NCERT Solutions for Class 12 Physics Chapter 13 Nuclei

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# NCERT Solutions for Class 12 Physics Chapter 13 Nuclei

Edited By Vishal kumar | Updated on Sep 13, 2023 08:46 AM IST | #CBSE Class 12th

## NCERT Solutions for Class 12 Physics Chapter 13 – Access and Download Free PDF

NCERT Solutions for Class 12 Physics Chapter 13 Nuclei serve as a crucial resource for achieving high scores in both board exams and competitive ones like JEE and NEET. On this NCERT solution page, you'll discover comprehensive and detailed class 12 nuclei ncert solutions to the entire exercise, ranging from questions 13.1 to 13.22 (exercise questions) and 13.23 to 13.31 (additional exercise questions). Additionally, these class 12 physics chapter 13 exercise solutions are conveniently available in PDF format, allowing students to access them offline, free from any internet constraints.

Do you know that the size of an atom is 10,000 times the size of a nucleus? But the nucleus contains 99.9% of the mass of an atom. We know that an atom has a structure. Does the nucleus also have a structure? If so what are the constituents and how they are arranged? All these questions are answered in Nuclei Class 12 NCERT text book.

Nuclei Class 12 chapter comes under modern Physics and you can expect at least one question for the board exam from NCERT Class 12 Physics chapter 13. Learning the class 12 physics chapter nuclei ncert solutions is important to score well in the board exam. The questions in NCERT Solutions for Class 12 Physics Chapter 13 Nuclei are divided into two parts namely exercise and additional exercise. Nuclei Class 12 NCERT solutions download PDF option is available to read the solution offline.

Free download physics chapter 13 class 12 ncert solutions pdf for CBSE exam.

## NCERT Solutions for Class 12 Physics Chapter 13 Nuclei

NCERT solutions for class 12 physics chapter 13 nuclei: Exercise Solution

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

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 x% and (100-x)%

$\\10.811=\frac{10.01294\times x+11.00931\times(100-x)}{100} \\x=19.89\\ 100-x=80.11$

Therefore the abundance of $_{5}^{10}\textrm{B}$ is 19.89% and that of $_{5}^{11}\textrm{B}$ is 80.11%

The atomic masses of the three isotopes are 19.99 u(m 1 ), 20.99 u(m 2 ) and 21.99u(m 3 )

Their respective abundances are 90.51%(p 1 ), 0.27%(p 2 ) and 9.22%(p 3 )

$\\m= \frac{19.99\times 90.51+20.99\times 0.27+21.99\times 9.22}{100}\\m=20.1771u$

The average atomic mass of neon is 20.1771 u.

m n = 1.00866 u

m p = 1.00727 u

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

E b =0.10844 $\times$ 931.5

E b =101.01186 MeV

Therefore binding energy of a Nitrogen nucleus is 101.01186 MeV.

m H = 1.007825 u

m n = 1.008665 u

The atomic mass of $_{26}^{56}\textrm{Fe}$ is m=55.934939 u

Mass defect

$\Delta m=(56-26)\times$ $m_H+26\times m_p - m$

$\Delta m=30\times1.008665+26\times1.007825 - 55.934939$

$\Delta$ m=0.528461

Now 1u is equivalent to 931.5 MeV

E b =0.528461 $\times$ 931.5

E b =492.2614215 MeV

Therefore the binding energy of a $_{26}^{56}\textrm{Fe}$ nucleus is 492.2614215 MeV.

Average binding energy

$=\frac{492.26}{56}MeV=8.79 MeV$

m H = 1.007825 u

m n = 1.008665 u

The atomic mass of $_{83}^{209}\textrm{Bi}$ is m=208.980388 u

Mass defect

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

$\Delta$ m=1.760877 u

Now 1u is equivalent to 931.5 MeV

E b =1.760877 $\times$ 931.5

E b =1640.2569255 MeV

Therefore the binding energy of a $_{83}^{209}\textrm{Bi}$ nucleus is 1640.2569255 MeV.

$Average\ binding\ energy=\frac{1640.25}{208.98}=7.84MeV$

Mass of the coin is w = 3g

Total number of Cu atoms in the coin is n

$\\n=\frac{w\times N_{A}}{Atomic\ Mass}\\ n=\frac{3\times 6.023\times 10^{23}}{62.92960}$

n=2.871 $\times$ 10 22

m H = 1.007825 u

m n = 1.008665 u

Atomic mass of $_{29}^{63}\textrm{Cu}$ is m=62.92960 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

$\Delta$ m=0.591935 u

Now 1u is equivalent to 931.5 MeV

E b =0.591935 $\times$ 931.5

E b =551.38745 MeV

Therefore binding energy of a $_{29}^{63}\textrm{Cu}$ nucleus is 551.38745 MeV.

The nuclear energy that would be required to separate all the neutrons and protons from each other is

n $\times$ E b =2.871 $\times$ 10 22 $\times$ 551.38745

=1.5832 $\times$ 10 25 MeV

=1.5832 $\times$ 10 25 $\times$ 1.6 $\times$ 10 -19 $\times$ 10 6 J

=2.5331 $\times$ 10 9 kJ

Q.13.6 (i) Write nuclear reaction equations for

The nuclear reaction equations for the given alpha decay

$_{88}^{226}\textrm{Ra}\rightarrow _{86}^{222}\textrm{Rn}+_{2}^{4}\textrm{He}$

Q.13.6 (ii) Write nuclear reaction equations for

The nuclear reaction equations for the given alpha decay is

$_{94}^{242}\textrm{Pu}\rightarrow _{92}^{238}\textrm{U}+_{2}^{4}\textrm{He}$

Q.13.6 (iii) Write nuclear reaction equations for

The nuclear reaction equations for the given beta minus decay is

$_{15}^{32}\textrm{P}\rightarrow _{16}^{32}\textrm{S}+e^{-}+\bar{\nu }$

Q.13.6 (iv) Write nuclear reaction equations for

The nuclear reaction equation for the given beta minus decay is

$_{83}^{210}\textrm{Bi}\rightarrow _{84}^{210}\textrm{Po}+e^{-}+\bar{\nu }$

Q.13.6 (v) Write nuclear reaction equations for

The nuclear reaction for the given beta plus decay will be

$_{6}^{11}\textrm{C}\rightarrow _{5}^{11}\textrm{P}+e^{+}+\nu$

Q.13.6 (vi) Write nuclear reaction equations for

nuclear reaction equations for

$\beta ^{+} -\: decay\; of\; _{43}^{97}\textrm{Tc}\ is$

$_{43}^{97}\textrm{Tc}\rightarrow _{42}^{97}\textrm{Mo}+e^{+}+\nu$

Q.13.6 (vii) Write nuclear reaction equations for

Electron capture of $_{54}^{120}\textrm{Xe}$

The nuclear reaction for electron capture of $_{54}^{120}\textrm{Xe}$ is

$_{54}^{120}\textrm{Xe}+e^{-}\rightarrow _{53}^{120}\textrm{I}+\nu$

(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.

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

$N=N_{0}e^{-\lambda t}$

$\\t=-\frac{1}{\lambda }ln\frac{N}{N_{0}}\\ t=-\frac{1}{\lambda }ln0.6\\ t=\frac{0.51}{\lambda }$

but $\lambda = \frac{0.693}{T_{1/2}}$

Therefore

$\\t=0.51\times \frac{T_{1/2}}{0.693}\\ \\t=0.735T_{1/2}$

$\\t\approx 4217$

The age of the Indus-Valley civilisation calculated using the given specimen is approximately 4217 years.

Required activity=8.0 mCi

1 Ci=3.7 $\times$ 10 10 decay s -1

8.0 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

$\lambda =\frac{0.693}{T_{1/2}}$

$\lambda =\frac{0.693}{5.3\times 365\times 24\times 3600}$

$\lambda =4.14\times 10^{-9}\ s^{-1}$

$\\\frac{\mathrm{d} N}{\mathrm{d} t}=-N\lambda \\ N=-\frac{\mathrm{d} N}{\mathrm{d} t}\times \frac{1}{\lambda }\\ N=-(-2.96\times 10^{8})\times \frac{1}{4.14\times 10^{-9}}\\ N=7.15\times 10^{16}\ atoms$

Mass of those many atoms of Cu will be

$w=\frac{7.15\times 10^{16}\times 60}{6.023\times 10^{23}}$

$w=7.12\times10^{-6} g$

7.12 $\times$ 10 -6 g of $_{27}^{60}\textrm{Co}$ is necessary to provide a radioactive source of 8.0 mCi strength.

T 1/2 =28 years

$\\\lambda =\frac{0.693}{28\times 365\times 24\times 3600}\\ \lambda =7.85\times 10^{-10} \ decay\ s^{-1}$

The number of atoms in 15 mg of $_{38}^{90}\textrm{Sr}$ is

$N=\frac{15\times 10^{-3}\times 6.023\times 10^{23}}{90}$

N=1.0038 $\times$ 10 20

The disintegration rate will be

$\frac{\mathrm{d} N}{\mathrm{d} t}=-N\lambda$

=-1.0038 $\times$ 10 20 $\times$ 7.85 $\times$ 10 -10

=-7.88 $\times$ 10 10 s -1

The disintegration rate is therefore 7.88 $\times$ 10 10 decay s -1 .

The nuclear radii are directly proportional to the cube root of the mass number.

The ratio of the radii of the given isotopes is therefore

$\left ( \frac{197}{107} \right )^{1/3}=1.23$

Mass defect is $\Delta$ m

$\Delta m=m(_{88}^{226}\textrm{Ra})-m(_{86}^{222}\textrm{Rn})-m(_{2}^{4}\textrm{He})$

$\Delta$ m=226.02540-222.0175-4.002603

$\Delta$ m=0.005297 u

1 u = 931.5 MeV/c 2

Q-value= $\Delta$ m $\times$ 931.5

=4.934515 MeV

By using Linear Momentum Conservation and Energy Conservation

The kinetic energy of alpha particle =

$\frac{mass\ of\ nucleus\ after\ decay}{mass\ of\ nucleus\ before\ decay}\times Q-value$

= $\frac{222.01750}{226.0254}\times 4.934515$

=4.847 MeV

Mass defect is $\Delta$ m

$\Delta m=m(_{86}^{222}\textrm{Rn})-m(_{84}^{216}\textrm{Po})-m(_{2}^{4}\textrm{He})$

$\Delta$ m=220.01137-216.00189-4.002603

$\Delta$ m=0.006877 u

1 u = 931.5 MeV/c 2

Q-value= $\Delta$ m $\times$ 931.5

=6.406 MeV

By using Linear Momentum Conservation and Energy Conservation

The kinetic energy of alpha particle =

$\frac{mass\ of\ nucleus\ after\ decay}{mass\ of\ nucleus\ before\ decay}\times Q-value$

= $\frac{216.00189}{220.01138}\times 6.406$

=6.289 MeV

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 a maximum energy of the emitted positron.

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 and therefore the maximum energy of the emitted electron is equal to the Q value.

$(i) _{1}^{1}\textrm{H}+_{1}^{3}\textrm{H}\rightarrow _{1}^{2}\textrm{H}+_{1}^{2}\textrm{H}$ the following

Atomic masses are given to be

$\\\Delta m=m(_{1}^{1}\textrm{H})+m(_{1}^{3}\textrm{H})-2m(_{1}^{2}\textrm{H})\\ \Delta m=1.007825+3.0016049-2\times 2.014102\\ \Delta m=-0.00433$

The above negative value of mass defect implies there will be a negative Q value and therefore the reaction is endothermic

Atomic masses are given to be

$\\\Delta m=2m(_{6}^{12}\textrm{C})-m(_{10}^{20}\textrm{Ne})-m(_{2}^{4}\textrm{He})\\ \Delta m=2\times 12.00000-19.992439-4.002603\\ \Delta m=0.004958$

The above positive value of mass defect implies Q value would be positive and therefore the reaction is exothermic

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 negative and therefore the fission is not energetically possible

Number of atoms present in 1 kg(w) of $_{94}^{239}\textrm{Pu}$ =n

$\\n=\frac{w\times N_{A}}{mass\ number\ of\ Pu}\\ n=\frac{1000\times 6.023\times 10^{23}}{239}\\n=2.52\times 10^{24}$

Energy per fission (E)=180 MeV

Total Energy released if all the atoms in 1 kg $_{94}^{239}\textrm{Pu}$ undergo fission = E $\times$ n

=180 $\times$ 2.52 $\times$ 10 24

=4.536 $\times$ 10 26 MeV

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

The energy liberated on the fusion of two atoms of deuterium= 3.27 MeV

Number of fusion reactions in 2 kg of deuterium = N A $\times$ 500

The energy liberated by 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

Time 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

For a head-on collision of two deuterons, the closest distances between their centres will be d=2 $\times$ r

d=2 $\times$ 2.0

d=4.0 fm

d=4 $\times$ 10 -15 m

charge on each deuteron = charge of one proton=q =1.6 $\times$ 10 -19 C

The maximum electrostatic potential energy of the system during the head-on collision will be E

$\\=\frac{q^{2}}{4\pi \epsilon _{0}d}\\ =\frac{9\times 10^{9}\times (1.6\times 10^{-19})^{2}}{4\times 10^{-15}}\ J\\ =\frac{9\times 10^{9}\times (1.6\times 10^{-19})^{2}}{4\times 10^{-15}\times 1.6\times 10^{-19}}\ eV\\=360\ keV$

The above basically means to bring 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.

Mass of an element with mass number A will be about A u. The density of its nucleus, therefore, would be

$\\d=\frac{m}{v}\\ d=\frac{A}{\frac{4\pi }{3}R^{3}}\\d=\frac{A}{\frac{4\pi }{3}(R_{0}A^{1/3})^{3}}\\d=\frac{3}{4\pi R{_{0}}^{3}}$

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

Show that if $\beta ^{+}$ emission is energetically allowed, electron capture is necessarily allowed but not vice–versa.

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 have to positive.

## NCERT solutions for class 12 physics chapter 13 nuclei: Additional Exercise Solution

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

$\\24.312=\frac{78.99\times 23.98504+x\times 24.98584+(100-x)\times 25.98259}{100}\\ x\approx 9.3\\ y=21.01-x\\ y=21.01-9.3\\ y=11.71$

The abundances of $_{12}^{25}\textrm{Mg}$ and $_{12}^{26}\textrm{Mg}$ are 9.3% and 11.71% respectively

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

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

Let initially there be N 1 atoms of $_{15}^{32}\textrm{P}$ and N 2 atoms of $_{15}^{33}\textrm{P}$ and let their decay constants be $\lambda _{1}$ and $\lambda _{2}$ respectively

Since initially the activity of $_{15}^{33}\textrm{P}$ is 1/9 times that of $_{15}^{32}\textrm{P}$ we have

$N_{1 } \lambda_{1}=\frac{N_{2}\lambda _{2}}{9}$ (i)

Let after time t the activity of $_{15}^{33}\textrm{P}$ be 9 times that of $_{15}^{32}\textrm{P}$

$N_{1 } \lambda_{1}e^{-\lambda _{1}t}=9N_{2}\lambda _{2}e^{-\lambda _{2}t}$ (ii)

Dividing equation (ii) by (i) and taking the natural log of both sides we get

$\\-\lambda _{1}t=ln81-\lambda _{2}t \\t=\frac{ln81}{\lambda _{2}-\lambda _{1}}$

where $\lambda _{2}=0.048/ day$ and $\lambda _{1}=0.027/ day$

t comes out to be 208.5 days

$_{88}^{223}\textrm{Ra}\rightarrow _{82}^{209}\textrm{Pb}+_{6}^{14}\textrm{C}$

$\\\Delta m=m(_{88}^{223}\textrm{Ra})-m(_{82}^{209}\textrm{Pb})-m(_{6}^{14}\textrm{C})\\ =223.01850-208.98107-14.00324 \\=0.03419u$

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

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, m N 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

$_{1}^{2}\textrm{H}+_{1}^{3}\textrm{H}\rightarrow _{2}^{4}\textrm{He}+n$

(a) Calculate the energy released in MeV in this reaction from the data:

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

(b) 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.)

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 $K=2\times \frac{3kT}{2}$

Therefore the temperature required to initiate the reaction is

$\\T=\frac{K}{3k}\\ =\frac{5.76\times 10^{-14}}{3\times 1.38\times 10^{-23}}\\=1.39\times 10^{9}\ K$

$m(^{198}Au)=197.968233\; u$

$m(^{198}Hg)=197.966760 \; u$

$\gamma _{1}$ decays from 1.088 MeV to 0 V

Frequency of $\gamma _{1}$ is

$\\\nu _{1}=\frac{1.088\times 10^{6}\times 1.6\times 10^{-19}}{6.62\times 10^{-34}}\\ \nu _{1}=2.637\times 10^{20}\ Hz$ Plank'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

(a) $_{1}^{1}\textrm{H}$ $_{1}^{1}\textrm{H}+_{1}^{1}\textrm{H}+_{1}^{1}\textrm{H}+_{1}^{1}\textrm{H}\rightarrow _{2}^{4}\textrm{He}$

The above fusion reaction releases the energy of 26 MeV

Number of Hydrogen atoms in 1.0 kg of Hydrogen is 1000N A

Therefore 250N A such reactions would take place

The energy released in the whole process is E 1

$\\=250\times 6.023\times 10^{23}\times 26\times 10^{6}\times 1.6\times 10^{-19}\\=6.2639\times 10^{14}\ J$

(b) The energy released in fission of one $_{92}^{235}\textrm{U}$ atom is 200 MeV

Number of $_{92}^{235}\textrm{U}$ atoms present in 1 kg of $_{92}^{235}\textrm{U}$ is N

$\\N=\frac{1000\times 6.023\times 10^{23}}{235}\\ N=2.562\times 10^{24}$

The energy released on fission of N atoms is E 2

$\\E=2.562\times 10^{24}\times 200\times 10^{6}\times 1.6\times 10^{-19}\\ E=8.198\times 10^{13}J$

$\frac{E_{1}}{E_{2}}=\frac{6.2639\times 10^{14}}{8.198\times 10^{13}}\approx 8$

Let the amount of energy to be produced using nuclear power per year in 2020 is 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)

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

This chapter revolves around the topic of 'Nuclei,' and class 12 physics chapter 13 exercise solutions are vital for several reasons. Firstly, understanding nuclear physics is crucial as it forms the basis for numerous advanced concepts. Secondly, scoring in this chapter can be relatively easier compared to some other physics chapters, provided one grasps the fundamentals. The exercise solutions provide a comprehensive breakdown of each question, aiding students in mastering this essential topic effectively and efficiently, even within a limited time frame.

## NCERT solutions for class 12 physics chapter-wise

 NCERT solutions for class 12 physics chapter 1 Electric Charges and Fields NCERT solutions for class 12 physics chapter 2 Electrostatic Potential and Capacitance NCERT solutions for class 12 physics chapter 3 Current Electricity NCERT solutions for class 12 physics chapter 4 Moving Charges and Magnetism NCERT solutions for class 12 physics chapter 5 Magnetism and Matter NCERT solutions for class 12 physics chapter 6 Electromagnetic Induction NCERT solutions for class 12 physics chapter 7 Alternating Current NCERT solutions for class 12 physics chapter8 Electromagnetic Waves NCERT solutions for class 12 physics chapter 9 Ray Optics and Optical Instruments NCERT solutions for class 12 physics chapter 10 Wave Optics Solutions NCERT solutions for class 12 physics chapter 11 Dual nature of radiation and matter NCERT solutions for class 12 physics chapter 12 Atoms NCERT solutions for class 12 physics chapter 13 Nuclei NCERT solutions for class 12 physics chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits

### Class 12 Physics Chapter Nuclei NCERT Solutions: Important Formulas and Diagrams

The following formulas will be helpful to understand the NCERT solutions for Class 12 Physics chapter 13 Nuclei

• Radii of the nuclei, $R=R_0A^{\frac{1}{3}}$

Where A is the mass number and $R_0=1.2fm$

• Mass defect: $\Delta M=(Zm_p+(A-Z)m_n)-M$

Here Z is the atomic number, M is the mass of the nucleus and A is the mass number. This equation tells that the mass of the nucleus is always less than the mass of their constituents.

$E_b=\Delta Mc^2$

Where c is the speed of light.

• Another main concept of NCERT is the law of radioactive decay. This is given by $N=N_0e^{-\lambda t}$

Where N is the number of nuclei at any time t, $N_0$ is the number of nuclei at any time $t_0$ and lambda is disintegration constant.

• The half-life of a radionuclide is given by $T_{\frac{1}{2}}=\frac{ln2}{\lambda}$

Where lambda is the disintegration constant.

After completing all these topics try to do NCERT class 12 chapter 13 exercises. If you are unable to solve or have any doubts refer to the solutions of NCERT class 12 physics chapter 13 nuclei provided below.

## Significance of NCERT solutions for class 12 physics chapter 13 nuclei:

• About 6% of questions are expected from the chapters atoms and NCERT Class 12th Physics Nuclei for CBSE board exams.
• The NCERT Class 12 Physics solutions chapter 13 will help to score well in this chapter.
• The topic of radioactive decay and half-life is important for the board and competitive exams like NEET and JEE Main.
• Sometimes same questions which are discussed in the solutions of NCERT class 12 Physics chapter 13 solutions come in the CBSE12th exam.

## Key Features of Physics Chapter 13 Class 12 NCERT Solutions

1. Comprehensive Coverage: These class 12 nuclei ncert solutions encompass all topics and questions found in Chapter 13, ensuring a thorough understanding of nuclear physics.

2. Detailed Explanations: Each nuclei class 12 ncert solutions offers comprehensive, step-by-step explanations, making complex nuclear physics concepts accessible to students.

3. Clarity and Simplicity: The class 12 physics chapter nuclei ncert solutions are presented in clear and straightforward language, ensuring ease of understanding.

4. Practice Questions: Exercise questions are included for practice and self-assessment, enhancing students' problem-solving skills.

5. Exam Preparation: These physics chapter 13 class 12 ncert solutions are essential for board exam preparation and provide valuable support for competitive exams.

6. Foundation for Advanced Study: The concepts explored in this chapter serve as the foundation for more advanced studies in nuclear physics and related fields.

7. Free Access: These solutions are available for free, ensuring accessibility to all students

## NCERT solutions subject wise

NCERT Exemplar Class 12 Solutions

 NCERT Exemplar Class 12 Chemistry Solutions NCERT Exemplar Class 12 Mathematics Solutions NCERT Exemplar Class 12 Biology Solutions NCERT Exemplar Class 12 Physics Solutions

### Frequently Asked Question (FAQs)

1. What is the weightage of the chapter nuclei for CBSE board exam

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.

2. Is the chapter Nuclei important for NEET and JEE Main

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.

3. What is the composition of the nucleus according to nuclei ncert solutions?

The nucleus is made up of protons, which are positively charged particles, and neutrons, which are neutral particles.

4. According to nuclei class 12 what is isotopes?

Isotopes are atoms of the same element that have the same number of protons but different numbers of neutrons in their nuclei.

5. How nuclei class 12 ncert solutions is important for Board?

nuclei 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.

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### Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

I hope this was helpful!

Good Luck

Hello dear,

If you was not able to clear 1st compartment and now you giving second compartment so YES, you can go for your improvement exam next year but if a student receives an improvement, they are given the opportunity to retake the boards as a private candidate the following year, but there are some requirements. First, the student must pass all of their subjects; if they received a compartment in any subject, they must then pass the compartment exam before being eligible for the improvement.

As you can registered yourself as private candidate for giving your improvement exam of 12 standard CBSE(Central Board of Secondary Education).For that you have to wait for a whole year which is bit difficult for you.

Positive side of waiting for whole year is you have a whole year to preparing yourself for your examination. You have no distraction or something which may causes your failure in the exams. In whole year you have to stay focused on your 12 standard examination for doing well in it. By this you get a highest marks as a comparison of others.

Believe in Yourself! You can make anything happen

All the very best.

Hello Student,

I appreciate your Interest in education. See the improvement is not restricted to one subject or multiple subjects  and  we cannot say if improvement in one subject in one year leads to improvement in more subjects in coming year.

You just need to have a revision of all subjects what you have completed in the school. have a revision and practice of subjects and concepts helps you better.

All the best.

If you'll do hard work then by hard work of 6 months you can achieve your goal but you have to start studying for it dont waste your time its a very important year so please dont waste it otherwise you'll regret.

Yes, you can take admission in class 12th privately there are many colleges in which you can give 12th privately.

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9

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3 Jobs Available
##### Content Writer

Content writing is meant to speak directly with a particular audience, such as customers, potential customers, investors, employees, or other stakeholders. The main aim of professional content writers is to speak to their targeted audience and if it is not then it is not doing its job. There are numerous kinds of the content present on the website and each is different based on the service or the product it is used for.

2 Jobs Available
##### Reporter

Individuals who opt for a career as a reporter may often be at work on national holidays and festivities. He or she pitches various story ideas and covers news stories in risky situations. Students can pursue a BMC (Bachelor of Mass Communication), B.M.M. (Bachelor of Mass Media), or MAJMC (MA in Journalism and Mass Communication) to become a reporter. While we sit at home reporters travel to locations to collect information that carries a news value.

2 Jobs Available
##### Linguist

Linguistic meaning is related to language or Linguistics which is the study of languages. A career as a linguistic meaning, a profession that is based on the scientific study of language, and it's a very broad field with many specialities. Famous linguists work in academia, researching and teaching different areas of language, such as phonetics (sounds), syntax (word order) and semantics (meaning).

Other researchers focus on specialities like computational linguistics, which seeks to better match human and computer language capacities, or applied linguistics, which is concerned with improving language education. Still, others work as language experts for the government, advertising companies, dictionary publishers and various other private enterprises. Some might work from home as freelance linguists. Philologist, phonologist, and dialectician are some of Linguist synonym. Linguists can study French, German, Italian

2 Jobs Available
##### Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available
##### Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product.

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available
##### Production Manager

Production Manager Job Description: A Production Manager is responsible for ensuring smooth running of manufacturing processes in an efficient manner. He or she plans and organises production schedules. The role of Production Manager involves estimation, negotiation on budget and timescales with the clients and managers.

Resource Links for Online MBA

3 Jobs Available

A Team Leader is a professional responsible for guiding, monitoring and leading the entire group. He or she is responsible for motivating team members by providing a pleasant work environment to them and inspiring positive communication. A Team Leader contributes to the achievement of the organisation’s goals. He or she improves the confidence, product knowledge and communication skills of the team members and empowers them.

2 Jobs Available
##### Quality Systems Manager

A Quality Systems Manager is a professional responsible for developing strategies, processes, policies, standards and systems concerning the company as well as operations of its supply chain. It includes auditing to ensure compliance. It could also be carried out by a third party.

2 Jobs Available
##### Merchandiser

A career as a merchandiser requires one to promote specific products and services of one or different brands, to increase the in-house sales of the store. Merchandising job focuses on enticing the customers to enter the store and hence increasing their chances of buying a product. Although the buyer is the one who selects the lines, it all depends on the merchandiser on how much money a buyer will spend, how many lines will be purchased, and what will be the quantity of those lines. In a career as merchandiser, one is required to closely work with the display staff in order to decide in what way a product would be displayed so that sales can be maximised. In small brands or local retail stores, a merchandiser is responsible for both merchandising and buying.

2 Jobs Available
##### Procurement Manager

The procurement Manager is also known as  Purchasing Manager. The role of the Procurement Manager is to source products and services for a company. A Procurement Manager is involved in developing a purchasing strategy, including the company's budget and the supplies as well as the vendors who can provide goods and services to the company. His or her ultimate goal is to bring the right products or services at the right time with cost-effectiveness.

2 Jobs Available
##### Production Planner

Individuals who opt for a career as a production planner are professionals who are responsible for ensuring goods manufactured by the employing company are cost-effective and meets quality specifications including ensuring the availability of ready to distribute stock in a timely fashion manner.

2 Jobs Available
##### Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available
##### ITSM Manager

ITSM Manager is a professional responsible for heading the ITSM (Information Technology Service Management) or (Information Technology Infrastructure Library) processes. He or she ensures that operation management provides appropriate resource levels for problem resolutions. The ITSM Manager oversees the level of prioritisation for the problems, critical incidents, planned as well as proactive tasks.

3 Jobs Available
##### Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack

3 Jobs Available
##### Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available
##### Computer System Analyst

Individuals in the computer systems analyst career path study the hardware and applications that are part of an organization's computer systems, as well as how they are used. They collaborate closely with managers and end-users to identify system specifications and business priorities, as well as to assess the efficiency of computer systems and create techniques to boost IT efficiency. Individuals who opt for a career as a computer system analyst support the implementation, modification, and debugging of new systems after they have been installed.

2 Jobs Available
##### Test Manager

A Test Manager is a professional responsible for planning, coordinating and controlling test activities. He or she develops test processes and strategies to analyse and determine test methods and tools for test activities. The test manager jobs involve documenting tests that have been carried out, analysing and evaluating software quality to determine further recommended procedures.

2 Jobs Available
##### Azure Developer

A career as Azure Developer comes with the responsibility of designing and developing cloud-based applications and maintaining software components. He or she possesses an in-depth knowledge of cloud computing and Azure app service.

2 Jobs Available
##### Deep Learning Engineer

A Deep Learning Engineer is an IT professional who is responsible for developing and managing data pipelines. He or she is knowledgeable about analyzing and storing data collected from various sources.  A Career as a Deep Learning Engineer needs to help the  data scientists and analysts to create effective data sets.

2 Jobs Available