NCERT Solutions for Class 12 Physics Chapter 13 Nuclei

Q. Given the mass of iron nucleus as 55.85 u and A = 56. Find the nuclear density ?

Ans mFe = 55.85

Mass of iron nucleus = 55.85 × 1.67 × 10–22 kg

u = 9.27 × 10– 26 kg

Radius of iron nucleus

r0 = 1.2 × 10–15 m

r = 1.2 × 10–15 × (56)1/3

The density of matter in neutron stars (an astrophysical object) is comparable to this density. This shows that matter in these objects has been compressed to such an extent that they resemble a big nucleus.

Q. Obtain approximately the ratio of the nuclear radii of the gold isotope

$$^{197}_{79}\textbf{Au}\space\textbf{and the silver isotope}^{107}_{47}\textbf{Ag}$$

  • Ans. Given, nuclear radius of the gold isotope
  • 79Au197 = RAu
  • Nuclear radius of the silver isotope 47Ag107 ∝ RAg
  • Mass number of gold (AAu) = 197
  • Mass number of silver (AAg) = 107
  • The ratio of the radii of the two nuclei is related with their mass numbers by

$$\frac{\text{R}_{Au}}{\text{R}_{Ag}}=\bigg(\frac{\text{A}_{Au}}{\text{A}_{Ag}}\bigg)^{\frac{1}{3}}\\=\bigg(\frac{197}{107}\bigg)^{\frac{1}{3}}\\=1.2256\\\text{Therefore, the ratio of the nuclear radii of the gold and silver isotopes is about 1.23.}$$

$$\textbf{Q. The three stable isotopes of neon :}_{10}^{20}\textbf{Ne},\space_{10}^{21}\textbf{Ne and}\space_{10}^{22}$$

Ne have respective abundances of 90.51 %, 0.27 % and 9.22 %. The atomic masses of the three isotopes are 19.99 u, 20.99 u and 21.99 u, respectively. Obtain the average atomic mass of neon.

$$\textbf{Ans.}\space\text{Atomic mass of}\space_{10}^{20}\space\text{Ne},m_1=19.99\space u\\\text{Abundance of}\space_{10}^{20}\text{Ne},η_1=90.51\%\\\text{Atomic mass of}\space_{10}^{21}\text{Ne},m_2=20.99u\\\text{Abundance of}\space _{10}^{21}\text{Ne},η_{2}=0.27\%\\\text{Atomic mass of}\space_{10}^{22}\text{Ne},m_3=21.99u\\\text{Abundance of}\space_{10}^{22}\space\text{Ne},η_{3}=9.22\%\\\text{The average atomic mass of neon is}\\\text{m}=\frac{m_1n_1+m_2n_2+m_3n_3}{n_1+n_2+n_3}\\=\frac{19.99×90.51+20.99×0.27+21.99×9.22}{90.51+0.27+9.22}\\=20.1771u$$

Q. Calculate the energy equivalent of 1 g of substance.

Ans. Energy, E =10– 3 × (3 × 108)2 J ( ∵ E = mc2)

E = 10–3 × 9 × 1016

= 9 × 1013 J

Thus, if one gram of matter is converted to , energy, there is a release of enormous amount of energy.

Q. Find the energy equivalent of one atomic mass unit, first in Joules and then in MeV. Using this, express the mass defect of 8 16 in MeV/c2

Ans Given, 1 u = 1.6605 × 10–27 kg

To convert it into energy units, we multiple it by c2 and find that energy equivalent

= 1.6605 × 10–27× (2.9979 × 108)2 kg m2/s2

= 1.4924 × 10–10 J

$$\frac{1.4924×10^{-10}}{1.602×10^{-19}}\text{eV}\\=0.9315×10^{9}\text{eV}\\=931.5\text{Mev}\\\text{or}\space\text{1u}=931.5\text{Mev/c}^{2}\\\text{For}\space_{8}^{16}\text{O},\Delta \text{M}=0.13691,\\u=0.13691×931.5\text{MeV/c}^{2}\\=127.5\text{MeV/c}^{2}\\\text{The energy needed to separate}\space_{8}^{16}\text{O\space into its constituents is thus 127.5 MeV/c}^{2}.$$

$$\textbf{Q. Suppose, we think of fission of a}\space_{26}^{56}\space\textbf{Fe}\textbf{nucleus into two equal fragments,}\space_{13}^{28}\space\textbf{Al}.\\\textbf{Is the fission energetically possible ? Argue by working out Q of the process. Given :}\\\space \textbf{m}\bigg(\space_{26}^{56}\space\textbf{Fe}\bigg)=55.93494\textbf{u}\\\textbf{and}\space \textbf{m}\bigg (\space _{13}^{28}\textbf{Al} \bigg)=27.981914$$

$$\textbf{Ans}.\text{The fission of}\space_{26}^{56}\text{Fe can be given by}\\\space_{26}^{56}\text{Fe}\xrightarrow{}2\space_{13}^{28}\text{Al}\\\text{From data given in question}\\\text{Atomic mass of m}\bigg(\space_{26}^{56}\text{Fe}\bigg)=55.93494\space u\\\text{Atomic mass of m}\bigg(\space_{13}^{28}\text{Al}\bigg)=27.98191u\\\text{The Q-value of this nuclear reaction is given by}\\\text{Q}=\bigg[m\bigg(\space_{26}^{56}\text{Fe}\bigg)-2m\bigg(\space_{13}^{28}\text{Al}\bigg)\bigg]c^{2}\\=[55.93494-2×27.98191]c^{2}\\=(-0.02888 c^{2})u\\\text{But}\space 1u = 931.5 \text{MeV/c}^{2}\\\therefore \text{Q}=-0.02888×931.5\\=26.902\text{MeV}$$

The Q-value of the fission is negative. Therefore, the fission is not possible energetically. For an energetically-possible fission reaction, the Q-value must be positive.

Q. Suppose India had a target of producing by 2020 AD, 200,000 MW of electric power, 10% of which was to be obtained from nuclear power plants. Suppose we are given that, on an average, the efficiency of utilisation (i.e., conversion to electric energy) of thermal energy produced in a reactor was 25%. How much amount of fissionable uranium would our country need per year by 2020 ? Take the heat energy per fission of to be about 200 MeV.

Ans. Given, amount of electric power to be generated,

P = 2 × 105 MW

10% of this amount has to be obtained from nuclear power plants.

$$\text{P}=\frac{10}{100}×2×10^{5}\\\therefore\text{Amount of nuclear power,}\\=2×10^{4}\text{MW}\\=2×10^{4}×10^{6}\text{J/s}\\=2×10^{10}×60×60×24×365\space\text{J/y}\\\text{Heat energy released per fission of a}\space^{235}\text{U}\text{nucleus, E = 200 MeV}\\\text{Efficiency of a reactor =}25\%\\\text{Hence, the amount of energy converted into the electrical energy per fission is calculated by}\\=\frac{25}{100}×200\text{MeV}=50\text{MeV}\\=50×1.6×10^{\normalsize -19}×10^{6}\\=8×10^{\normalsize -12}\text{J}\\\text{Number of atoms required for fission per year :}\\=\frac{2×10^{10}×60×60×24×365}{8×10^{\normalsize -12}}\\=78840 × 10^{24} \text{atoms}\\\text{1 mole, i.e., 235 g of}\space\text{U}^{235}\text{contains}\space6.023×10^{23}\text{atoms}.\\\therefore\text{Mass of}\space6.023×10^{23}\text{atoms of}\space\text{U}^{235}=235g=235×10^{\normalsize -3}\text{Kg}\\\therefore\text{Mass of}\space78840×10^{24}\space\text{atoms of}\space\text{U}^{235}=\frac{235×10^{\normalsize -3}}{6.023×10^{23}}×78840×10^{24}\\=3.076×10^{4}\text{Kg}\\\text{Thus, the mass of uranium needed per year is}3.076 × 10^{4}\text{kg}.$$

$$\textbf{Q. The fission properties of}\space_{94}^{239}\space\textbf{Pu}\textbf{are very similar to those of}\space_{94}^{235}\textbf{U}.$$

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

$$\space_{94}^{239}\textbf{Pu}\space\textbf{undergo fission ?}$$

$$\textbf{Ans.}\space\text{Given, Average energy released per fission of}\space_{94}^{239}\space\text{E}_{av}=180\text{MeV}.\\\text{Amount of pure}\space_{94}\text{Pu}^{239}(m)=1\text{kg}=1000\text{g}\\\text{Avogadro number}(\text{N}_{\text{A}})=6.023×10^{23}\\\text{Mass number of}\space_{94}^{239}\text{Pu}=239\text{g}\\\text{1 mole of}\space_{94}\text{Pu}^{239}\text{contains N}_{A}\space\text{atoms}.\\\therefore\text{Mg of}\space_{94}\text{Pu}^{239}\text{contains}\bigg(\frac{N_A}{\text{Mass number}}×m\bigg)\text{atom}\\=\frac{6.023×10^{23}}{239}×1000\\=2.52×10^{24}\text{atoms}\\\therefore\text{Total energy released during the fission of 1 kg of}\space_{94}^{239}\text{Pu is given by}\\\text{E}=\text{E}_{av}×2.52×10^{24}\\=180 × 2.52 × 10^{24}\text{MeV}\\=4.536×10^{26}\text{MeV}\\\text{Therefore}\space4.536×10^{26}\text{MeV is released if all the atoms in 1 kg of pure}\space_{94}\text{Pu}^{239}\text{undergo fission.}$$

Q. 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

$$\textbf{Ans}.\space\text{The given fusion reaction is :}\\\space_{1}^{2}\text{H}+_{1}^{2}\text{H}\xrightarrow{}_{2}^{3}\text{H}+n+3.27\space\text{MeV}\\\text{Amount of deuterium,}\\m=2\text{kg}\\\text{1 mole, i.e.,}\text{2 g of deuterium contains}\space6.023×10^{23}\space\text{atoms}.\\\therefore2.0\text{Kg of deuterium contains =}\frac{6.023×10^{23}}{2}×2000\\=6.023×10^{26}\text{atoms}\\\text{When two atoms of deuterium fuse, 3.27 MeV energy is released.}\\\therefore\text{Total energy per nucleus released in the fusion reaction :}\\\text{E}=\frac{3.27}{2}×6.023×10^{26}\text{MeV}\\=\frac{3.27}{2}×6.023×10^{26}×1.6×10^{-19}×10^{6}\text{J}\\=1.576×10^{14}\text{J} $$

$$\text{Power of the electric lamp,}\\\text{P}=100\text{W}=100\space\text{J/s}\\\text{Hence, the energy consumed by the lamp per second = 100 J}\\\text{The total time for which the electric lamp will glow is calculated by}\\\text{t}=\frac{1.576×10^{14}}{100}s\\=\frac{1.576×10^{14}}{100×60×60×24×365}\approx4.9×10^{4}\text{years}.$$

$$\textbf{Q. Consider the D–T reaction (deuterium-tritium fusion)}\\\space ^{2}_{1}\text{H}+^{3}_{1}\text{H}\xrightarrow{}\space^{4}_{2}\text{He}+n\\\textbf{(i) Calculate the energy released in MeV in this reaction from the data :}\\m\bigg(\space^{2}_{1}\text{H}\bigg)=2.014102\space u\\m\bigg(\space^{3}_{1}\text{H}\bigg)=3.016049\space u$$

(ii) Consider the radius of both deuterium and tritium to be approximately 2.0 fm.What is the kinetic energy needed to overcome the coulomb repulsion between the two nuclei ? To what temperature must the gas be heated to initiate the reaction [Hint : Kinetic energy required for one fusion event = average thermal kinetic energy available with the interacting particles = 2(3kT/2) k = Boltzman’s constant, T = absolute temperature.]

$$\textbf{Ans.}\space\text{(i) Take the D-T nuclear reaction :}\\\space_{1}^{2}\text{H}+_{1}^{3}\text{H}\xrightarrow{}_{2}^{4}\text{H}+n\\\text{Data given in question is as follows}\\\text{Mass of}\space_{1}^{2}\text{H},m_1=2.014102\text{u}\\\text{Mass of}\space_{1}^{3}\text{H},m_2=3.016049\text{u}\\\text{Mass of}\space_{2}^{4}\space\text{He},m_3=4.002603\text{u}\\\text{Mass of}\space_{0}^{1}n,m_4=1.008665\text{u}\\\text{Q-value of the given reaction is :}\\\text{Q}=[m_1+m_2-m_3-m_4]c^{2}\\=[2.014102+3.016049-4.002603=1.008665]c^{2}\\=[0.018883\space c^{2}]\text{u}\\\text{But}\space 1u=931.5\text{MeV/C}^{2}\\\therefore\text{Q}=0.018883×931.5\\=17.59\text{MeV}$$

$$\text{(ii) Radius of deuterium and tritium,}\\\text{r}\approx2.0\text{fm}=2×10^{\normalsize-15} m\\\text{Distance between the two nuclei at the moment when they touch each other,}\\d = A + A = 4 × 10^{\normalsize–15} m\\\text{Charge on the deuterium nucleus = e}\\\text{Charge on the tritium nucleus = e}\\\text{Hence, the repulsive potential energy between the two nuclei is given by}\\\text{U}=\frac{e^{2}}{4\pi\epsilon_0(d)}\\\text{Here},\space\epsilon_{0}=\text{Permittivity of free space}\\\frac{1}{4\pi\epsilon_0}=9×10^{9}\text{Nm}^{2}\text{C}^{\normalsize-2}\\\therefore\text{U}=\frac{9×10^{9}×(1.6×10^{\normalsize-19})^{2}}{4×10^{\normalsize-15}} =5.76×10^{\normalsize-14}\text{J}\\=\frac{5.76×10^{\normalsize-14}}{1.6×10^{\normalsize-19}}\text{eV}=3.6×10^{5}\text{eV}\\=360\space\text{keV}\\\text{Therefore,} 5.76 × 10^{\normalsize–14} J \text{or 360 keV of kinetic energy (KE) is needed to overcome the Coulomb repulsion between the two nuclei.}\\\text{K.E.}=2×\frac{3}{2}\text{kT}\\\text{Here, k = Boltzmann constant}\\=1.38×10^{\normalsize-23}m^{2}\text{kg}s^{\normalsize-2}\text{k}^{\normalsize-1}\\\text{T}=\text{Temperature required for triggering the reaction}\\\therefore\space\text{T}=\frac{\text{KE}}{3k}\\=\frac{5.76×10^{\normalsize-14}}{3×1.38×10^{\normalsize-23}}\\=1.39×10^{9}\text{K}\\\text{Thus, the gas must be heated to a temperature of} 1.39 × 10^9 \text{K to initiate the reaction.}$$

$$\textbf{Q. From the relation R}=\textbf{R}_0\textbf{A}^{1/3},\textbf{where}\space \textbf{R}=\textbf{R}_0A^{1/3},\textbf{where R}_0\textbf{is a constant and A is the mass number of a nucleus,}\\\textbf{ show that the nuclear matter density is nearly constant (i.e., independent of A).}\\\textbf{Ans}.\space\text{Nuclear radius},\text{R}=\text{R}_0A^{\frac{1}{3}}\\\text{Here},\text{R}_0=\text{Constant}\\\text{A}=\text{Mass number of the nucleus}\\\text{Nuclear matter density,}\\(ρ)=\frac{Mass of the nucleus}{Volume of the nucleus}\\\text{Let m be the average mass of nucleus.}\\\text{Therefore, mass of the nucleus = mA}\\\therefore\space(ρ)=\frac{mA}{\frac{4}{3}\pi\text{R}^{3}}=\frac{3mA}{4\pi(R_0A^{1/3})^{3}}\\=\frac{3mA}{4\pi R_0^{3}A}=\frac{3m}{4\pi R_{0}^{3}}\\\text{Therefore, the nuclear matter density is independent of A. It is nearly constant.}$$