NCERT Solutions for Class 12 Physics Chapter 13 Nuclei

# 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.

##### PTE Registrations 2024

Register now for PTE & Unlock 10% OFF : Use promo code: 'C360SPL10'. Limited Period Offer!

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
JEE Main Highest Scoring Chapters & Topics
Just Study 40% Syllabus and Score upto 100%

### 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

### Also Check NCERT Books and NCERT Syllabus here:

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.

## Upcoming School Exams

#### National Means Cum-Merit Scholarship

Application Date:01 August,2024 - 16 September,2024

Exam Date:19 September,2024 - 19 September,2024

Exam Date:20 September,2024 - 20 September,2024

Exam Date:26 September,2024 - 26 September,2024

Application Date:30 September,2024 - 30 September,2024

Edx
1113 courses
Coursera
804 courses
Udemy
394 courses
Futurelearn
222 courses
IBM
85 courses

## Explore Top Universities Across Globe

University of Essex, Colchester
Wivenhoe Park Colchester CO4 3SQ
University College London, London
Gower Street, London, WC1E 6BT
The University of Edinburgh, Edinburgh
Old College, South Bridge, Edinburgh, Post Code EH8 9YL
University of Bristol, Bristol
Beacon House, Queens Road, Bristol, BS8 1QU
University of Nottingham, Nottingham
University Park, Nottingham NG7 2RD

### Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

Hi,

The Medhavi National Scholarship Program, under the Human Resources & Development Mission (HRDM), offers financial assistance to meritorious students through a scholarship exam. To be eligible, candidates must be between 16 and 40 years old as of the last date of registration and have at least passed the 10th grade from a recognized board. Higher qualifications, such as 11th/12th grade, graduation, post-graduation, or a diploma, are also acceptable.

The scholarships are categorized based on the marks obtained in the exam: Type A for those scoring 60% or above, Type B for scores between 50% and 60%, and Type C for scores between 40% and 50%. The cash scholarships range from Rs. 2,000 to Rs. 18,000 per month, depending on the exam and the marks obtained.

Since you already have a 12th-grade qualification with 84%, you meet the eligibility criteria and can apply for the Medhavi Scholarship exam. Preparing well for the exam can increase your chances of receiving a higher scholarship.

Yuvan 01 September,2024

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

Hello student,

If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:

• No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
• You have to appear for the 2025 12th board exams.
• Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
• Aim to register before late October to avoid extra fees.
• Schools might not offer classes for private students, so focus on self-study or coaching.

Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

I hope this answer helps you. If you have more queries then feel free to share your questions with us, we will be happy to help you.

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