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Consider the giant power of nuclear power plants or how the stars shine-all this is thanks to the small nucleus situated in the centre of atoms. Despite being nothing even close to 10,000 times smaller than the atom, the nucleus possesses practically all the mass of an atom. Chapter 13 takes us on an interesting journey into the nucleus, and some of the key highlights pertaining to the matter include the atomic masses, nuclear binding energy, nuclear forces, radioactivity and nuclear energy, as important as it is to both theoretical and applied physics.
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NCERT Solutions of Class 12 Physics Chapter 13 are important to students studying in the CBSE board and those attempting to crack the competitive exams like JEE and NEET. The NCERT solutions, offered in PDF format, are easy to understand, with step-by-step explanations of all exercises and additional questions, which makes complex findings easier to understand. Practising with these solutions can assist in reinforcing conceptual comprehension and confidence in the exam.
The Nuclear Physics chapter 13 NCERT Solutions presented the answers to all the questions with step-by-step explanations. These solutions will make the students grasp tricky topics without difficulty and can be used both in the CBSE board exams as well as JEE and NEET exams. The PDF version can also be downloaded so that you can have an easy offline reference and a quick revision anytime.
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Class 12 Physics Chapter 13 Nuclei: Exercise Solutions offer step-by-step answers to all textbook questions, covering topics like nuclear structure, binding energy, and radioactivity. These solutions help students strengthen concepts, practice effectively, and prepare for board exams as well as competitive exams like JEE and NEET.
Answer:
$
\begin{aligned}
& \mathrm{m}_{\mathrm{n}}=1.00866 \mathrm{u} \\
& \mathrm{~m}_{\mathrm{p}}=1.00727 \mathrm{u}
\end{aligned}
$
Atomic mass of Nitrogen $m=14.00307 u$
Mass defect $\Delta m=7 \times m_n+7 \times m_p-m$
$
\Delta m=7 \times 1.00866+7 \times 1.00727-14.00307
$
$\Delta m=0.10844$
Now 1u is equivalent to 931.5 MeV
$
\begin{aligned}
& E_b=0.10844 \times 931.5 \\
& E_b=101.01186 \mathrm{MeV}
\end{aligned}
$
Therefore binding energy of a Nitrogen nucleus is 101.01186 MeV
$m (_{26}^{56}\textrm{Fe})=55.934939\; \; u$
Answer:
$
\begin{aligned}
& m_H=1.007825 u \\
& m_n=1.008665 u
\end{aligned}
$
Atomic mass of ${ }_{26}^{56} \mathrm{Fe}$ is $\mathrm{m}=55.934939 \mathrm{u}$
Mass defect
$
\begin{aligned}
& \Delta m=(56-26) \times m_H+26 \times m_p-m \\
& \Delta m=30 \times 1.008665+26 \times 1.007825-55.934939 \\
& \Delta m=0.528461
\end{aligned}
$
Now 1 u is equivalent to 931.5 MeV
$
\begin{aligned}
& E_b=0.528461 \times 931.5 \\
& E_b=492.2614215 \mathrm{MeV}
\end{aligned}
$
Therefore binding energy of a ${ }_{26}^{56} \mathrm{Fe}$ nucleus is 492.2614215 MeV .
Average binding energy
$
=\frac{492.26}{56} \mathrm{MeV}=8.79 \mathrm{MeV}
$
$m(_{83}^{209}\textrm{Bi})=208.980388\; \; u$
Answer:
$
m_H=1.007825 u
$
$
m_n=1.008665 u
$
Atomic mass of ${ }_{83}^{209} \mathrm{Bi}$ is $\mathrm{m}=208.980388 \mathrm{u}$
Mass defect
$
\begin{aligned}
& \Delta m=(209-83)+83 \times m_H-m \\
& \Delta m=126 \times 1.008665+83 \times 1.007825-208.980388 \\
& \Delta m=1.760877 u
\end{aligned}
$
Now 1u is equivalent to 931.5 MeV
$
\begin{aligned}
& E_b=1.760877 \times 931.5 \\
& E_b=1640.2569255 \mathrm{MeV}
\end{aligned}
$
Therefore binding energy of a ${ }_{83}^{209} \mathrm{Bi}$ nucleus is 1640.2569255 MeV .
Average binding energy $=\frac{1640.25}{208.98}=7.84 \mathrm{MeV}$
Answer:
Mass of the coin is $w=3 g$
Total number of Cu atoms in the coin is n
$
\begin{aligned}
n & =\frac{w \times N_A}{\text { Atomic Mass }} \\
n & =\frac{3 \times 6.023 \times 10^{23}}{62.92960} \\
\mathrm{n} & =2.871 \times 10^{22} \\
\mathrm{~m}_{\mathrm{H}} & =1.007825 \mathrm{u} \\
\mathrm{~m}_{\mathrm{n}} & =1.008665 \mathrm{u}
\end{aligned}
$
Atomic mass of ${ }_{29}^{63} \mathrm{Cu}$ is $\mathrm{m}=62.92960 \mathrm{u}$
Mass defect $\Delta m=(63-29) \times m_n+29 \times m_H-m$
$\Delta m=34 \times 1.008665+29 \times 1.007825-62.92960$
$\Delta \mathrm{m}=0.591935 \mathrm{u}$
Now 1u is equivalent to 931.5 MeV
$
E_b=0.591935 \times 931.5
$
$\mathrm{E}_{\mathrm{b}}=551.38745 \mathrm{MeV}$
Therefore binding energy of a ${ }_{29}^{63} \mathrm{Cu}$ nucleus is 551.38745 MeV .
The nuclear energy that would be required to separate all the neutrons and protons from each other is
$
\begin{aligned}
& n \times E_b=2.871 \times 10^{22} \times 551.38745 \\
& =1.5832 \times 10^{25} \mathrm{MeV} \\
& =1.5832 \times 10^{25} \times 1.6 \times 10^{-19} \times 10^6 \mathrm{~J} \\
& =2.5331 \times 10^9 \mathrm{~kJ}
\end{aligned}
$
Answer:
The nuclear radii are directly proportional to the cube root of the mass number.
The ratio of the radii of the given isotopes is
$\left ( \frac{197}{107} \right )^{1/3}=1.23$
$m(_{1}^{2}\textrm{H})=2.014102\; u$
$m(_{1}^{3}\textrm{H})=3.0016049\; u$
$m(_{6}^{12}\textrm{H})=12.000000\; u$
$m(_{10}^{20}\textrm{Ne})=19.992439\; u$
Answer:
$\\\Delta m=m(_{1}^{1}\textrm{H})+m(_{1}^{3}\textrm{H})-2m(_{1}^{2}\textrm{H})\\ $
$\Delta m=1.007825+3.0016049-2\times 2.014102\\ $
$\Delta m=-0.00433$
The above negative value of mass defect implies there will be a negative Q value and therefore the reaction is endothermic
$_{6}^{12}\textrm{C}+_{6}^{12}\textrm{C}\rightarrow _{10}^{20}\textrm{Ne}+_{2}^{4}\textrm{He}$
$m(_{1}^{2}\textrm{H})=2.014102\; u$
$m(_{1}^{3}\textrm{H})=3.0016049\; u$
$m(_{6}^{12}\textrm{H})=12.000000\; u$
$m(_{10}^{20}\textrm{Ne})=19.992439\; u$
Answer:
$\\\Delta m=2m(_{6}^{12}\textrm{C})-m(_{10}^{20}\textrm{Ne})-m(_{2}^{4}\textrm{He})\\ $
$\Delta m=2\times 12.00000-19.992439-4.002603\\ $
$\Delta m=0.004958$
The above positive value of mass defect implies the Q value would be positive and therefore the reaction is exothermic.
Answer:
The reaction will be $_{26}^{56}\textrm{Fe}\rightarrow _{13}^{28}\textrm{Al}+_{13}^{28}\textrm{Al}$
The mass defect of the reaction will be
$\\\Delta m=m(_{26}^{56}\textrm{Fe})-2m( _{13}^{28}\textrm{Al})\\ $
$\Delta m=55.93494-2\times 27.98191\\ $
$\Delta m=-0.02888u$
Since the mass defect is negative, the Q value will also be negative and therefore, the fission is not energetically possible.
Answer:
Number of atoms present in $1 \mathrm{~kg}(\mathrm{w})$ of ${ }_{94}^{239} \mathrm{Pu}=\mathrm{n}$
$
\begin{aligned}
n & =\frac{w \times N_A}{\text { mass number of Pu }} \\
n & =\frac{1000 \times 6.023 \times 10^{23}}{239} \\
n & =2.52 \times 10^{24}
\end{aligned}
$
Energy per fission $(E)=180 \mathrm{MeV}$
Total Energy released if all the atoms in $1 \mathrm{~kg}{ }_{94}^{239} \mathrm{Pu}$ undergo fission $=\mathrm{E} \times \mathrm{n}$
$
\begin{aligned}
& =180 \times 2.52 \times 10^{24} \\
& =4.536 \times 10^{26} \mathrm{MeV}
\end{aligned}
$
$_{1}^{2}\textrm{H}+_{1}^{2}\textrm{H}\rightarrow _{2}^{3}\textrm{He}+n+3.27\; MeV$
Answer:
The energy liberated on the fusion of two atoms of deuterium is 3.27 MeV
Number of fusion reactions in 2 kg of deuterium = N A $\times$ 500
The energy liberated by the fusion of 2.0 kg of deuterium atoms E
$\\=3.27\times 10^{6}\times 1.6\times 10^{-19}\times 6.023\times 10^{23}\times 500\\=1.576\times 10^{14}\ J$
Power of lamp (P)= 100 W
The time for which the lamp would glow using E amount of energy is
$\\T=\frac{E}{P}\\ =\frac{1.576\times 10^{14}}{100\times 3600\times 24\times 365}$
=4.99 $\times$ $10^4$ years
Answer:
For a head-on collision of two deuterons, the closest distances between their centres will be $d=2 \times r$
$
\begin{aligned}
& d=2 \times 2.0 \\
& d=4.0 \mathrm{fm} \\
& d=4 \times 10^{-15} \mathrm{~m}
\end{aligned}
$
charge on each deuteron $=$ charge of one proton $=\mathrm{q}=1.6 \times 10^{-19} \mathrm{C}$
The maximum electrostatic potential energy of the system during the head-on collision will be E
$
\begin{aligned}
& =\frac{q^2}{4 \pi \epsilon_0 d} \\
& =\frac{9 \times 10^9 \times\left(1.6 \times 10^{-19}\right)^2}{4 \times 10^{-15}} J \\
& =\frac{9 \times 10^9 \times\left(1.6 \times 10^{-19}\right)^2}{4 \times 10^{-15} \times 1.6 \times 10^{-19}} \mathrm{eV} \\
& =360 \mathrm{keV}
\end{aligned}
$
The above basically means that bringing two deuterons from infinity to each other would require 360 keV of work to be done or would require 360 keV of energy to be spent.
Answer:
The mass of an element with mass number A will be about A u. The density of its nucleus, therefore, would be
$
\begin{aligned}
& d=\frac{m}{v} \\
& d=\frac{A}{\frac{4 \pi}{3} R^3} \\
& d=\frac{A}{\frac{4 \pi}{3}\left(R_0 A^{1 / 3}\right)^3} \\
& d=\frac{3}{4 \pi R_0{ }^3}
\end{aligned}
$
As we can see, the above density comes out to be independent of mass number $A$ and $R_0$ is constant, so matter density is nearly constant
Class 12 Physics Chapter 13 Nuclei: Additional Questions are designed to enhance conceptual clarity and problem-solving skills beyond the NCERT textbook. These questions cover nuclear forces, energy, and decay processes, making them highly useful for board exams and entrance tests like JEE/NEET.
Answer:
Mass of the two stable isotopes and their respective abundances are $6.01512 \; u$ and $7.01600 \; u$ and $7.5\; ^{o}/_{o}$ and $92.5\; ^{o}/_{o}$ .
$m=\frac{6.01512\times7.5+7.01600\times92.5}{100}$
m=6.940934 u
Answer:
The atomic mass of boron is 10.811 u
Mass of the two stable isotopes are $10.01294 u$ and $11.00931 u$ respectively
Let the two isotopes have abundances $\mathrm{x} \%$ and $(100-\mathrm{x}) \%$
$
\begin{aligned}
& 10.811=\frac{10.01294 \times x+11.00931 \times(100-x)}{100} \\
& x=19.89 \\
& 100-x=80.11
\end{aligned}
$
Therefore the abundance of ${ }_5^{10} \mathrm{~B}$ is $19.89 \%$ and that of ${ }_5^{11} \mathrm{~B}$ is $80.11 \%$
Answer:
The atomic masses of the three isotopes are $19.99 u\left(m_1\right), 20.99 u\left(m_2\right)$ and $21.99 u\left(m_3\right)$
Their respective abundances are $90.51 \%\left(p_1\right), 0.27 \%\left(p_2\right)$ and $9.22 \%\left(p_3\right)$
$
\begin{aligned}
& m=\frac{19.99 \times 90.51+20.99 \times 0.27+21.99 \times 9.22}{100} \\
& m=20.1771 u
\end{aligned}
$
The average atomic mass of neon is 20.1771 u .
3. (i) Write nuclear reaction equations for
$\; \alpha -decay\; of\; _{88}^{226}\textrm{Ra}$
Answer:
The nuclear reaction equations for the given alpha decay
$_{88}^{226}\textrm{Ra}\rightarrow _{86}^{222}\textrm{Rn}+_{2}^{4}\textrm{He}$
3. (ii) Write nuclear reaction equations for
$\; \alpha -decay\; of\; _{94}^{242}\textrm{Pu}$
Answer:
The nuclear reaction equations for the given alpha decay is
$_{94}^{242}\textrm{Pu}\rightarrow _{92}^{238}\textrm{U}+_{2}^{4}\textrm{He}$
3. (iii) Write nuclear reaction equations for
$\; \beta ^{-} -\: decay\; of\; _{15}^{32}\textrm{P}$
Answer:
The nuclear reaction equations for the given beta minus decay is
$_{15}^{32}\textrm{P}\rightarrow _{16}^{32}\textrm{S}+e^{-}+\bar{\nu }$
3.(iv) Write nuclear reaction equations for
$\; \beta ^{-} -\: decay\; of\; _{83}^{210}\textrm{Bi}$
Answer:
The nuclear reaction equation for the given beta minus decay is
$_{83}^{210}\textrm{Bi}\rightarrow _{84}^{210}\textrm{Po}+e^{-}+\bar{\nu }$
3.(v) Write nuclear reaction equations for
$\; \beta ^{+} -\: decay\; of\; _{6}^{11}\textrm{C}$
Answer:
The nuclear reaction for the given beta plus decay will be
$_{6}^{11}\textrm{C}\rightarrow _{5}^{11}\textrm{P}+e^{+}+\nu$
13.6 (vi) Write nuclear reaction equations for
$\; \beta ^{+} -\: decay\; of\; _{43}^{97}\textrm{Tc}$
Answer:
nuclear reaction equations for
$\beta ^{+} -\: decay\; of\; _{43}^{97}\textrm{Tc}\ is$
$_{43}^{97}\textrm{Tc}\rightarrow _{42}^{97}\textrm{Mo}+e^{+}+\nu$
Electron capture of $_{54}^{120}\textrm{Xe}$
Answer:
The nuclear reaction for electron capture of $_{54}^{120}\textrm{Xe}$ is
$_{54}^{120}\textrm{Xe}+e^{-}\rightarrow _{53}^{120}\textrm{I}+\nu$
Answer:
(a) The activity is proportional to the number of radioactive isotopes present
The number of half years in which the number of radioactive isotopes reduces to x% of its original value is n.
$n=log_{2}(\frac{100}{x})$
In this case
$n=log_{2}(\frac{100}{3.125})=log_{2}32=5$
It will take 5T years to reach 3.125% of the original activity.
(b) In this case
$n=log_{2}(\frac{100}{1})=log_{2}100=6.64$
It will take 6.64T years to reach 1% of the original activity.
Answer:
Since we know that activity is proportional to the number of radioactive isotopes present in the sample.
$
\frac{R}{R_0}=\frac{N}{N_0}=\frac{9}{15}=0.6
$
Also
$
\begin{aligned}
& N=N_0 e^{-\lambda t} \\
& t=-\frac{1}{\lambda} \ln \frac{N}{N_0} \\
& t=-\frac{1}{\lambda} \ln 0.6 \\
& t=\frac{0.51}{\lambda}
\end{aligned}
$
but $\lambda=\frac{0.693}{T_{1 / 2}}$
Therefore
$
\begin{aligned}
& t=0.51 \times \frac{T_{1 / 2}}{0.693} \\
& t=0.735 T_{1 / 2} \\
& t \approx 4217
\end{aligned}
$
The age of the Indus-Valley civilisation, calculated using the given specimen, is approximately 4217 years.
Answer:
Required activity $=8.0 \mathrm{mCi}$
$1 \mathrm{Ci}=3.7 \times 10^{10}$ decay s $\mathrm{s}^{-1}$
$8.0 \mathrm{mCi}=8 \times 10^{-3} \times 3.7 \times 10^{10}=2.96 \times 10^8$ decay s ${ }^{-1}$
$T_{1 / 2}=5.3$ years
$
\lambda=\frac{0.693}{T_{1 / 2}}
$
$
\lambda=\frac{0.693}{5.3 \times 365 \times 24 \times 3600}
$
$
\lambda=4.14 \times 10^{-9} s^{-1}
$
$
\frac{\mathrm{d} N}{\mathrm{~d} t}=-N \lambda
$
$
N=-\frac{\mathrm{d} N}{\mathrm{~d} t} \times \frac{1}{\lambda}
$
$
N=-\left(-2.96 \times 10^8\right) \times \frac{1}{4.14 \times 10^{-9}}
$
$
N=7.15 \times 10^{16} \text { atoms }
$
Mass of those many atoms of Cu will be
$
\begin{aligned}
& w=\frac{7.15 \times 10^{16} \times 60}{6.023 \times 10^{23}} \\
& w=7.12 \times 10^{-6} g
\end{aligned}
$
$7.12 \times 10^{-6} \mathrm{~g}$ of ${ }_{27}^{60} \mathrm{Co}$ is necessary to provide a radioactive source of 8.0 mCi strength.
Answer:
$T_{1 / 2}=28$ years
$
\begin{aligned}
\lambda & =\frac{0.693}{28 \times 365 \times 24 \times 3600} \\
\lambda & =7.85 \times 10^{-10} \text { decay } \mathrm{s}^{-1}
\end{aligned}
$
The number of atoms in 15 mg of ${ }_{38}^{90} \mathrm{Sr}$ is
$
\begin{aligned}
& N=\frac{15 \times 10^{-3} \times 6.023 \times 10^{23}}{90} \\
& N=1.0038 \times 10^{20}
\end{aligned}
$
The disintegration rate will be
$
\begin{aligned}
& \frac{\mathrm{d} N}{\mathrm{~d} t}=-N \lambda \\
& =-1.0038 \times 10^{20} \times 7.85 \times 10^{-10} \\
& =-7.88 \times 10^{10} \mathrm{~s}^{-1}
\end{aligned}
$
The disintegration rate is therefore $7.88 \times 10^{10}$ decay $\mathrm{s}^{-1}$.
8.(a) Find the Q-value and the kinetic energy of the emitted $\alpha$ -particle in the $\alpha$ -decay of
$(a)\; _{88}^{226}\textrm{Ra}$
Given $m(_{88}^{226}\textrm{Ra})=226.02540\; u,$ $m(_{86}^{222}\textrm{Rn})=222.01750\; u,$
$m(_{86}^{222}\textrm{Rn})=220.01137\; u,$ $m(_{84}^{216}\textrm{Po})=216.00189\; u,$
Answer:
Mass defect is $\Delta m$
$
\begin{aligned}
& \Delta m=m\left({ }_{88}^{226} \mathrm{Ra}\right)-m\left({ }_{86}^{222} \mathrm{Rn}\right)-m\left({ }_2^4 \mathrm{He}\right) \\
& \Delta \mathrm{m}=226.02540-222.0175-4.002603 \\
& \Delta \mathrm{~m}=0.005297 \mathrm{u} \\
& 1 \mathrm{u}=931.5 \mathrm{MeV} / \mathrm{c}^2 \\
& \text { Q-value }=\Delta \mathrm{m} \times 931.5 \\
& =4.934515 \mathrm{MeV}
\end{aligned}
$
By using Linear Momentum Conservation and Energy Conservation
Kinetic energy of alpha particle =
$
\begin{aligned}
& \frac{\text { mass of nucleus after decay }}{\text { mass of nucleus before decay }} \times Q-\text {value } \\
& =\frac{222.01750}{226.0254} \times 4.934515 \\
& =4.847 \mathrm{MeV}
\end{aligned}
$
8.(b) Find the Q-value and the kinetic energy of the emitted $\alpha$ -particle in the $\alpha$ -decay of
$(b)\; _{86}^{220}\textrm{Rn}$
Given $m(_{88}^{226}\textrm{Ra})=226.02540\; u,$ $m(_{86}^{222}\textrm{Rn})=222.01750\; u,$
$m(_{86}^{222}\textrm{Rn})=220.01137\; u,$ $m(_{84}^{216}\textrm{Po})=216.00189\; u,$
Answer:
Mass defect is $\Delta m$
$
\begin{aligned}
& \Delta m=m\left({ }_{88}^{226} \mathrm{Ra}\right)-m\left({ }_{86}^{222} \mathrm{Rn}\right)-m\left({ }_2^4 \mathrm{He}\right) \\
& \Delta \mathrm{m}=226.02540-222.0175-4.002603 \\
& \Delta \mathrm{~m}=0.005297 \mathrm{u} \\
& 1 \mathrm{u}=931.5 \mathrm{MeV} / \mathrm{c}^2 \\
& \text { Q-value }=\Delta \mathrm{m} \times 931.5 \\
& =4.934515 \mathrm{MeV}
\end{aligned}
$
By using Linear Momentum Conservation and Energy Conservation
Kinetic energy of alpha particle =
$
\begin{aligned}
& \frac{\text { mass of nucleus after decay }}{\text { mass of nucleus before decay }} \times Q-\text { value } \\
& =\frac{222.01750}{226.0254} \times 4.934515 \\
& =4.847 \mathrm{MeV}
\end{aligned}
$
9. The radionuclide $^{11}C$ decays according to
Answer:
If we use atomic masses
$\\\Delta m=m(_{6}^{11}\textrm{C})-m(_{5}^{11}\textrm{B})-2m_{e}\\ $
$\Delta m=11.011434-11.009305-2\times 0.000548\\ $
$\Delta m=0.001033u$
Q-value= 0.001033 $\times$ 931.5=0.9622 MeV, which is comparable with the maximum energy of the emitted positron.
$(i) m (_{10}^{23}\textrm{Ne} ) = 22.994466 \; u$
$(ii) m (_{11}^{23}\textrm{Na} ) = 22.089770 \; u$
Answer:
The $\beta$ decay equation is
$_{10}^{23}\textrm{Ne}\rightarrow _{11}^{23}\textrm{Na}+e^{-}+\bar{\nu }+Q$
$\\\Delta m=m(_{10}^{23}\textrm{Ne})-_{11}^{23}\textrm{Na}-m_{e}\\ $
$\Delta m=22.994466-22.989770\\ $
$\Delta m=0.004696u$
(we did not subtract the mass of the electron, as it is cancelled because of the presence of one more electron in the sodium atom)
Q=0.004696 $\times$ 931.5
Q=4.3743 eV
The emitted nucleus is way heavier than the $\beta$ particle and the energy of the antineutrino is also negligible therefore, the maximum energy of the emitted electron is equal to the Q value.
Answer:
The amount of energy liberated on fission of 1 $_{92}^{235}\textrm{U}$ atom is 200 MeV.
The amount of energy liberated on fission of 1g $_{92}^{235}\textrm{U}$
$\\=\frac{200\times 10^{6} \times 1.6\times 10^{-19}\times 6.023\times 10^{23}}{235}\\=8.2\times 10^{10}\ Jg^{-1}$
Total Energy produced in the reactor in 5 years
$\\=1000\times 10^{6}\times 0.8\times 5\times 365\times 24\times 3600\\ =1.261\times 10^{17}\ J$
Mass of $_{92}^{235}\textrm{U}$ which underwent fission, m
$=\frac{1.261\times 10^{17}}{8.2\times 10^{10}}$
=1537.8 kg
The amount present initially in the reactor = 2m
=2 $\times$ 1537.8
=3075.6 kg
$e^{+}+_{z}^{A}\textrm{X}\rightarrow _{Z-1}^{A}\textrm{Y}+v$
Answer:
For the electron capture, the reaction would be
$_{Z}^{A}\textrm{X}+e^{-}\rightarrow _{Z-1}^{A}\textrm{Y}+\nu +Q_{1}$
The mass defect and q value of the above reaction would be
$\\\Delta m_{1}=m(_{Z}^{A}\textrm{X})+m_{e}-m(_{Z-1}^{A}\textrm{Y})\\ Q_{1}=([m(_{Z}^{A}\textrm{X})-m(_{Z-1}^{A}\textrm{Y})]+m_{e})c^{2}$
where m N $(_{Z}^{A}\textrm{X})$ and m N $(_{Z-1}^{A}\textrm{Y})$ are the nuclear masses of elements X and Y respectively
For positron emission, the reaction would be
$_{Z}^{A}\textrm{X}\rightarrow _{Z-1}^{A}\textrm{Y}+e^{+}+\bar{\nu }+Q_{2}$
The mass defect and q value for the above reaction would be
$\\\Delta m_{2}=m(_{Z}^{A}\textrm{X})-m(_{Z-1}^{A}\textrm{Y})-m_{e}\\ Q_{2}=([m(_{Z}^{A}\textrm{X})-m(_{Z-1}^{A}\textrm{Y})]-m_{e})c^{2}$
From the above values, we can see that if Q 2 is positive, Q 1 will also be positive but Q 1 being positive does not imply that Q 2 will also be positive.
Answer:
Let the abundances of $_{12}^{25}\textrm{Mg}$ and $_{12}^{26}\textrm{Mg}$ be x and y respectively.
x+y+78.99=100
y=21.01-x
The average atomic mass of Mg is 24.312 u
$\begin{aligned} & 24.312=\frac{78.99 \times 23.98504+x \times 24.98584+(100-x) \times 25.98259}{100} \\ & x \approx 9.3 \\ & y=21.01-x \\ & y=21.01-9.3 \\ & y=11.71\end{aligned}$
The abundances of $_{12}^{25}\textrm{Mg}$ and $_{12}^{26}\textrm{Mg}$ are 9.3% and 11.71% respectively
$m(_{20}^{40}\textrm{Ca})=39.962591\; u$
$m(_{20}^{41}\textrm{Ca})=40.962278 \; u$
$m(_{13}^{26}\textrm{Al})=25.986895 \; u$
$m(_{13}^{27}\textrm{Al})=26.981541 \; u$
Answer:
The reaction showing the neutron separation is
$_{20}^{41}\textrm{Ca}+E\rightarrow _{20}^{40}\textrm{Ca}+_{0}^{1}\textrm{n}$
$\\E=(m(_{20}^{40}\textrm{Ca})+m(_{0}^{1}\textrm{n})-m(_{20}^{41}\textrm{Ca}))c^{2}\\ $
$E=(39.962591+1.008665-40.962278)c^{2}\\ $
$E=(0.008978)u\times c^{2}$
But 1u=931.5 MeV/$c^{2}$
Therefore E=(0.008978) $\times$ 931.5
E=8.363007 MeV
Therefore to remove a neutron from the $_{20}^{41}\textrm{Ca}$ nucleus 8.363007 MeV of energy is required
$m(_{20}^{40}\textrm{Ca})=39.962591\; u$
$m(_{20}^{41}\textrm{Ca})=40.962278 \; u$
$m(_{13}^{26}\textrm{Al})=25.986895 \; u$
$m(_{13}^{27}\textrm{Al})=26.981541 \; u$
Answer:
The reaction showing the neutron separation is
$_{13}^{27}\textrm{Al}+E\rightarrow _{13}^{26}\textrm{Al}+_{0}^{1}\textrm{n}$
$\\E=(m(_{13}^{26}\textrm{Ca})+m(_{0}^{1}\textrm{n})-m(_{13}^{27}\textrm{Ca}))c^{2}\\ $
$\\E=(25.986895+1.008665-26.981541)c^{2} \\$
$\\E=(0.014019)u\times c^{2}$
But 1u=931.5 MeV/$c^{2}$
Therefore E=(0.014019) $\times$ 931.5
E=13.059 MeV
Therefore to remove a neutron from the $_{13}^{27}\textrm{Al}$ nucleus 13.059 MeV of energy is required
Answer:
Let initially there be $\mathrm{N}_1$ atoms of ${ }_{15}^{32} \mathrm{P}$ and $\mathrm{N}_2$ atoms of ${ }_{15}^{33} \mathrm{P}$ and let their decay constants be $\lambda_1$ and $\lambda_2$ respectively Since initially the activity of ${ }_{15}^{33} \mathrm{P}$ is $1 / 9$ times that of ${ }_{15}^{32} \mathrm{P}$ we have
$
N_1 \lambda_1=\frac{N_2 \lambda_2}{9}
$
Let after time $t$ the activity of ${ }_{15}^{33} \mathrm{P}$ be 9 times that of ${ }_{15}^{32} \mathrm{P}$
$
N_1 \lambda_1 e^{-\lambda_1 t}=9 N_2 \lambda_2 e^{-\lambda_2 t}
$
Dividing equation (ii) by (i) and taking the natural $\log$ of both sides we get
$
\begin{aligned}
& -\lambda_1 t=\ln 81-\lambda_2 t \\
& t=\frac{\ln 81}{\lambda_2-\lambda_1}
\end{aligned}
$
where $\lambda_2=0.048 /$ day and $\lambda_1=0.027 /$ day
t comes out to be 208.5 days
16. Under certain circumstances, a nucleus can decay by emitting a particle more massive than an $\alpha$ -particle. Consider the following decay processes:
$_{88}^{223}\textrm{Ra}\rightarrow _{82}^{209}\textrm{Pb}+_{6}^{14}\textrm{C}$
$_{88}^{223}\textrm{Ra}\rightarrow _{86}^{219}\textrm{Rn}+_{2}^{4}\textrm{He}$
Calculate the Q-values for these decays and determine that both are energetically allowed.
Answer:
$_{88}^{223}\textrm{Ra}\rightarrow _{82}^{209}\textrm{Pb}+_{6}^{14}\textrm{C}$
$\\\Delta m=m(_{88}^{223}\textrm{Ra})-m(_{82}^{209}\textrm{Pb})-m(_{6}^{14}\textrm{C})\\ =223.01850-208.98107-14.00324 \\=0.03419u$
1 u = 931.5 MeV/$c^{2}$
Q=0.03419 $\times$ 931.5
=31.848 MeV
As the Q value is positive the reaction is energetically allowed
$_{88}^{223}\textrm{Ra}\rightarrow _{86}^{219}\textrm{Rn}+_{2}^{4}\textrm{He}$
$\\\Delta m=m(_{88}^{223}\textrm{Ra})-m(_{86}^{219}\textrm{Rn})-m(_{2}^{4}\textrm{He})\\ =223.01850-219.00948-4.00260 \\=0.00642u$
1 u = 931.5 MeV/$c^{2}$
Q=0.00642 $\times$ 931.5
=5.98 MeV
As the Q value is positive, the reaction is energetically allowed
$m(_{92}^{238}\textrm{U})=238.05079\; u$
$m(_{58}^{140}\textrm{Ce})=139.90543\; u$
$m(_{44}^{99}\textrm{Ru})= 98.90594\; u$
Answer:
The fission reaction given in the question can be written as
$_{92}^{238}\textrm{U}+_{0}^{1}\textrm{n}\rightarrow _{58}^{140}\textrm{Ce}+_{44}^{99}\textrm{Ru}+10e^{-}$
The mass defect for the above reaction would be
$\Delta m=m_{N}(_{92}^{238}\textrm{U})+m(_{0}^{1}\textrm{n})-m_{N}(_{58}^{140}\textrm{Ce})-m_{N}(_{44}^{99}\textrm{Ce})-10m_{e}$
In the above equation, mN represents nuclear masses
$\\\Delta m=m(_{92}^{238}\textrm{U})-92m_{e}+m(_{0}^{1}\textrm{n})-m(_{58}^{140}\textrm{Ce})+58m_{e}-m(_{44}^{99}\textrm{Ru})+44m_{e}-10m_{e} \\\Delta m=m(_{92}^{238}\textrm{U})+m(_{0}^{1}\textrm{n})-m(_{58}^{140}\textrm{Ce})-m(_{44}^{99}\textrm{Ru})\\ \Delta m=238.05079+1.008665-139.90543-98.90594\\ \Delta m=0.247995u$
but 1u =931.5 MeV/$c^{2}$
Q=0.247995 $\times$ 931.5
Q=231.007 MeV
Q value of the fission process is 231.007 MeV
18. (a)Consider the D–T reaction (deuterium-tritium fusion)
$m(_{1}^{2}\textrm{H})=2.014102\; u$
$m(_{1}^{3}\textrm{H})=3.016049\; u$
Answer:
The mass defect of the reaction is
$\\\Delta m=m(_{1}^{2}\textrm{H})+m(_{1}^{3}\textrm{H})-m(_{2}^{4}\textrm{He})-m(_{0}^{1}\textrm{n})\\ $
$\Delta m=2.014102+3.016049-4.002603-1.008665\\ $
$\Delta m=0.018883u$
1u = 931.5 MeV/$c^{2}$
Q=0.018883 $\times$ 931.5=17.59 MeV
18. (b) Consider the D–T reaction (deuterium-tritium fusion)
$_{1}^{2}\textrm{H}+_{1}^{3}\textrm{H}\rightarrow _{2}^{4}\textrm{He}+n$
Answer:
To initiate the reaction, both the nuclei would have to come in contact with each other.
Just before the reaction, the distance between their centres would be 4.0 fm.
The electrostatic potential energy of the system at that point would be
$\\U=\frac{q^{2}}{4\pi \epsilon _{0}d}\\ $
$U=\frac{9\times 10^{9}(1.6\times 10^{-19})^{2}}{4\times 10^{-15}}\\$
$U=5.76\times 10^{-14}J$
The same amount of Kinetic Energy K, would be required to overcome the electrostatic forces of repulsion to initiate the reaction
It is given that $K=2\times \frac{3kT}{2}$
Therefor,e the temperature required to initiate the reaction is
$\\T=\frac{K}{3k}\\ =\frac{5.76\times 10^{-14}}{3\times 1.38\times 10^{-23}}\\=1.39\times 10^{9}\ K$
Answer:
$\gamma _{1}$ decays from 1.088 MeV to 0 V
Frequency of $\gamma _{1}$ is
$\\\nu _{1}=\frac{1.088\times 10^{6}\times 1.6\times 10^{-19}}{6.62\times 10^{-34}}\\ $
$\nu _{1}=2.637\times 10^{20}\ Hz$
Planck's constant, h=6.62 $\times$ $10 ^{-34}$ Js,
$E=h\nu$
Similarly, we can calculate frequencies of $\gamma _{2}$ and $\gamma _{3}$
$\\\nu _{2}=9.988\times 10^{19}\ Hz\\ $
$\nu _{3}=1.639\times 10^{20}\ Hz$
The energy of the highest level would be equal to the energy released after the decay
Mass defect is
$\\\Delta m=m(_{79}^{196}\textrm{U})-m(_{80}^{196}\textrm{Hg})\\ $
$\Delta m=197.968233-197.966760\\$
$\Delta m=0.001473u$
We know 1u = 931.5 MeV/$c^{2}$
Q value= 0.001473 $\times$ 931.5=1.3721 MeV
The maximum Kinetic energy of $\beta _{1}^{-}$ would be 1.3721-1.088=0.2841 MeV
The maximum Kinetic energy of $\beta _{2}^{-}$ would be 1.3721-0.412=0.9601 MeV
Answer:
(a) ${ }_1^1 \mathrm{H}_1^1 \mathrm{H}+{ }_1^1 \mathrm{H}+{ }_1^1 \mathrm{H}+{ }_1^1 \mathrm{H} \rightarrow{ }_2^4 \mathrm{He}$
The above fusion reaction releases the energy of 26 MeV
Number of Hydrogen atoms in 1.0 kg of Hydrogen is $1000 \mathrm{~N}_{\mathrm{A}}$
Therefore $250 \mathrm{~N}_{\mathrm{A}}$ such reactions would take place
The energy released in the whole process is $\mathrm{E}_1$
$
\begin{aligned}
& =250 \times 6.023 \times 10^{23} \times 26 \times 10^6 \times 1.6 \times 10^{-19} \\
& =6.2639 \times 10^{14} J
\end{aligned}
$
(b) The energy released in fission of one ${ }_{92}^{235} \mathrm{U}$ atom is 200 MeV
Number of ${ }_{92}^{235} \mathrm{U}$ atoms present in 1 kg of ${ }_{92}^{235} \mathrm{U}$ is N
$
\begin{aligned}
& N=\frac{1000 \times 6.023 \times 10^{23}}{235} \\
& N=2.562 \times 10^{24}
\end{aligned}
$
The energy released on fission of N atoms is $\mathrm{E}_2$
$
\begin{aligned}
& E=2.562 \times 10^{24} \times 200 \times 10^6 \times 1.6 \times 10^{-19} \\
& E=8.198 \times 10^{13} J \\
& \frac{E_1}{E_2}=\frac{6.2639 \times 10^{14}}{8.198 \times 10^{13}} \approx 8
\end{aligned}
$
Answer:
Let the amount of energy produced using nuclear power per year in 2020 be E
$E=\frac{200000\times 10^{6}\times 0.1\times 365\times 24\times 3600}{0.25}\ J$
(Only 10% of the required electrical energy is to be produced by Nuclear power and only 25% of therm-nuclear is successfully converted into electrical energy)
The amount of Uranium required to produce this much energy is M
$=\frac{200000\times 10^{6}\times 0.1\times 365\times 24\times 3600\times 235}{0.25\times 200\times 10^{6}\times 1.6\times 10^{-19}\times 6.023\times 10^{23}\times 1000}$ (N A =6.023 $\times$ $10^{ 23}$ , Atomic mass of Uranium is 235 g)
=3.076 $\times$ $10^{4}$ kg
Class 12 Physics Chapter 13 Nuclei: Higher Order Thinking Skills (HOTS) Questions encourage students to apply concepts like nuclear binding energy, radioactivity, and nuclear reactions in complex scenarios. These advanced problems help in developing analytical skills and are very useful for competitive exam preparation.
Q1:
The energy released in the fusion of $2 \mathrm{~kg}$ of hydrogen deep in the sun is $\mathrm{E}_{\mathrm{H}}$ and the energy released in the fission of $2 \mathrm{~kg}$ of ${ }^{235} \mathrm{U}$ is $E_U$. The ratio $\frac{E_H}{E_U}$ is approximately :
(Consider the fusion reaction as $4{ }_1^1 \mathrm{H}+2 \mathrm{e}^{-} \rightarrow{ }_2^4 \mathrm{He}+2 \mathrm{v}+6 \gamma+26.7 \mathrm{MeV}$, energy released in the fission reaction of ${ }^{235} \mathrm{U}$ is $200 \mathrm{MeV}$ per fission nucleus and $\mathrm{N}_{\mathrm{A}}=6.023 \times 10^{23}$ )
Answer:
In each fusion reaction, $4{ }_1^1 \mathrm{H}$ nuclei are used.
Energy released per Nuclei of ${ }_1^1 \mathrm{H}=\frac{26.7}{4} \mathrm{MeV}$
$\therefore$ Energy released by $2 \mathrm{~kg}$ hydrogen $\left(\mathrm{E}_{\mathrm{H}}\right.$ )
$=\frac{2000}{1} \times \mathrm{N}_{\mathrm{A}} \times \frac{26.7}{4} \mathrm{MeV}$
$\therefore$ Energy released by $2 \mathrm{~kg}$ Cranium $\left(\mathrm{E}_{\mathrm{v}}\right)$
$=\frac{2000}{235} \times \mathrm{N}_{\mathrm{A}} \times 200 \mathrm{MeV}$
So,
$\frac{E_H}{E_V}=235 \times \frac{26.7}{4 \times 200}=7.84$
$\therefore$ Approximately close to 7.62
Q2:
The disintegration energy $Q$ for the nuclear fission of ${ }^{235} \mathrm{U} \rightarrow{ }^{140} \mathrm{Ce}+{ }^{94} \mathrm{Zr}+\mathrm{n}$ is _____ $\qquad$ $\mathrm{MeV}$.
Given atomic masses of
$\begin{aligned}
& { }^{235} \mathrm{U}: 235.0439 \mathrm{u} ;{ }^{140} \mathrm{Ce} ; 139.9054 \mathrm{u}, \\
& { }^{94} \mathrm{Zr}: 93.9063 \mathrm{u} ; \mathrm{n}: 1.0086 \mathrm{u},
\end{aligned}
$
Value of $\mathrm{c}^2=931 \mathrm{MeV} / \mathrm{u}$
Answer:
${ }^{235} \mathrm{U} \rightarrow{ }^{140} \mathrm{Ce}+{ }^{94} \mathrm{Zr}+\mathrm{n}$
Disintegration energy
$\begin{aligned}
\mathrm{Q} & =\left(\mathrm{m}_{\mathrm{R}}-\mathrm{m}_{\mathrm{p}}\right) \cdot \mathrm{c}^2 \\
\mathrm{~m}_{\mathrm{R}} & =235.0439 \mathrm{u} \\
\mathrm{m}_{\mathrm{p}} & =139.9054 \mathrm{u}+93.9063 \mathrm{u}+1.0086 \mathrm{u} \\
& =234.8203 \mathrm{u} \\
\mathrm{Q} & =(235.0439 \mathrm{u}-234.8203 \mathrm{u}) \mathrm{c}^2 \\
& =0.2236 \mathrm{c}^2 \\
& =0.2236 \times 931 \\
\mathrm{Q} & =208.1716
\end{aligned}
$
Q3:
The half-life of a radioactive isotope is 5.5 h. If there are initially $48 \times 12^{32}$ atoms of this isotope, the number of atoms of the isotope remaining after 22 h is -
Answer:
Use, $\mathrm{t}_{1 / 2}=\frac{\mathrm{t}}{\log \left(\frac{1}{2}\right)\left[\frac{\mathrm{N}(\mathrm{t})}{\mathrm{N}_0}\right]}$
$
\begin{aligned}
& 5.5=\frac{22}{\log \left(\frac{1}{2}\right) \frac{x}{48 \times 10^{32}}} \\
& \left(\frac{1}{2}\right) \frac{x}{48 \times 10^{32}}=\frac{1}{10000} \\
& \frac{x}{48 \times 10^{32}}=\frac{1}{4} \\
& x=12 \times 10^{32}
\end{aligned}
$
Q4:
The $\mathrm{k}_{\alpha}$ radiation of $\mathrm{M}_0(\mathrm{z}=42)$ has a wavelength of $0.71 \dot{\mathrm{~A}}$, the wave length of the corresponding radiation of $\mathrm{Cu}(z=29)$
Answer:
From Moseley's law for $\mathrm{K}_\alpha$ - line, we have
$
\begin{gathered}
\frac{1}{\lambda} \alpha(z-1)^2 \\
\therefore \frac{\lambda_{c u}}{\lambda m_0}=\frac{\left(z_{m_0}-1\right)^2}{\left(z_{c u}-1\right)^2}=\frac{(41)^2}{(28)^2} \\
\lambda m_0=0.71Å, \\
\therefore \lambda_{c u}=(0.71 Å) \times \frac{(41)^2}{(28)^2} \\
=1.52 Å
\end{gathered}
$
Q5:
The count rate from $100 \mathrm{~cm}^2$ of a radioactive liquid is c. Some of the liquid is now discarded. The count rate of the remaining liquid is found to be $\left(\frac{\mathrm{c}}{10}\right)$ after three half-lives. The volume of the remaining liquid in $\mathrm{cm}^2$ is:-
Answer:
Initial count rate (C R) for $1 \mathrm{~cm}^3$ of
$
\text { liquid }=\frac{c}{100}
$
After 3 half -lives, langle $R$ for $1 \mathrm{~cm}^3$ of liquid
$
=\frac{1}{8} \times \frac{\mathrm{c}}{100}
$
If $u$ is the volume of remaining liquid, then
$
\begin{aligned}
\mathrm{v} \times \frac{\mathrm{c}}{800} & =\frac{c}{10} \\
\therefore \quad \frac{\mathrm{v}}{80} & =1 \\
\mathrm{v} & =80
\end{aligned}
$
Class 12 Physics Chapter 13 Nuclei deals with the study of the atomic nucleus, its properties, and the forces that hold it together. The topics cover atomic masses, binding energy, nuclear force, radioactivity, and nuclear energy, which are crucial for understanding both natural processes and modern technology.
Class 12 Physics Chapter 13 Nuclei formulas include key equations related to nuclear radius, binding energy, decay laws, half-life, and activity. These formulas are essential for solving numerical problems in board exams as well as JEE and NEET.
Understand fundamental nuclear terms:
Atomic number (Z), Mass number (A), Neutrons (N = A – Z)
Nucleus structure: Protons and neutrons (nucleons)
Emphasise half-life, decay, reaction equations, and energy calculations
Beyond the NCERT, students preparing for JEE/NEET should focus on advanced concepts of the Nuclei chapter such as mass defect and binding energy curve analysis, nuclear reactions (fission & fusion), Q-value calculations, and detailed radioactive decay series. These topics strengthen problem-solving ability and are frequently tested in competitive exams.
Also Check:
Frequently Asked Questions (FAQs)
Yes the NCERT chapter Nuclei are important for both the exams. Both in NEET and JEE main syllabus the chapter Nuclei is present and 1 or 2 questions from the chapter can be expected for the exams. The questions discussed in the NCERT Solutions for the chapter Nuclei will give a better idea on how to use the formulas and give a better understanding of the concepts discussed.
NCERT solutions are important for the Board exam as they provide clear explanations, help in solving questions, cover all important topics, provide a structured approach to solving problems, and are designed with the exam pattern in mind, helping in exam-oriented preparation.
The nucleus is made up of protons, which are positively charged particles, and neutrons, which are neutral particles.
Isotopes are atoms of the same element that have the same number of protons but different numbers of neutrons in their nuclei.
For CBSE board exam from NCERT class, 12 chapters 13 around 4 to 6 marks questions can be expected. All topics of the NCERT syllabus for the chapter Nuclei should be covered for the CBSE board exam.
On Question asked by student community
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Yes, if you’re not satisfied with your marks even after the improvement exam, many education boards allow you to reappear as a private candidate next year to improve your scores. This means you can register independently, study at your own pace, and take the exams without attending regular classes. It’s a good option to improve your results and open up more opportunities for higher studies or careers. Just make sure to check the specific rules and deadlines of your education board so you don’t miss the registration window. Keep your focus, and you will do better next time.
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Yes, in the case that you appeared for the 2025 improvement exam and your roll number is different from what was on the previous year’s marksheet, the board will usually release a new migration certificate. This is because the migration certificate will reflect the most recent exam details, roll number and passing year. You can apply to get it from your board using the process prescribed by them either online or through your school/college.
Yes, if you miss the 1st CBSE exam due to valid reasons, then you can appear for the 2nd CBSE compartment exam.
From the academic year 2026, the board will conduct the CBSE 10th exam twice a year, while the CBSE 12th exam will be held once, as per usual. For class 10th, the second phase exam will act as the supplementary exam. Check out information on w hen the CBSE first exam 2026 will be conducted and changes in 2026 CBSE Board exam by clicking on the link .
If you want to change your stream to humanities after getting a compartment in one subject in the CBSE 12th Board Exam , you actually have limited options to qualify for your board exams. You can prepare effectively and appear in the compartment examination for mathematics again. If you do not wish to continue with the current stream, you can take readmission in the Humanities stream and start from Class 11th again, and continue studying for two more years to qualify for the 12th examination.
The GUJCET Merit List is prepared based on the Class 12th marks and GUJCET marks received by the students. CBSE students who are not from the Gujarat board can definitely compete with GSEB students, as their eligibility is decided based on the combined marks scored by them in GUJCET and the 12th board. The weightage of the GUJCET score is 40% and the weightage of the class 12 scores is 60%.
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