NCERT Solutions for Class 12 Physics Chapter 13 - Nuclei
$$\textbf{1. (a) Two stable isotopes of lithium,}\space^{6}_{3}\space\text{Li and}\space^{7}_{3}\textbf{Li have respective abundance of 7.5}\% \text{and} 92.5\%.\\\textbf{These isotopes have masses 6.01512m and 7.01600u, respectively. Find the atomic mass of lithium.}\\\textbf{(b) Boron has two stable isotopes,}\space^{10}_{5}\textbf{B and}^{11}_{5}\textbf{B}.\\\textbf{Their respective masses are 10.01294u}\\\textbf{ and 11.00931u and the atomic mass of boron is 10.811u.}\\\textbf{ Find the abundances of}\space^{10}_{5}\textbf{B and}\space ^{11}_{5}\textbf{B}.$$
Sol. (a) Given, abundance percent of 6Li = 7.5%
Abundance percent of 7Li = 92.5%
Atomic mass of 6Li = 6.01512u
Atomic mass of 7Li = 7.01600u
Atomic mass is given by
$$=\frac{6.01512×7.5+7.01600×92.5}{7.5+92.5}\\=\frac{45.1134+648.98}{100}=6.941\text{u}$$
(b) Given, mass of 10B = 10.01294u
Mass of 11B = 11.00931u
Atomic mass of boron = 10.811u
Let the abundance of 10B be x%
So, the abundance of 11B be (100 – x)%
Atomic mass is given by,
$$10.811=\frac{x×10.01294+(100-x)×11.00931}{(x+100-x)}$$
Abundance of 10B, x = 19.9%
Abundance of 11B, (100 – x) = 100 – 19.9 = 80.1%
Thus, the abundance of 10B is 19.9% and the abundance of 11B is 80.1%.
$$\textbf{2. The three stable isotopes of neon,}\space^{20}_{10}\textbf{Ne},\space^{21}_{10}\textbf{Ne}\space\textbf{and}\space^{22}_{10}\textbf{Ne}\\\text{ have respective abundances of 90.51}\%, \textbf{0.27}\% \text{and} \textbf{9.22}\%.\\\textbf{ The atomic masses of the three isotopes are 19.99u, 20.99u and 21.99u,} \\\textbf{respectively.}\\\textbf{ Obtain the average atomic mass of neon.}$$
Sol. Given, abundance percent of Ne20 = 90.51%
Abundance percent of 21Ne = 0.27%
Abundance percent of 22Ne = 9.22%
Mass of 20Ne = 19.99u
Mass of 21Ne = 20.99u
Mass of 22Ne = 21.99u
Average atomic mass (m)
= Weighted average of all isotopes
$$=\frac{90.51×19.99+0.27×20.99+9.22×21.99}{90.51+0.27+9.22}\\=\frac{1809.29+5.67+202.75}{100}=\frac{2017.7}{100}=20.18\text{u}$$
Thus, the average atomic mass of neon is 20.18u.
3. Obtain the binding energy (in MeV) of a nitrogen nucleus
$$\bigg(\space^{14}_7\textbf{N}\bigg),\textbf{given m}\bigg(\space^{14}_{7}\textbf{N}\bigg)=14.00307u.$$
Sol. Given, mass of proton, mp = 1.007834,
Mass of neutron, mn = 1.00867u
147Nnucleus contains 7 protons and 7 neutrons.
Mass defect (Δm) = mass of nucleons – mass of nucleus
= 7mp + 7mn – mN
= 7 × 1.00783 + 7 × 1.00867 – 14.00307
= 7.05481 + 7.06069 – 14.00307
= 0.11243u
Binding energy of nitrogen nucleus
= Δm × 931MeV
= 0.11243 × 931 MeV
= 104.67 MeV
Thus, the binding energy is 104.67 MeV.
$$\textbf{4. Obtain the binding energy of the nuclei}\space^{56}_{26}\text{Fe}\space\textbf{and}\space^{209}_{83}\textbf{Bi in units of MeV from the following data:}\\\textbf{m}\bigg(\space^{56}_{26}\bigg)\textbf{Fe}=55.934939 u, m\bigg(\space^{209}_{83}\textbf{Bi}\bigg)=208.980388u.$$
Sol. Given, mass of proton mp = 1.00783u
Mass of neutron, mn = 1.00867u
$$\textbf{(i) For}\space^{56}_{26}\space\textbf{Fe}$$
$$^{56}_{26}\space\text{Fe contains 26 protons and 30 neutrons}\\\text{Mass defect (Δm)}=\text{mass of nucleons – mass of nucleus of}\space^{56}_{26}\text{Fe}$$
Mass defect (Δm) = mp + 30mn – mn
= 26 × 1.00783 + 30 × 1.00867 – 55.934939
= 26.20345 + 30.25995 – 55.934939
= 0.528461u
Total binding energy = Δm × 931 MeV
= 0.528461 × 931.5 = 492.26 MeV
Average binding energy per nucleon of $$^{56}_{26}\text{Fe}$$
$$=\frac{\text{Binding energy}}{\text{Total number of nucleons}}\\=\frac{492.26}{56}=8.790\space\text{MeV}\\\textbf{(ii) For}\space^{209}_{83}\textbf{Bi}$$
= 83 × mp + 126 × mn – mN
= 83 × 1.007825 + 126 × 1.008665
– 208.980388
= 83.649475 + 127.091790 – 208.980388
= 1.760877u
Binding energy = Δm × 931MeV
= 1.760877 ×931.5 = 1640.26 MeV
$$\text{Average binding energy per nucleoon of}\space^{209}_{83}\text{Bi}\\=\frac{\text{Bindingenergy}}{\text{Total number of nucleons}}\\=\frac{1640.26}{209}=7.848\space\text{MeV}$$
Thus, the binding energy per nucleon of Fe is more than Bi.
5. A given coin has a mass of 3.0g. 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 62.92960u).
Sol. Given, mass of coin = 3g
$$\text{Number of atoms in 1g of Cu =}\frac{6.023×10^{23}}{63}\\\text{Number of atoms in 3g of Cu =}\frac{6.023×10^{23}}{63}×3$$
Number of protons in Cu atom = 29
Number of neutrons in Cu atom = 63 – 29 = 34
Mass defect, Δm = 29 × mp + 34 × mn – mCu
= 29 × 1.00783 + 34 × 1.00867 – 62.9260 = 0.59225u
∴ Total mass defect = 0.59225 × 2.868 × 1022
= 1.6985 × 1022u
Binding energy = Mass defect × 931 MeV
= 1.6985 × 1022 × 931
= 1.58 × 1025 MeV
Thus, the energy required to separate all the neutrons and protons is 1.58 ×1025 MeV i.e., equal to binding energy.
6. Write nuclear reaction equations for $$\textbf{(a)}\space\alpha \textbf{-decay of}\space^{226 }_{88}\textbf{Ra},\\\textbf{(b)}\space\alpha\textbf{-decay of}\\\textbf{(c)} \beta\textbf{-decay of}\space^{32}_{15}\textbf{P},\\\textbf{(d)}\space\beta-\textbf{decay of}\space^{210}_{83}\space\textbf{Bi}\\\textbf{(e)}\space\beta^{\normalsize+}\textbf{-decay of}\space^{11}_{6}\textbf{C}\\\textbf{(f)}\space\beta^{\normalsize+}\textbf{-decay of}\space^{97}_{43}\text{Tc},\\\textbf{(g)}\space\textbf{Electron capture of}\space^{120}_{54}\space\textbf{Xe}$$
Sol. We know that
1. in a-decay, the mass number is reduced by 4 and atomic number is reduced by 2.
2. in b-decay, the mass number remains constant but atomic number is increased by 1.
3. in a g-decay, the mass number and atomic number both remain same. The following equations are given:
$$\text{(a)}\space^{226}_{88}\text{Ra}\xrightarrow{-\alpha}^{222}_{88}\space\text{Rn}+^{4}_{2}\space\text{He}\\\text{(b)}\space^{242}_{94}\text{Pu}\xrightarrow{\alpha}\space^{238}_{92}\text{U}+^{4}_{2}\text{He}\\\text{(c)}\space^{32}_{15}\text{P}\xrightarrow{-(-\beta)}\space^{32}_{16}\text{S}+ ^{0}_{\normalsize-1}e+\bar{v}$$
i.e., β–decay is accompanied by release of antineutrino.
$$\text{(d)\space}^{210}_{83}\text{Bi}\xrightarrow{-(-\beta)}\space^{210}_{84}\text{X} + ^{0}_{\normalsize-1}e+\bar{v}\\\text{(e)}\space^{11}_{6}\text{C}\xrightarrow{-(+\beta)}\space^{11}_{5}\text{B} + e^{\normalsize+}+ v\\\beta ^{\normalsize+} \text{decay of}\space^{11}\text{C}_{6}\space\text{is accompanied by the release neutrino.}\\\text{(f)}\space^{97}_{43}\text{TC}\xrightarrow{-(+\beta)}\space^{97}_{42}\text{X}×e^{\normalsize+}+v\\\text{(g)}\space^{120}_{54}\text{Xe}+^{0}_{\normalsize-1}e\xrightarrow{}^{120}_{53}\space\text{X}$$
7. A radioactive isotope has a half-lie of T years. How long will it take the activity to reduce to:
(a) 3.125% and
(b) 1% of its original value?
Sol. Given, half-life T1/2 = T yr
(a) N = 3.125 of N0
$$\therefore\space\frac{\text{N}}{\text{N}_{0}}=\frac{3.125}{100}=\frac{1}{32}\\\text{We know}\space\frac{\text{N}}{\text{N}_{0}}=\bigg(\frac{1}{2}\bigg)^{n}\\\therefore\space\frac{1}{32}=\bigg(\frac{1}{2}\bigg)^{n}\\\Rarr\space\bigg(\frac{1}{2}\bigg)^{5}=\bigg(\frac{1}{2}\bigg)^{n}$$
or n = 5
So, time t = n × T1/2 = 5T
After 5 half-time periods activity reduces to 3.125% of initial activity.
(b) Given, N = 1% of N0
$$\therefore\space\frac{\text{N}}{\text{N}_{0}}=\frac{1}{100}\\\text{We know that}\\\frac{\text{N}}{\text{N}_{0}}=e^{-\lambda t}\\\therefore\space\frac{1}{100}=e^{-\lambda t}\\\text{Taking log on both the sides, we get log}_e^1 – \text{log}_e 100 = – \lambda \text{t log}_e\text{e}\\– 2.303 × 2 = –\lambda t\\\text{or}\space t=\frac{4.606}{\lambda}\\\text{Also,}\space\lambda=\frac{0.693}{\text{T}_{1/2}}\\t=\frac{4.606. \text{T}_{1/2}}{0.693}=6.65\text{T}$$
8. The normal activity of living carbon containing matter is found to be about 15 decays per minute for every gram of carbon. This activity arises from the small proportion of radioactivity
$$^{14}_{6}\textbf{C}$$
614Cpresent with the stable carbon isotope $$^{12}_{6}\textbf{C}.$$
When the organism is dead, its interaction with the atmosphere (which maintains the above equilibrium activity) ceases and its activity begins to drop. From the known half-life (5730 year) of$$^{14}_{6}\space\textbf{C}.$$ and the measured activity the age of the specimen can be approximately estimated. This is the principle of $$^{14}_{6}\space\textbf{C}.$$dating used in archaeology. Suppose a specimen from Mohenjodaro gives an activity of 9 decays per minute per gram of carbon. Estimate the approximate age of the Indus-Valley civilization.
Sol. Given, normal activity, A0 = 15 decay/min
Present activity,
A = 9 decay/min
T1/2 = 5730 year
$$\frac{\text{A}}{\text{A}_{0}}=e^{-\lambda t}\\=\frac{9}{15}=e^{-\lambda t}\\\text{or\space}\frac{3}{5}=e^{-\lambda t}\\\text{or}\space e^{\lambda t}=\frac{5}{3}$$
Taking log on both the sides, we get
λt loge e = loge 5 – loge 3
or λt = 2.303 (0.69 – 0.47)
λt=0.5109
$$\bigg(\because\lambda=\frac{0.693}{\text{T}_{1/2}}\bigg)\\\therefore\space t=\frac{0.5066×T_{1/2}}{0.693}\\=\frac{0.5066×5730}{0.693}=4224.47\space\text{year}$$
Thus, the approximate age of Indus-Valley civilization is 4224 year.
$$\textbf{9. Obtain the amount of}\space^{60}_{27}\space\textbf{CO necessary to provide a radioactive source of 8.0 mCi strength. The half-life of}\\\space^{60}_{27}\space\textbf{Co is 5.3 year.}\\\textbf{Sol.}\space\text{Activity,}\frac{\text{dN}}{dt}=8 \text{mCi}$$
= 8 × 10–3 × 3.7 × 1010 = 3.7×107 disintegration/s
(∵1 Ci = 3.7 × 1010 disintegration/s)
$$\text{Half-life of}\space^{60}_{27}\space\text{Co},\text{T}_{1/2}=5.3\text{year}\\= 5.3 × 365 × 24 × 60 × 60\\= 1.67 × 10^{8}\text{s}\\\text{We know that}\\\lambda=\frac{0.693}{\text{T}_{1/2}}=\frac{0.693}{1.67×10^{8}}\\\text{= 4.14 × 10}^{\normalsize–9}/s\\\text{Activity,}\space\frac{\text{dN}}{dt}=\lambda\text{N}\\\text{or}\space\text{N}=\frac{\text{dN/dt}}{\lambda}=\frac{8×3.7×10^{7}}{4.14×10^{\normalsize-9}}$$
= 7.133 × 1016
By using the concept of Avogadro number:
$$\text{Mass of 6.023 × 10}^{23} \text{atoms of}\space^{60}_{27}\text{Co}=60\text{(g)}\\\text{Mass of 7.133 × 10}^{16} \text{atoms of}\space^{60}_{27}\text{Co}\\=\frac{60×7.133×10^{16}}{6.023×10^{23}}\\\text{Mass m = 7.12 × 10}^{\normalsize–6}\text{g}\\\text{Thus, the required mass of}\space^{60}_{27}\text{Co is 7.12 × 10}^{\normalsize–6}\text{g.}$$
10. The half-life of $$^{90}_{38}\text{Sr}$$ is 28 year. What is the disintegration rate of 15 mg of this isotope?
$$\textbf{Sol.}\space\text{Given, half life of}\space^{90}_{38}\space\text{Sr , T}_{1/2}=28\text{year}$$
= 28 × 365 × 24 × 60 × 60s
According to Avogadro number concept:
90g of Sr contains = 6.023 × 1023 atom
$$\text{15 mg of Sr contains =}\frac{6.023×10^{23}×15×10^{\normalsize-3}}{90}\\\text{Number of atoms, N = 1.0038 × 10}^{20}\\\text{Activity,}\space\frac{\text{dN}}{dt}=\lambda N\\\text{or}\space\frac{\text{dN}}{dt}=\frac{0.6931}{\text{T}_{1/2}}.\text{N}\\=\frac{0.6931×1.0038×10^{20}}{28×365×24×60×60}\\\bigg(\because\lambda=\frac{0.,693}{\text{T}_{1/2}}\bigg)\\\frac{\text{dN}}{dt}=7.877×10^{10}\text{disintegration/s}$$
= 7.877 × 1010 Bq
$$\textbf{11. Obtain approximately the ratio of the nuclear radii of the gold isotope}\space^{197}_{79}\textbf{AU}\\\textbf{and the silver isotope}\space^{107}_{79}\space\textbf{Ag}.$$
Sol. Radius of nuclei, R = R0A1/3.
where A is the mass number of nucleus
∴ R ∝A1/3
$$\therefore\space\frac{\text{R}_{\text{gold}}}{\text{R}_{\text{silver}}}=\bigg(\frac{\text{A}_{gold}}{\text{A}_{silver}}\bigg)^{1/3}=\bigg(\frac{197}{107}\bigg)^{1/3}=1.225$$
= 1.23
12. Find the Q-value and the kinetic energy of the emitted α-particle in the α-decay of (a)
$$^{226}_{88}\textbf{Ra}\space\textbf{and (b)}\space^{220}_{86}\textbf{Rn}\\\textbf{Given, m}\bigg(\space^{226}_{88} \textbf{Ra}\bigg)\textbf{=226.0.2540}\textbf{u},\\\textbf{m}\bigg(\space^{222}_{86}\textbf{Rn}\bigg)\textbf{=222.01750}\textbf{u.}\\\textbf{m}_\alpha=4.00260\textbf{u, m}\bigg(\space^{226}_{86}\textbf{Rn}\bigg)\textbf{=220.01137u.}\\\textbf{m}\bigg(\space^{216}_{84}\textbf{PO}\bigg)\textbf{=216.00189u}$$
Sol. (a) The process of a-decay of $$^{226}_{88}\text{Ra}$$ can be expressed as
$$^{226}_{88}\text{Ra}\xrightarrow{}^{222}_{86}\text{Rn} + ^{4}_{2}\text{He} + \text{Q}\\\text{Q-value}=[\text{m}\bigg(\space^{226}_{88}\text{Ra}\bigg)-m\bigg(\space^{222}_{86}\text{Rn}\bigg)-m_\alpha]×931.5\text{MeV}$$
= [226.02540 – 222.01750 – 4.00260) × 931.5
= 0.0053 × 931.5 = 4.94 MeV
Kinetic energy of emitted α-particle
$$=\bigg(\frac{\text{A}-4}{\text{A}}\bigg).\text{Q}\\=\frac{226-4}{226}×4.94$$
= 4.85 MeV
$$\textbf{(b) The process of} \alpha \textbf{-decay of}\space^{226}_{88}\textbf{Ra can be expressed as:}\\\\^{220}_{86}\textbf{Rn}\xrightarrow{}^{216}_{84}\textbf{PO} + ^{4}_{2}\textbf{He}\\\text{Q-value}\\=[m\bigg(\space^{220}_{86}\text{Rn}\bigg)-m\bigg(\space^{216}_{84}\text{PO}\bigg)-m_{\alpha}]×931.5\space\text{MeV}$$
= [220.1137 – 216.00189 – 4.00260] × 931.5
= 6.41 MeV
Kinetic energy of emitted α-particle
$$=\frac{(\text{A}-4)Q}{\text{A}}\\=\frac{220-4}{220}×6.41=6.29\space\text{MeV}$$
13. The radionuclide 11C decays according to
$$^{11}_{6}\textbf{C}\xrightarrow{}^{11}_{5}\textbf{B}+e^{4}+v:\textbf{T}_{1/2}=20.3\space\textbf{min}$$
The maximum energy of the emitted positron is 0.960 MeV.
Given, the mass values
$$m\bigg(\space^{11}_{6}\text{C}\bigg)=11.01143\textbf{u}\space\textbf{and}\space\textbf{m}\bigg(\space^{11}_{6}\textbf{B}\bigg)\textbf{= 11.009305u}$$
Calculate Q and compare it with the maximum energy of the positron emitted.
Sol. Mass of e = 0.000548u
$$\text{The mass defect,}\Delta m=\bigg[m\bigg(\space^{11}_{6}\text{C}\bigg)-m\bigg(\space^{11}_{5}\text{B}\bigg)-m_{e}\bigg]$$
where, the masses used are those of nuclei and not of atoms. If we use atomic masses, we have to add 6 me in case of 11C and 5me in case of 11B.
$$\text{As}\space^{11}_{6}\text{C}\space\text{atom is made up of}\space^{11}_{6}\text{C nucleus and 6 protons.}\\\therefore\space\text{Mass of}\space^{11}_{6}\text{C nucleus}\\=\text{Mass of}\space^{11}_{6}\text{C atom - mass of 6 electrons}\\\text{= 11.011434u – 6m}_e\\\text{Similarly mass of}\space^{11}_{5}\text{B nucleus}$$
= 11.00930 – 5me
Q = [(11.011434 – 6me) – (11.009305 – 5me) – me]
$$\Delta m=\bigg[m\bigg(\space^{11}_{6}\text{C}\bigg)-m\bigg(\space^{11}_{5}\text{B}\bigg)-2m_{e}\bigg]$$
= 11.011434 – 11.009305 – 2 × 0.000548
= 0.001033
Q = Binding energy = ∆m × 931
= 0.001033 × 931 = 0.9617 MeV
The daughter nucleus is too heavy compared to e+ and V. So it carries negligible energy (Ed = 0). It the kinetic energy (Ev) carried by the neutrino is minimum (i.e. zero, the position carries maximum energy and this is practically all energy Q, hence maximum Ee ≈ Q.)
$$\textbf{14. The nucleus}\space^{23}_{10}\textbf{Ne decays by} \beta– \textbf{emission. Write down the} \beta– \textbf{decay equation and}\\\textbf{determine the maximum kinetic energy of the electrons emitted. Given that}\\\textbf{m}\bigg(\space^{23}_{10}\textbf{Ne}\bigg)\textbf{=22.994466u}\\\textbf{m}\bigg(\space^{23}_{11} \textbf{Na}\bigg)\textbf{=22.989770u}\\\textbf{Sol.}\space\text{The}\beta\text{-decay equation of}\space^{23}_{10}\text{Ne is given by}\\\\^{23}_{10}\textbf{Ne}\xrightarrow{-\beta}\space^{23}_{11}\text{Na}^{\normalsize+}\space_{\normalsize-1}e^{0}+ \bar v+\text{Q}\\\text{Mass defect}\space∆m =\bigg(\space^{23}_{10}\text{Ne}\bigg)-m\bigg(\space^{23}_{11}\text{Na}\bigg)$$
= 22.994466 – 22.989770
= 0.004696u
Q = ∆m × 931 = 0.004696 × 931
= 4.372 MeV
The maximum kinetic energy of the electron of the emitted β-particle is equal to the Q-value.
Ee = Q = 4.37 MeV
2310Nanucleus is too heavier than electron-neutron, practically whole of the energy released is carried by electron-neutrino pair. When neutrino gets zero energy, the electron will carry the maximum energy i.e. 4.374 MeV.
15. The Q-value of a nuclear reaction A + b → C + d is defined by Q = [mA + mb – mC – md]c2, 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.
$$\textbf{(a)}\space^{1}_{1}\textbf{H}+^{3}_{1}\textbf{H}]\xrightarrow{}^{2}_{1}\textbf{H}+^{2}_{1}\textbf{H}\\\textbf{(b)}\space^{12}_{6}\textbf{C}+^{12}_{6}\textbf{C}\xrightarrow{}^{20}_{10}\textbf{Ne + }^{4}_{2}\textbf{He}$$
Atomic masses are given to be
$$\textbf{m}\bigg(\space^{2}_{1}\textbf{H}\bigg)\textbf{= 2.014102u}\\\textbf{m}\bigg(\space^{3}_{1}\text{H}\bigg)\textbf{= 3.016049u}\\\textbf{m}\bigg(\space^{12}_{6}\textbf{C}\bigg)\textbf{= 12.000000u}\\\textbf{m}\bigg(\space^{20}_{10}\textbf{Ne}\bigg)\textbf{= 19.992439u.}$$
Sol. The given reaction
$$\text{(a)}\space^{1}_{1}\text{H}+^{3}_{1}\text{H}\xrightarrow{}^{2}_{1}\text{H}+^{2}_{1}\text{H}\\\text{Mass defect}\space\Delta m=m\bigg(\space^{1}_{1}\text{H}\bigg) + m\bigg(\space^{3}_{1}\text{H}\bigg) + 2m\bigg(\space^{2}_{1}\text{H}\bigg) $$
= 1.007825 + 3.016049 – 2(2.014102)
= – 0.00433u
Q-value of the reaction
Q = ∆m × 931 = – 0.00433 × 931
= – 4.031 MeV
As the energy is negative so, the reaction is endothermic.
(b) The given reaction
$$^{12}_{6}\text{C}+^{12}_{6}\text{C}\xrightarrow{}^{20}_{10}\text{Ne} + ^{4}_{2}\text{He}\\\Delta m=2m\bigg(\space^{12}_{6}\text{C}\bigg)-m\bigg(\space^{20}_{10}\text{Ne}\bigg)-m\bigg(\space^{4}_{2}\text{He}\bigg)$$
= 2 × 12 – 19.992439 – 4.002603
= 0.00495u
Q = ∆m × 931 = 0.00495 × 931
= 4.62 MeV
Since, the energy is positive, the reaction is exothermic.
$$\textbf{16. Suppose, we think of fission of a}\space^{56}_{26}\textbf{Fe nucleus into two equal fragments,}\space^{28}_{13}\textbf{Al.}\\\textbf{Is the fission energetically possible? Argue by working out Q of the process. Given}\space \textbf{m}\bigg(\space^{56}_{26}\textbf{Fe}\bigg)\textbf{=55.93494u}\space\textbf{and m}\bigg(\space^{28}_{13}\textbf{AI}\bigg) = 27.98191u.$$
Sol. The given reaction for the decay process
$$^{56}_{26}\textbf{Fe}\xrightarrow{}2^{28}_{13}\text{AI}\\\text{Mass defect} \Delta m =m\bigg(\space^{56}_{26}\textbf{Fe}\bigg)-2m\bigg(\space^{28}_{13}\text{AI}\bigg)$$
= 55.93494 – 2(27.98191)
= – 0.02888u
Q = ∆m × 931 = – 26.88728 MeV
Because the energy is negative so, the fission is not possible.
$$\textbf{17. The fission properties of}\space^{239}_{94}\textbf{Pu are very similar to those of}\space^{235}_{92}\textbf{U.}\\\textbf{The average energy released per fission is 180 MeV.}\\ \textbf{How much energy, in MeV, is released if all the atoms in 1 kg of pure}\space^{239}_{94}\textbf{Pu}\space\textbf{undergo fission}$$
Sol. The number of atoms in 239g of
$$^{239}_{94}\text{Pu}=6.023 ×10^{23}\\\text{Number of atoms in 1 kg of}\space^{239}_{94}\text{Pu}\\=\frac{6.023×10^{23×1000}}{239}$$
= 2.52 × 1024
The average energy released in one fission = 180 MeV
So, total energy released in fission of 1 kg of 23994Pu= 180 × 2.52 × 1024
= 4.53 × 1026 MW
19. How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as
$$^{2}_{1}\text{H}+^{2}_{1}\text{H}\xrightarrow{}^{3}_{2}\text{He}+n 3.27\textbf{MeV}$$
Sol. Let t be the time.
Number of atoms in 2g of deuterium = 6.023 × 1023
Number of atoms in 2kg of deuterium
$$=\frac{6.023×10^{23}×2×10^{3}}{2}=6.023×10^{26}\space\text{nuclei}$$
Energy released during fusion of two deuterium = 3.27 Mev
$$\therefore\space\text{Energy released by one deuterium =}\frac{3.27}{2}$$
= 1.635 MeV
Energy released in 6.023 × 1026 deuterium atoms
= 1.635 × 6.023 × 1026 = 9.848 × 1026 MeV
= 9.848 × 1026 × 1.6 × 10–13 = 15.75 × 1013J
Energy consumed by bulb in 1s = 100J
Since, 100 J energy used in time = 1s
$$15.75 × 10^{13}\space\text{J energy used in time =}\frac{1×15.75×10^{13}}{100}\\\text{= 15.75 × 10}^{11}s\\=\frac{15.75×10^{11}}{60×24×60×365}\text{yr}=4.99 × 10^4 \text{Year}\\\text{= 15.75 × 10}^{11}s\\=\frac{15.75×10^{11}}{60×24×60×365}\text{yr}=4.99×10^{4}\text{Year}$$
Thus, the bulbs glow for 4.99 × 104 year.
20. 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.]
Sol. Given, radius r = 2 fm = 2 × 10–15 m
For head on collision,
the distance between two deuterons
d = r
d = 2 × 10–15 = 2 × 10–15m
$$\text{Potential energy =}\frac{1}{4\pi\epsilon_{0}}.\frac{q_1q_2}{d}\\=\frac{9×10^{9}×1.6×10^{\normalsize-19}×1.6×10^{\normalsize-19}}{2×10^{\normalsize-15}}\\=\frac{5.76×10^{-14}}{1.6×10^{\normalsize-19}}=720000\text{eV}$$
According to the law of conservation of energy, this potential energy will be equal to kinetic energy of both deuteron.
i.e., Potential energy = 2 × Kinetic energy of each deuteron
$$\text{Kinetic energy of each deuteron =}\frac{720000}{2}$$
= 360000 eV
= 360 keV
Thus, the potential barrier is 360 keV.
21. From the relation R = R0A1/3, where R0 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).
Sol. Given, the expression of the radius of nucleus is given by R = R0A1/3, A is the mass number of nucleus.
$$\rho=\frac{\text{Mass of each neucleon × Number of neucleons}}{\frac{4}{3}\pi R^{3}}=\\\frac{m×A×3}{4\pi R^{3}}=\frac{\text{Am3}}{4\pi R_{0}^{3}\text{A}}\\=\frac{3m}{4\pi R_{0}^{3}}=\frac{3×1.66×10^{\normalsize-27}}{4×3.14×(1.1×10^{\normalsize-15})^{3}}$$
= 2.97 ×1017 kg/m3
Since, R0 is a constant, density is constant or independent of A.
22. For the β+ (positron) emission from a nucleus, there is another competing process known as electron capture (electron from an inner orbit say, the K-shell is captured by the nucleus and a neutrino is emitted).
$$\textbf{e}^{\normalsize+}+^{\textbf{A}}_{\textbf{Z}}X\xrightarrow{}^{\textbf{A}}_{\textbf{Z-1}}Y+\textbf{v}$$
Show that if β+ emission is energetically allowed, electron capture is necessarily allowed but not vice-versa.
$$\textbf{Sol.}\space\text{Consider positron emission}\\\\_\text{Z}X^{A}\xrightarrow{}_{Z-1}Y^{A}+_{1}e^{0}+ Q_1\space\text{...(i)}\\\text{Now consider electron capture}\\\\_{Z}X^{A+}\space_{\normalsize-1}e^{0}\xrightarrow{}_{Z-1}Y^{A}+v+Q_{2}\space\text{...(ii)}\\\text{The energy released in Eq. (i),}\\\text{Q}_{1}=[m_{N}(_zX^{A})-m_{N}(_{Z-1}Y^{A})-m_e]c^{2}\\=[m_N(_ZX^A) + Zm_e – m_N(_{Z–1}Y^A) –(Z –1) m_e – m_ec^2\\=[m_{N}(_ZX^A)-m_{N}(_{z-1}Y^A)-2m_e]c^{2}\space\text{...(iii)}\\\text{where, m}_e \text{= mass of electron}\\\text{Energy released in Eq. (ii),}\\\text{Q}_{2}=[m_{N}(_zX^A)+m_e-m_N(_{z-1}Y^{A})]c^{2}\\=[m_{N}(_ZX^A) + Zm_{e} + m_{e}-m_{N}(_{z-1}Y^{A})-(Z-1) m_{e}-m_{e}]c^{2}\\=[m_{N}(_{Z}X^A)-m_{N(Z-1)}Y^{A}]c^{2} $$
= [mN(ZXA) – mN(Z–1)YA]c2 …(iv)
Here, if Q1 > 0 then Q2 > 0
i.e., if positron emission is energetically allowed electron capture is necessarily allowed.
But if Q2 > 0 does not necessarily mean that Q1 > 0. Hence, converse is not true.
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