Exercises — Nuclei

1. 13 marksNCERT Exercises, Chapter 13

   Obtain the binding energy (in MeV) of a nitrogen nucleus ${}_7^{14}$N, given $m{(}_7^{14}\text{N})=14.00307$ u.

   Hint. Count the protons and neutrons, add their separate masses, and find how much mass is missing from the actual nucleus.

   Nitrogen-14 has $Z=7$ protons and $A-Z=14-7=7$ neutrons.

   The total mass of the constituents taken separately is:

   $7m_H+7m_n=7(1.007825)+7(1.008665)=7.054775+7.060655=14.115430\text{u}$

   The mass defect is the amount by which the assembled nucleus falls short of this:

   $\Delta m=14.115430-14.00307=0.11236\text{u}$

   Using $1\text{u}=931.5$ MeV/c${}^2$:

   $E_b=0.11236\times 931.5=104.7\text{MeV}$

   The binding energy per nucleon is $104.7/14=7.48$ MeV, which is a little below the 8.0 MeV plateau because nitrogen is a relatively light nucleus.

   Note that $m_H$ is the mass of the hydrogen atom, not the bare proton. Using atomic masses throughout is consistent, because the $Z$ electron masses included on the left cancel against those in the atomic mass on the right.

   ✦ Binding energy = 104.7 MeV

2. 24 marksNCERT Exercises, Chapter 13

   Obtain the binding energy of the nuclei ${}_{26}^{56}$Fe and ${}_{83}^{209}$Bi in units of MeV from the following data: $m{(}_{26}^{56}\text{Fe})=55.934939$ u, $m{(}_{83}^{209}\text{Bi})=208.980388$ u.

   Hint. Work out each mass defect separately, then divide by the mass number to compare stabilities.

   Iron-56 has $Z=26$ and $N=30$:

   $\Delta m=26(1.007825)+30(1.008665)-55.934939$

   $=26.203450+30.259950-55.934939=0.528461\text{u}$

   $E_b=0.528461\times 931.5=492.3\text{MeV},\frac{E_b}{A}=\frac{492.3}{56}=8.79\text{MeV}$

   Bismuth-209 has $Z=83$ and $N=126$:

   $\Delta m=83(1.007825)+126(1.008665)-208.980388$

   $=83.649475+127.091790-208.980388=1.760877\text{u}$

   $E_b=1.760877\times 931.5=1640.3\text{MeV},\frac{E_b}{A}=\frac{1640.3}{209}=7.85\text{MeV}$

   Bismuth has the larger total binding energy but iron is the more tightly bound nucleus, because binding energy per nucleon is what measures stability. Iron sits near the peak of the curve at about 8.8 MeV, which is why it is among the most stable nuclei and why both fission of heavy nuclei and fusion of light ones release energy by moving towards it.

   ✦ Fe-56: 492.3 MeV, or 8.79 MeV per nucleon. Bi-209: 1640.3 MeV, or 7.85 MeV per nucleon

3. 34 marksNCERT Exercises, Chapter 13

   A given coin has a mass of 3.0 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 ${}_{29}^{63}$Cu atoms (of mass 62.92960 u).

   Hint. Find the binding energy of one nucleus, then count how many nuclei the coin contains.

   Binding energy of one nucleus. Copper-63 has $Z=29$ and $N=34$:

   $\Delta m=29(1.007825)+34(1.008665)-62.92960$

   $=29.226925+34.294610-62.929600=0.591935\text{u}$

   $E_b=0.591935\times 931.5=551.4\text{MeV}$

   Number of nuclei in the coin. One mole of copper-63 has mass 62.92960 g, so:

   $N=\frac{3.0}{62.92960}\times 6.023\times 10^{23}=2.871\times 10^{22}\text{atoms}$

   Total energy required:

   $E=2.871\times 10^{22}\times 551.4=1.58\times 10^{25}\text{MeV}$

   Converting with $1\text{MeV}=1.6\times 10^{-13}$ J:

   $E=1.58\times 10^{25}\times 1.6\times 10^{-13}=2.53\times 10^{12}\text{J}$

   This is an enormous quantity, comparable to the energy released by several hundred tonnes of TNT, and it shows why nuclear binding energies dwarf chemical ones by a factor of about a million.

   ✦ About 1.58 x 10^25 MeV, that is 2.53 x 10^12 J

4. 42 marksNCERT Exercises, Chapter 13

   Obtain approximately the ratio of the nuclear radii of the gold isotope ${}_{79}^{197}$Au and the silver isotope ${}_{47}^{107}$Ag.

   Hint. Nuclear radius depends only on the mass number, through a cube-root relation.

   The nuclear radius depends only on the mass number, not on the atomic number:

   $R=R_0{A}^{1/3}$

   Taking the ratio, the constant $R_0$ cancels:

   $\frac{R_{\text{Au}}}{R_{\text{Ag}}}={\left(\frac{A_{\text{Au}}}{A_{\text{Ag}}}\right)}^{1/3}={\left(\frac{197}{107}\right)}^{1/3}=(1.8411{)}^{1/3}$

   $\frac{R_{\text{Au}}}{R_{\text{Ag}}}\approx 1.23$

   The gold nucleus is only about 23 per cent larger in radius despite having nearly twice the mass, because the cube root compresses the difference. This weak dependence is exactly what makes nuclear density constant, as Exercise 13.10 shows.

   ✦ The ratio is about 1.23

5. 54 marksNCERT Exercises, Chapter 13

   The Q value of a nuclear reaction $A+b\to C+d$ is defined by $Q=[m_A+m_b-m_C-m_d]c^2$, where the masses refer to the respective nuclei. Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic. (i) ${}_1^2$H + ${}_1^2$H $\to$ ${}_1^3$H + ${}_1^1$H (ii) ${}_6^{12}$C + ${}_6^{12}$C $\to$ ${}_{10}^{20}$Ne + ${}_2^4$He. Atomic masses: $m{(}_1^2\text{H})=2.014102$ u, $m{(}_1^3\text{H})=3.016049$ u, $m{(}_6^{12}\text{C})=12.000000$ u, $m{(}_{10}^{20}\text{Ne})=19.992439$ u.

   Hint. A positive Q means mass has been converted to energy; compare the total mass before and after.

   The sign of $Q$ decides the answer: positive $Q$ means energy is released (exothermic) and negative means energy must be supplied (endothermic).

   (i) ${}_1^2$H + ${}_1^2$H $\to$ ${}_1^3$H + ${}_1^1$H

   $Q=[2(2.014102)-3.016049-1.007825]\times 931.5$

   $=[4.028204-4.023874]\times 931.5=0.004330\times 931.5$

   $Q=+4.03\text{MeV}$

   Since $Q$ is positive, the reaction is exothermic.

   (ii) ${}_6^{12}$C + ${}_6^{12}$C $\to$ ${}_{10}^{20}$Ne + ${}_2^4$He

   $Q=[2(12.000000)-19.992439-4.002603]\times 931.5$

   $=[24.000000-23.995042]\times 931.5=0.004958\times 931.5$

   $Q=+4.62\text{MeV}$

   Again positive, so this reaction is also exothermic.

   In both cases the products are more tightly bound than the reactants, so a small amount of mass disappears and reappears as kinetic energy of the products.

   ✦ (i) Q = +4.03 MeV, exothermic (ii) Q = +4.62 MeV, exothermic

6. 63 marksNCERT Exercises, Chapter 13

   Suppose we think of fission of a ${}_{26}^{56}$Fe nucleus into two equal fragments, ${}_{13}^{28}$Al. Is the fission energetically possible? Argue by working out Q of the process. Given $m{(}_{26}^{56}\text{Fe})=55.93494$ u and $m{(}_{13}^{28}\text{Al})=27.98191$ u.

   Hint. Compute Q and read its sign; then explain the result using the binding energy per nucleon curve.

   Computing the Q value for the proposed split:

   $Q=[m{(}^{56}\text{Fe})-2m{(}^{28}\text{Al})]\times 931.5$

   $=[55.93494-2(27.98191)]\times 931.5=[55.93494-55.96382]\times 931.5$

   $=(-0.02888)\times 931.5=-26.9\text{MeV}$

   The Q value is negative, so the fission is not energetically possible. Energy would have to be supplied rather than released, and the process cannot occur spontaneously.

   Why this happens. Iron-56 sits at the peak of the binding energy per nucleon curve, at about 8.8 MeV per nucleon. Aluminium-28, being lighter, lies on the rising part of the curve at a lower value. Splitting iron therefore moves the nucleons to a less tightly bound configuration, which costs energy.

   This is exactly why fission releases energy only for heavy nuclei beyond the peak, such as uranium, and fusion only for light nuclei below it. Iron is the turning point at which neither process pays.

   ✦ Q = -26.9 MeV. The fission is not energetically possible, since iron-56 lies at the peak of the binding energy curve

7. 73 marksNCERT Exercises, Chapter 13

   The fission properties of ${}_{94}^{239}$Pu are very similar to those of ${}_{92}^{235}$U. The average energy released per fission is 180 MeV. How much energy, in MeV, is released if all the atoms in 1 kg of pure ${}_{94}^{239}$Pu undergo fission?

   Hint. Count the nuclei in one kilogram, then multiply by the energy released per fission.

   First find how many plutonium nuclei are present. One mole of ${}^{239}$Pu has a mass of 239 g, so 1 kg contains:

   $N=\frac{1000}{239}\times 6.023\times 10^{23}=4.184\times 6.023\times 10^{23}$

   $N=2.52\times 10^{24}\text{atoms}$

   Each fission releases 180 MeV, so the total energy is:

   $E=2.52\times 10^{24}\times 180=4.54\times 10^{26}\text{MeV}$

   Converting for scale, this is about $7.3\times 10^{13}$ J, roughly the energy of 17 kilotonnes of TNT. Burning a kilogram of coal releases around $3\times 10^7$ J, so nuclear fission of the same mass yields over two million times as much energy.

   ✦ E = 4.54 x 10^26 MeV

8. 84 marksNCERT Exercises, Chapter 13

   How long can an electric lamp of 100 W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as ${}_1^2$H + ${}_1^2$H $\to$ ${}_2^3$He + $n$ + 3.27 MeV.

   Hint. Note that each reaction consumes two deuterium nuclei, so the number of reactions is half the number of atoms.

   The number of deuterium atoms in 2.0 kg, taking the molar mass as 2.0 g, is:

   $N=\frac{2000}{2.0}\times 6.023\times 10^{23}=6.023\times 10^{26}\text{atoms}$

   Each reaction consumes two deuterium nuclei, so the number of fusion events is half of this:

   $\text{reactions}=\frac{6.023\times 10^{26}}{2}=3.01\times 10^{26}$

   Total energy released:

   $E=3.01\times 10^{26}\times 3.27=9.85\times 10^{26}\text{MeV}$

   $E=9.85\times 10^{26}\times 1.6\times 10^{-13}=1.58\times 10^{14}\text{J}$

   The lamp consumes 100 J every second, so:

   $t=\frac{E}{P}=\frac{1.58\times 10^{14}}{100}=1.58\times 10^{12}\text{s}$

   In years, using $1$ year $=3.154\times 10^7$ s:

   $t=\frac{1.58\times 10^{12}}{3.154\times 10^7}\approx 5.0\times 10^4\text{years}$

   Two kilograms of deuterium could keep the lamp burning for about fifty thousand years, which is why fusion is so attractive as an energy source.

   ✦ About 1.58 x 10^12 s, that is roughly 5.0 x 10^4 years

9. 93 marksNCERT Exercises, Chapter 13

   Calculate the height of the potential barrier for a head on collision of two deuterons. (Hint: The height of the potential barrier is given by the Coulomb repulsion between the two deuterons when they just touch each other. Assume that they can be taken as hard spheres of radius 2.0 fm.)

   Hint. The centre-to-centre separation when two spheres touch is twice the radius, not once.

   When two deuterons of radius $2.0$ fm just touch, their centres are separated by twice the radius:

   $r=2\times 2.0=4.0\text{fm}=4.0\times 10^{-15}\text{m}$

   Each deuteron carries a single positive charge $e$, so the Coulomb potential energy at this separation is:

   $U=\frac{1}{4\pi {\epsilon }_0}\frac{e^2}{r}=\frac{9\times 10^9\times (1.6\times 10^{-19}{)}^2}{4.0\times 10^{-15}}$

   $U=\frac{9\times 10^9\times 2.56\times 10^{-38}}{4.0\times 10^{-15}}=5.76\times 10^{-14}\text{J}$

   Converting to MeV:

   $U=\frac{5.76\times 10^{-14}}{1.6\times 10^{-13}}=0.36\text{MeV}=360\text{keV}$

   Since the barrier is shared between the two approaching deuterons, each needs about $0.18$ MeV of kinetic energy. This corresponds to a temperature of roughly $10^9$ K, which is why fusion requires the extreme conditions found in stellar cores.

   ✦ Height of the potential barrier = 0.36 MeV, about 360 keV

10. 103 marksNCERT Exercises, Chapter 13

    From the relation $R=R_0{A}^{1/3}$, where $R_0$ is a constant and $A$ is the mass number of a nucleus, show that the nuclear matter density is nearly constant (i.e. independent of $A$).

    Hint. Write both the mass and the volume in terms of A and see whether A survives the division.

    Nuclear density is mass divided by volume, so express each in terms of the mass number.

    Mass. A nucleus of mass number $A$ contains $A$ nucleons, each of mass approximately $m$, so:

    $M\approx Am$

    Volume. Treating the nucleus as a sphere of radius $R=R_0{A}^{1/3}$:

    $V=\frac{4}{3}\pi R^3=\frac{4}{3}\pi {\left(R_0{A}^{1/3}\right)}^3=\frac{4}{3}\pi R_0^{3}A$

    The cube of the cube root returns $A$ exactly, so the volume is directly proportional to $A$.

    Density. Dividing one by the other:

    $\rho =\frac{M}{V}=\frac{Am}{\frac{4}{3}\pi R_0^{3}A}=\frac{3m}{4\pi R_0^3}$

    The mass number $A$ cancels completely, so the density depends only on constants and is the same for every nucleus.

    Substituting $m\approx 1.67\times 10^{-27}$ kg and $R_0\approx 1.2\times 10^{-15}$ m gives $\rho \approx 2.3\times 10^{17}$ kg m${}^{-3}$ — about $10^{14}$ times the density of water. This constancy tells us nuclear matter is essentially incompressible and that nucleons are packed as closely as the nuclear force allows.

    ✦ Density = 3m/(4 pi R_0 cubed), independent of A, and about 2.3 x 10^17 kg per cubic metre

Solutions written by the tuition.in editorial team and checked against leph205.pdf (NCERT, Reprint 2026-27). Questions are referenced from the NCERT textbook for identification.