text{amu}$

   3. Find the mass defect

      Actual nuclear mass = $19.98170\ \text{amu}$

      Mass defect $\Delta m =$ (sum of nucleon masses) – (actual nuclear mass)

   $\Delta m = 20.1650 - 19.98170 = 0.18330\ \text{amu}$

   4. Convert mass defect to binding energy Using $1\ \text{amu} = 931\ \text{MeV}$:

   $\text{Binding energy} = 0.18330 \times 931\ \text{MeV}$

   Let’s compute:

   $0.18330 \times 900 = 164.97$

   $0.18330 \times 31 = 5.6823$

   Sum = $164.97 + 5.6823 = 170.6523\ \text{MeV}$

   (More precisely: $0.18330 \times 931 = 170.6523\ \text{MeV}$)

   5. Compute binding energy per nucleon …

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2. COMEDK 2026Set 2026-M1 markMCQ
   Q.A nucleus of uranium -235 absorbs a slow neutron and undergoes nuclear fission according to the reaction: ${ }_{92}^{235} U+{ }_0^1 n \rightarrow{ }_{56}^{141} B a+{ }_{36}^{92} K r+3{ }_0^1 n+Q$ If the average energy released per fission is 202 MeV , the energy released when 2.35 g of $U^{235}$ undergoes complete fission is approximately; [Given $1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}$, Avogadro number $=6.02 \times 10^{23}$ ] (A) $1.945 \times 10^{10} J$ (B) $19.45 \times 10^{11} J$ (C) $1.945 \times 10^{11} J$ (D) $19.45 \times 10^{10} \mathrm{~J}$

   ›Reveal solutionSolution

   [!TLDR]

   Count the nuclei in $2.35$ g of U-235, multiply by $202$ MeV and convert to joules to get $\approx1.945\times10^{11}$ J.

   Concept

   Nuclear fission energetics (CBSE Class-12 Nuclei): total energy released = (number of nuclei that fission) × (energy per fission). The number of nuclei comes from moles × Avogadro’s number.

   Solution

   Number of moles of U-235:

   $n=\frac{2.35}{235}=0.01\ \text{mol}.$

   Number of nuclei:

   $N=n\,N_A=0.01\times6.02\times10^{23}=6.02\times10^{21}.$

   Total energy in MeV:

   $E=N\times202=6.02\times10^{21}\times202=1.216\times10^{24}\ \text{MeV}.$

   Convert to joules using $1\ \text{MeV}=10^{6}\times1.6\times10^{-19}=1.6\times10^{-13}\ \text{J}$: …

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3. COMEDK 2026Set 2026-M1 markMCQ
   Q.Two deuterons are fused to form one alpha particle. If binding energy per nucleon of deuterium is 1.05 MeV and that of alpha particle is 7 MeV , what is the energy released in the formation of one alpha particle from the fusing of two deuterons? (A) 24.8 MeV (B) 23.8 MeV (C) 26.8 MeV (D) 28.3 MeV

   ›Reveal solutionSolution

   The energy released is the difference between the total binding energy of the products and that of the reactants. For two deuterons fusing into one alpha particle, the released energy is 23.8 MeV, corresponding to option (B).

   The key idea here is that binding energy is the energy required to break a nucleus into its individual nucleons. When lighter nuclei fuse into a heavier one, the difference in binding energy before and after the reaction is released as kinetic energy (or gamma rays). This is because the final nucleus is more tightly bound — its nucleons are in a lower energy state.

   We are given binding energy per nucleon, so we must multiply by the number of nucleons to get the total binding energy of each nucleus.

   ---

   1. Find the total binding energy of one deuteron.

      Deuterium (²H) has 2 nucleons. Binding energy per nucleon = 1.05 MeV.

      Total binding energy of one deuteron = $2 \times 1.05 = 2.10$ MeV.

   2. Find the total binding energy of two deuterons (the reactants).

      Since we start with two separate deuterons:

      Total binding energy of reactants = $2 \times 2.10 = 4.20$ MeV.

   3. Find the total binding energy of one alpha particle (the product).

      An alpha particle (⁴He) has 4 nucleons. Binding energy per nucleon = 7 MeV.

      Total binding energy of alpha particle = $4 \times 7 = 28.0$ MeV.

   4. Calculate the energy released.

      The energy released in fusion is the increase in binding energy:

   $\text{Energy released} = (\text{Total BE of products}) - (\text{Total BE of reactants})$

   $$$$

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4. COMEDK 2025Set 2025-A1 markMCQ
   Q.To get 300 MW electric power for half an hour, how much mass is to be completely converted into energy? (A) $6 \times 10^{-2} \mathrm{~kg}$ (B) $3 \times 10^{-6} \mathrm{~kg}$ (C) $6 \times 10^{-3} \mathrm{~kg}$ (D) $6 \times 10^{-6} \mathrm{~kg}$

   ›Reveal solutionSolution

   Energy delivered $=P t=5.4\times10^{11}\ \text{J}$; mass $=E/c^2=6\times10^{-6}\ \text{kg}$ — option (D).

   Energy required:

   $E=P\,t=(300\times10^{6}\ \text{W})\times(1800\ \text{s})=5.4\times10^{11}\ \text{J},$

   where half an hour $=1800\ \text{s}$.

   Mass–energy equivalence ($E=mc^2$, $c=3\times10^{8}\ \text{m/s}$): …

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5. COMEDK 2025Set 2025-A1 markMCQ
   Q.Fusion reaction is more energetic than fission reaction because (A) Uncontrolled chain reaction is taking place In the fusion reaction. (B) Fusion reaction is taking place at very high temperature (C) The energy released per unit mass of the fuel in fusion reaction is larger than the energy released per unit mass of the fuel in fission reaction. (D) In the fusion reaction lighter nuclei combine to form a heavier nucleus

   ›Reveal solutionSolution

   The key idea is that fusion releases more energy per unit mass of fuel than fission, making option (C) correct. The other options describe conditions or processes, not the fundamental reason for greater energy output.

   The question asks why a fusion reaction is more energetic than a fission reaction. The answer lies in the physics of nuclear binding energy and the mass defect, not in the temperature or the type of chain reaction.

   Concept and Intuition:

   The energy released in any nuclear reaction comes from the conversion of a tiny amount of mass into energy, as described by Einstein’s famous equation $E = mc^2$. The key measure is the binding energy per nucleon — the energy needed to hold a nucleus together. For light elements (like hydrogen isotopes), fusing them into a heavier nucleus (like helium) increases the binding energy per nucleon dramatically. For very heavy elements (like uranium), splitting them into medium-mass nuclei also increases binding energy per nucleon, but the gain per nucleon is smaller. This means fusion releases more energy per kilogram of fuel than fission does.

   Now, let’s evaluate each option step by step.

   1. Option (A): "Uncontrolled chain reaction is taking place in the fusion reaction."

      This is false. Fusion reactions (like in the sun or hydrogen bombs) are not chain reactions in the same sense as fission. A chain reaction involves neutrons causing subsequent fissions. Fusion requires extremely high temperatures and pressures to overcome electrostatic repulsion; it does not sustain itself via a chain mechanism. Even if it did, that wouldn’t explain why it’s more energetic — it would only describe how it proceeds.

   2. Option (B): "Fusion reaction is taking place at very high temperature."

      This is true — fusion requires temperatures of millions of degrees to give nuclei enough kinetic energy to overcome repulsion. However, high temperature is a condition for fusion to occur, not the reason it releases more energy. The energy output per reaction is determined by nuclear forces, not by the temperature of the environment.

   3. Option (C): "The energy released per unit mass of the fuel in fusion reaction is larger than the energy released per unit mass of the fuel in fission reaction." …

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6. COMEDK 2025Set 2025-A1 markMCQ
   Q.The nucleus of oxygen atom contains 8 protons and 8 neutrons. What is the mass defect in amu? [Given Mass of proton $=1.00727 \mathrm{amu}$ Mass of neutron $=1.00866 \mathrm{amu}$ and the mass of oxygen nucleus $=15.99053 \mathrm{amu}$. ] (A) 0.12691 amu (B) 0.13692 amu (C) 0.13691 amu (D) 0.12961 amu

   ›Reveal solutionSolution

   The mass defect is the difference between the sum of the individual masses of the protons and neutrons and the actual mass of the nucleus. For oxygen-16, this comes out to 0.13691 amu, which corresponds to option (C).

   ---

   Concept and Intuition

   The mass defect is a direct consequence of Einstein’s famous equation $E = mc^2$. When protons and neutrons bind together to form a nucleus, some of their mass is converted into binding energy — the energy that holds the nucleus together. So the nucleus always weighs less than the sum of its individual parts. That missing mass is the mass defect. To find it, we simply subtract the actual nuclear mass from the total mass of the separate nucleons.

   ---

   Step-by-Step Solution

   1. Identify the composition of the oxygen nucleus

      Oxygen has 8 protons and 8 neutrons (since its atomic number is 8 and mass number is 16). So we have:

      • Number of protons = 8
      • Number of neutrons = 8

   2. Calculate the total mass of the separate nucleons

      Mass of one proton = $1.00727 \, \text{amu}$

      Mass of one neutron = $1.00866 \, \text{amu}$

      Total mass of protons = $8 \times 1.00727 = 8.05816 \, \text{amu}$

      Total mass of neutrons = $8 \times 1.00866 = 8.06928 \, \text{amu}$

      Sum = $8.05816 + 8.06928 = 16.12744 \, \text{amu}$

   3. Subtract the actual nuclear mass

      Given mass of oxygen nucleus = $15.99053 \, \text{amu}$ …

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7. COMEDK 2025Set 2025-E1 markMCQ
   Q.Radium having mass number 200 and binding energy per nucleon 5.6 MeV , splits into two fragments Cadmium of mass number 112 and Hassium of mass number 108. If the binding energy per nucleon for Cadmium and Hassium is approximately 8.0 MeV , then the energy $Q$ released per fission will be: (A) 598 MeV (B) 176 MeV (C) 640 MeV (D) 475 MeV

   ›Reveal solutionSolution

   Energy released equals the gain in total binding energy: $Q=B_{\text{products}}-B_{\text{parent}}=(112+108)\times 8.0-200\times 5.6=1760-1120=640\ \text{MeV}$ — option (C).

   Concept

   In fission, energy is released because the fragments are more tightly bound than the parent. The energy released equals the increase in the total binding energy of the system:

   $Q=B_{\text{fragments}}-B_{\text{parent}}.$

   Step-by-step solution

   1. Binding energy of the parent (Radium, $A=200$). $B_{\text{parent}}=200\times 5.6=1120\ \text{MeV}.$ …

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8. COMEDK 2025Set 2025-M1 markMCQ
   Q.If the binding energy per nucleon in $3 \mathrm{Li}^7$ and ${ }_2 \mathrm{He}^4$ nuclei are respectively 5.60 MeV and 7.06 MeV , then energy of $p$ in the reaction $p+{ }_3 \mathrm{Li}^7 \rightarrow 2{ }_2 \mathrm{He}^4$ is (A) 12.28 MeV (B) 13.28 MeV (C) 28.28 MeV (D) 17.28 MeV

   ›Reveal solutionSolution

   The key idea is to use the difference in binding energies to find the energy released in the reaction, then apply conservation of energy to find the proton’s kinetic energy. The correct answer is 17.28 MeV.

   Concept and Intuition

   Binding energy is the energy needed to break a nucleus into its individual protons and neutrons. When a reaction rearranges nucleons into more tightly bound nuclei (higher binding energy per nucleon), the excess binding energy is released as kinetic energy. Here, a proton plus lithium-7 yields two alpha particles (helium-4). Since alpha particles have a higher binding energy per nucleon, the reaction is exothermic. The energy released (the Q-value) equals the difference in total binding energy before and after. That released energy plus the proton’s initial kinetic energy must equal the total kinetic energy of the two alpha particles. But the problem asks for the proton’s energy, assuming the alpha particles are produced at rest? Actually, the question likely means: what must the proton’s kinetic energy be so that the reaction can occur? In many such problems, the proton’s energy is the Q-value itself if the products are at rest, but here we need to check.

   Let’s work it out step by step.

   1. Write the reaction and identify the nuclei

      Reaction: $p + {}_3^7\mathrm{Li} \rightarrow 2\; {}_2^4\mathrm{He}$

      The proton is ${}_1^1\mathrm{H}$. Lithium-7 has 3 protons and 4 neutrons. Helium-4 has 2 protons and 2 neutrons.

   2. Find total binding energy before the reaction

      • For ${}_3^7\mathrm{Li}$: binding energy per nucleon = 5.60 MeV, so total binding energy = $7 \times 5.60 = 39.20$ MeV.
      • For a free proton, binding energy is 0 (it’s a single nucleon). Total binding energy before = $39.20 + 0 = 39.20$ MeV.

   3. Find total binding energy after the reaction

      • For each ${}_2^4\mathrm{He}$: binding energy per nucleon = 7.06 MeV, so total per alpha = $4 \times 7.06 = 28.24$ MeV.
      • Two alphas: total = $2 \times 28.24 = 56.48$ MeV.

   4. Calculate the energy released (Q-value)

      The increase in binding energy is the energy released:

      $Q = 56.48 - 39.20 = 17.28$ MeV. …

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9. COMEDK 2025Set 2025-M1 markMCQ
   Q.In a nuclear fusion reaction, two nuclei, $A$ and $B$ fuse to produce a nucleus $C$, releasing an amount of energy $\Delta \mathrm{E}$ in the process. If the mass defects of the three nuclei are $\Delta M_A, \Delta M_B$ and $\Delta M_C$ respectively, then which of the following relations is true? ( $c$ is the speed of light). (A) $\Delta M_A+\Delta M_B=\Delta M_C+\frac{\Delta E}{c^2}$ (B) $\Delta M_A-\Delta M_B=\Delta M_C+\frac{\Delta E}{c^2}$ (C) $\Delta M_A-\Delta M_B=\Delta M_C-\frac{\Delta E}{c^2}$ (D) $\Delta M_A+\Delta M_B=\Delta M_C-\frac{\Delta E}{c^2}$

   ›Reveal solutionSolution

   Writing each nuclear mass as $M = (\text{nucleon masses}) - \Delta M$ and applying $\Delta E=(M_A+M_B-M_C)c^2$, the nucleon terms cancel to give $\Delta M_A+\Delta M_B = \Delta M_C - \dfrac{\Delta E}{c^2}$ — option (D).

   Concept. The mass defect of a nucleus is $\Delta M = (\text{sum of free nucleon masses}) - (\text{actual nuclear mass})$, i.e. the mass equivalent of its binding energy. So the actual mass is $M = \sum(\text{nucleons}) - \Delta M$.

   Step 1 — Nucleon conservation.

   In $A+B\to C$ the total number of protons and neutrons is conserved, so the summed free-nucleon mass of $A$ and $B$ equals that of $C$. Call it $\Sigma$.

   Step 2 — Express the actual masses.

   $M_A + M_B = \Sigma - (\Delta M_A + \Delta M_B), \qquad M_C = \Sigma - \Delta M_C.$

   Step 3 — Energy released.

   $\Delta E = (M_A + M_B - M_C)c^2.$

   Substituting and cancelling $\Sigma$: …

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10. COMEDK 2024Set 2024-A1 markMCQ
    Q.A nucleus with mass number 190 initially at rest emits an alpha particle. If the $\mathrm{Q}$ value of the reaction is $4.5 \mathrm{~MeV}$, the kinetic energy of the alpha particle is (A) 4 MeV (B) 3.2 MeV (C) 0.43 MeV (D) 4.4 MeV

    ›Reveal solutionSolution

    In a nuclear decay where the parent nucleus is initially at rest, the Q‑value is shared between the alpha particle and the recoil daughter nucleus in inverse proportion to their masses. For mass numbers 190 (parent) and 4 (alpha), the alpha gets about 97% of the Q‑value, so its kinetic energy is roughly 4.4 MeV, corresponding to option (D).

    ---

    Concept & Intuition

    When a stationary nucleus emits an alpha particle, momentum must be conserved — the daughter nucleus recoils in the opposite direction with equal and opposite momentum. The Q‑value (the energy released) becomes the total kinetic energy of the two products. Because kinetic energy depends on both mass and velocity, the lighter alpha particle carries away most of the energy. The exact split follows from combining conservation of momentum and energy.

    ---

    Step‑by‑step solution

    1. Set up the reaction Let the parent nucleus have mass number $A = 190$. It emits an alpha particle (mass number 4) and becomes a daughter nucleus of mass number $A_d = 190 - 4 = 186$. The Q‑value is the total kinetic energy released:

    $Q = K_\alpha + K_d = 4.5\ \text{MeV}.$

    2. Apply conservation of momentum Initially, the parent is at rest, so total momentum is zero. After decay:

    $m_\alpha v_\alpha = m_d v_d \quad \Rightarrow \quad v_d = \frac{m_\alpha}{m_d} v_\alpha.$

    (Masses are proportional to mass numbers, so we can use $m_\alpha = 4u$, $m_d = 186u$, where $u$ is the atomic mass unit.)

    3. Express kinetic energies in terms of $v_\alpha$

    $K_\alpha = \frac12 m_\alpha v_\alpha^2, \qquad K_d = \frac12 m_d v_d^2 = \frac12 m_d \left(\frac{m_\alpha}{m_d} v_\alpha\right)^2 = \frac12 \frac{m_\alpha^2}{m_d} v_\alpha^2.$

    4. Find the ratio of kinetic energies

    $\frac{K_d}{K_\alpha} = \frac{\frac12 \frac{m_\alpha^2}{m_d} v_\alpha^2}{\frac12 m_\alpha v_\alpha^2} = \frac{m_\alpha}{m_d} = \frac{4}{186} = \frac{2}{93}.$

    So $K_d = \frac{2}{93} K_\alpha$.

    5. Use the Q‑value equation $$$$

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11. COMEDK 2024Set 2024-E1 markMCQ
    Q.The binding energy per nucleon for $\mathrm{C}^{12}$ is $7.68 \mathrm{~MeV}$ and that for $\mathrm{C}^{13}$ is $7.47 \mathrm{~MeV}$. The energy required to remove a neutron from $\mathrm{C}^{13}$ is (A) $7.92 \times 10^{-13} \mathrm{~MeV}$ (B) $4.95 \times 10^{-13} \mathrm{eV}$ (C) $7.92 \times 10^{-13} \mathrm{~J}$ (D) $7.92 \times 10^{-19} \mathrm{~J}$

    ›Reveal solutionSolution

    The energy to remove a neutron from $\mathrm{C}^{13}$ is the difference between its total binding energy and that of $\mathrm{C}^{12}$, giving $4.95 \, \mathrm{MeV}$, which converts to $7.92 \times 10^{-13} \, \mathrm{J}$. The correct option is (C).

    Concept & Intuition

    The binding energy per nucleon tells us how tightly each nucleon is held on average. To remove a neutron from $\mathrm{C}^{13}$, we must supply enough energy to overcome the binding of that neutron. This is the neutron separation energy. A neat way to find it: compare the total binding energy of $\mathrm{C}^{13}$ with that of $\mathrm{C}^{12}$. The difference is exactly the energy needed to pluck one neutron away, because the leftover nucleus is $\mathrm{C}^{12}$. No need to look up masses — the given data is enough.

    Step-by-step reasoning

    1. Find total binding energies

       For $\mathrm{C}^{12}$:

       Binding energy per nucleon = $7.68 \, \mathrm{MeV}$, and it has 12 nucleons.

       Total binding energy $B_{12} = 12 \times 7.68 = 92.16 \, \mathrm{MeV}$.

       For $\mathrm{C}^{13}$:

       Binding energy per nucleon = $7.47 \, \mathrm{MeV}$, and it has 13 nucleons.

       Total binding energy $B_{13} = 13 \times 7.47 = 97.11 \, \mathrm{MeV}$.

    2. Interpret the removal process

       Removing a neutron from $\mathrm{C}^{13}$ leaves $\mathrm{C}^{12}$. The energy required is the difference in total binding energy:

    $S_n = B_{13} - B_{12} = 97.11 - 92.16 = 4.95 \, \mathrm{MeV}.$

    This makes sense: the extra neutron in $\mathrm{C}^{13}$ is less bound (lower per-nucleon average) than the average in $\mathrm{C}^{12}$, so the separation energy is less than the average binding energy.

    3. Convert to joules The answer choices are in joules or weird multiples, so convert: $1 \, \mathrm{MeV} = 1.602 \times 10^{-13} \, \mathrm{J}$. $$$$

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12. COMEDK 2024Set 2024-M1 markMCQ
    Q.Find the binding energy of the tritium nucleus: [Given: mass of $1 \mathrm{H}^3=3.01605 \mathrm{~u} ; \mathrm{~m}_{\mathrm{p}}=1.00782 \mathrm{~u} ; \mathrm{~m}_{\mathrm{n}}=1.00866 \mathrm{~u}$.] (A) 8.5 MeV (B) 8.5 J (C) 0.00909 MeV (D) 0.00909 eV

    ›Reveal solutionSolution

    The binding energy is the energy equivalent of the mass defect — the difference between the sum of the masses of the constituent nucleons and the actual nuclear mass. For tritium, this comes out to about 8.5 MeV, so the correct option is (A).

    Concept & Intuition

    The nucleus of tritium ($^3_1\text{H}$) contains 1 proton and 2 neutrons. If you add up the masses of these three separate nucleons, you get a number larger than the measured mass of the tritium nucleus. That missing mass — the mass defect — has been converted into the energy that holds the nucleus together. By Einstein’s $E = mc^2$, we convert that mass difference into energy units (MeV). The result is the binding energy.

    Step-by-step solution

    1. Identify the nucleon composition

       Tritium has atomic number $Z = 1$ (one proton) and mass number $A = 3$.

       Number of neutrons = $A - Z = 3 - 1 = 2$.

       So the nucleus consists of 1 proton and 2 neutrons.

    2. Calculate the total mass of the separate nucleons

       Given:

       $m_p = 1.00782 \, \text{u}$

       $m_n = 1.00866 \, \text{u}$

       Total mass = $1 \times m_p + 2 \times m_n$

    $= 1.00782 + 2 \times 1.00866 = 1.00782 + 2.01732 = 3.02514 \, \text{u}$

    3. Find the mass defect Actual mass of tritium nucleus = $3.01605 \, \text{u}$ (given). Mass defect $\Delta m$ = (sum of nucleon masses) – (actual nuclear mass)

    $\Delta m = 3.02514 - 3.01605 = 0.00909 \, \text{u}$

    4. Convert mass defect to energy The standard conversion: $1 \, \text{u} = 931.5 \, \text{MeV}/c^2$. Binding energy $E_b = \Delta m \times 931.5 \, \text{MeV}$

    $E_b = 0.00909 \times 931.5 \approx 8.47 \, \text{MeV}$

    Rounding gives 8.5 MeV. …

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