NCERT Solutions for Class 12 Physics Chapter 13

NCERT Solutions for Class 12 Physics Chapter 13 - Nuclei

CBSE board examiners always look for well-managed, stepwise detailed, accurate, and clear answers. Most of the questions are further extensions of basic NCERT questions; that’s why students look for mostly Chapter 13 which are accurate and solved with simplification. This chapter includes multiple numbers and makes the calculation process a little bit difficult. Thus, in this chapter, Chapter 13 with easy methods and less calculation are most important.


Class 12 Physics Chapter 13 Chapter 13 has many NCERT questions and examples for problem-solving and practising. Thus, there are high chances that a number of new questions might be given in the examination, keeping these questions as the base.

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Physics Class 12 Chapter 13 part-1

Access NCERT Solutions for Class 12 Physics Chapter 13 – Nuclei

1. Questions and Answers

(a) Two stable isotopes of lithium ${}_{3}^{6}Li $and ${}_{3}^{7}Li$ have respective abundances of $7.5%$ and $92.5%$These isotopes have masses $6.01512u$and$7.01600u$, respectively. Find the atomic mass of lithium.

Ans: We are given the following information:

Mass of ${}_{3}^{6}Li$ lithium isotope, ${{m}_{1}}=6.01512u$

Mass of ${}_{3}^{7}Li$ lithium isotope, ${{m}_{2}}=7.01600u$

Abundance of${}_{3}^{6}Li$, ${{n}_{1}}=7.5%$

Abundance of  ${}_{3}^{7}Li$, ${{n}_{2}}=92.5%$

The atomic mass of lithium atom is given by,

$m=\frac{{{m}_{1}}{{n}_{1}}+{{m}_{2}}{{n}_{2}}}{{{n}_{1}}+{{n}_{2}}}$

Substituting the given values, we get,

$m=\frac{6.01512\times 7.5+7.01600\times 92.5}{92.5+7.5}$

$\therefore m=6.940934u$

Therefore, we found the atomic mass of lithium atoms to be $6.940934u$. 


(b) Boron has Two Stable Isotopes ${}_{5}^{10}B$ and${}_{5}^{11}B$. Their respective masses are $10.01294u$and$11.00931u$, and the atomic mass of boron is $10.811u$. Find the abundances of ${}_{5}^{10}B$ and ${}_{5}^{11}B$ .

Ans: We are given:

Mass of ${}_{5}^{10}B$ Boron isotope, ${{m}_{1}}=10.01294u$

Mass of ${}_{5}^{11}B$ lithium isotope, ${{m}_{2}}=11.00931u$

Abundance of${}_{5}^{10}B$, ${{n}_{1}}=x%$

Abundance of  ${}_{5}^{11}B$, ${{n}_{2}}=\left( 100-x \right)%$

We know the atomic mass of boron to be, $m=10.811u$

The atomic mass of lithium atom is given by,

$m=\frac{{{m}_{1}}{{n}_{1}}+{{m}_{2}}{{n}_{2}}}{{{n}_{1}}+{{n}_{2}}}$

Substituting the given values, we get,

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

$\Rightarrow 1081.11=10.01294x+1100.931-11.00931x$

$\therefore x=\frac{19.821}{0.99637}=19.89%$

And, $100-x=80.11%$

Therefore, we found the abundance of ${}_{5}^{10}B$and ${}_{5}^{11}B$to be $19.89%$ and $80.11%$respectively. 


2. The three stable isotopes of neon: ${}_{10}^{20}Ne,\text{ }{}_{10}^{21}Ne\text{ and }{}_{10}^{22}Ne$ have respective abundances of \[90.51%,\text{ }0.27%\text{ }and\text{ }9.22%.\]The atomic masses of the three isotopes are\[\mathbf{19}.\mathbf{99}\text{ }\mathbf{u},\text{ }\mathbf{20}.\mathbf{99}\text{ }\mathbf{u}\text{ }\mathbf{and}\text{ }\mathbf{21}.\mathbf{99}\text{ }\mathbf{u}\], respectively. Obtain the average atomic mass of neon.

Ans: We are given that:

Atomic mass of  ${}_{10}^{20}Ne$, ${{m}_{1}}=19.99u$

Abundance of ${}_{10}^{20}Ne$, ${{\eta }_{1}}=90.51%$

Atomic mass of${}_{10}^{21}Ne$, ${{m}_{2}}=20.99u$

Abundance of ${}_{10}^{21}Ne$, ${{\eta }_{2}}=0.27%$

Atomic mass of ${}_{10}^{22}Ne$, ${{m}_{3}}=21.99u$

Abundance of ${}_{10}^{22}Ne$, ${{\eta }_{3}}=9.22%$

The average atomic mass of neon could be given as,

$m=\frac{{{m}_{1}}{{\eta }_{1}}+{{m}_{2}}{{\eta }_{2}}+{{m}_{3}}{{\eta }_{3}}}{{{\eta }_{1}}+{{\eta }_{2}}+{{\eta }_{3}}}$

Substituting the given values, we get, 

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

$\therefore m=20.1771u$

The average atomic mass of neon is thus found to be $20.177u.$


3. Obtain the binding energy (in MeV) of a nitrogen nucleus $\left( {}_{7}^{14}N \right)$, given $m\left( {}_{7}^{14}N \right)=14.00307u$

Ans: We are given:

Atomic mass of nitrogen $\left( {}_{7}{{N}^{14}} \right)$, $m=14.00307u$

A nucleus of ${}_{7}{{N}^{14}}$nitrogen contains 7 protons and 7 neutrons.

Hence, the mass defect of this nucleus would be, $\Delta m=7{{m}_{H}}+7{{m}_{n}}-m$

Where,

Mass of a proton, ${{m}_{H}}=1.007825u$

Mass of a neutron, ${{m}_{n}}=1.008665u$

Substituting these values into the above equation, we get, 

$\Delta m=7\times 1.007825+7\times 1.008665-14.00307$

$\Rightarrow \Delta m=7.054775+7.06055-14.00307$

$\therefore \Delta m=0.11236u$

But we know that, $1u=931.5MeV/{{c}^{2}}$

$\Rightarrow \Delta m=0.11236\times 931.5MeV/{{c}^{2}}$

Now, we could give the binding energy as, 

${{E}_{b}}=\Delta m{{c}^{2}}$

Where, $c=\text{speed of light =3}\times \text{1}{{\text{0}}^{8}}m{{s}^{-2}}$

Now, ${{E}_{b}}=0.11236\times 931.5\left( \frac{MeV}{{{c}^{2}}} \right)\times {{c}^{2}}$

$\therefore {{E}_{b}}=104.66334MeV$

Therefore, we found the binding energy of a Nitrogen nucleus to be $104.66334MeV$. 


4. Obtain the binding energy of the nuclei ${}_{26}^{56}Fe$ and ${}_{83}^{209}Bi$ in units of MeV from the following data: $m\left( {}_{26}^{56}Fe \right)=55.934939u$, $m\left( {}_{83}^{209}Bi \right)=208.980388u$

Ans: We are given the following:

Atomic mass of ${}_{26}^{56}Fe$, ${{m}_{1}}=55.934939u$

${}_{26}^{56}Fe$ nucleus has 26 protons and $56-26=30$neutrons

Hence, the mass defect of the nucleus would be, $\Delta m=26\times {{m}_{H}}+30\times {{m}_{n}}-{{m}_{1}}$

Where, Mass of a proton,${{m}_{H}}=1.007825u$

Mass of a neutron,${{m}_{n}}=1.008665u$

Substituting these values into the above equation, we get, 

$\Delta m=26\times 1.007825+30\times 1.008665-55.934939$

$\Rightarrow \Delta m=26.20345+30.25995-55.934939$

$\therefore \Delta m=0.528461u$

But we have, $1u=931.5MeV/{{c}^{2}}$

$\Delta m=0.528461\times 931.5MeV/{{c}^{2}}$

The binding energy of this nucleus could be given as,

${{E}_{b1}}=\Delta m{{c}^{2}}$

Where, c = Speed of light

$\Rightarrow {{E}_{b1}}=0.528461\times 931.5\left( \frac{MeV}{{{c}^{2}}} \right)\times {{c}^{2}}$

$\therefore {{E}_{b1}}=492.26MeV$

Now, we have the average binding energy per nucleon to be, 

$B.E=\frac{492.26}{56}=8.79MeV$

Also, atomic mass of ${}_{83}^{209}Bi$, ${{m}_{2}}=208.980388u$

We know that, ${}_{83}^{209}Bi$ nucleus has 83 protons and $209-83=126\text{ neutrons}$

Where,

Mass of a proton, ${{m}_{H}}=1.007825u$

Mass of a neutron, ${{m}_{n}}=1.008665u$

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

$\Rightarrow \Delta m'=83.649475+127.091790-208.980388$

$\therefore \Delta m'=1.760877u$

But we know, $1u=931.5MeV/{{c}^{2}}$

Hence, the binding energy of this nucleus could be given as,

${{E}_{b2}}=\Delta m'{{c}^{2}}=1.760877\times 931.5\left( \frac{MeV}{{{c}^{2}}} \right)\times {{c}^{2}}$

$\therefore {{E}_{b2}}=1640.26MeV$

Average binding energy per nucleon is found to be $=\frac{1640.26}{209}=7.848MeV$

Hence, the average binding energy per nucleon is found to be $7.848MeV$. 


5. A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of ${}_{29}^{63}Cu$ atoms (of mass\[\mathbf{62}.\mathbf{92960u}\]).

Ans: We are given:

Mass of a copper coin, $m'=3g$

Atomic mass of ${}_{29}C{{u}^{63}}$atom, \[m=62.92960u\]

The total number of ${}_{29}^{63}Cu$ atoms in the coin, $N=\frac{{{N}_{A}}\times m'}{\text{Mass number}}$

Where,\[{{N}_{A}}=\text{ }Avogadro\text{ }number=6.023\times \text{ }{{10}^{23}}atoms/g\]

Mass number = 63g

$\Rightarrow N=\frac{6.023\times {{10}^{23}}\times 3}{63}=2.868\times {{10}^{22}}$

${}_{29}C{{u}^{63}}$nucleus has 29 protons and \[\left( 63-29 \right)=34\]neutrons

Mass defect of this nucleus would be, $\Delta m'=29\times {{m}_{H}}+34\times {{m}_{n}}-m$

Where,

Mass of a proton, ${{m}_{H}}=1.007825u$

Mass of a neutron, ${{m}_{n}}=1.008665u$$\Delta m'=29\times 1.007825+34\times 1.008665-62.9296=0.591935u$

Mass defect of all the atoms present in the coin would be, 

$\Delta m=0.591935\times 2.868\times {{10}^{22}}=1.69766958\times {{10}^{22}}u$

But we have, $1u=931.5MeV/{{c}^{2}}$

$\Rightarrow \Delta m=1.69766958\times {{10}^{22}}\times 931.5MeV/{{c}^{2}}$

Hence, the binding energy of the nuclei of the coin could be given as:

${{E}_{b}}=\Delta m{{c}^{2}}=1.69766958\times {{10}^{22}}\times 931.5\left( \frac{MeV}{{{c}^{2}}} \right)\times {{c}^{2}}$

$\therefore {{E}_{b}}=1.581\times {{10}^{25}}MeV$

But, $1MeV=1.6\times {{10}^{-13}}J$

$\Rightarrow {{E}_{b}}=1.581\times {{10}^{25}}\times 1.6\times {{10}^{-13}}$

$\therefore {{E}_{b}}=2.5296\times {{10}^{12}}J$

This much energy is needed to separate all the neutrons and protons from the given coin.

6. Write the Nuclear Reactions For:

(a) $\alpha -\text{decay of }{}_{88}^{226}Ra$

Ans: We know that, $\alpha $is basically a nucleus of Helium $\left( {}_{2}H{{e}^{4}} \right)$and $\beta $is an electron $\left( {{e}^{-}}\text{ for }{{\beta }^{-}}\text{ and }{{\text{e}}^{+}}\text{ for }{{\beta }^{+}} \right)$. In every $\alpha $-decay, there is a loss of 2 protons and 2 neutrons. In every ${{\beta }^{+}}$-decay, there is a loss of 1 proton and a neutrino is emitted from the nucleus. In every ${{\beta }^{-}}$-decay, there is a gain of 1 proton and an antineutrino is emitted from the nucleus.

For the given case, the nuclear reaction would be, 

${}_{88}R{{a}^{226}}\to {}_{86}R{{n}^{222}}+{}_{2}H{{e}^{4}}$


(b) $\alpha -\text{decay of }{}_{94}^{242}Pu$

Ans: ${}_{94}^{242}Pu\to {}_{92}^{238}U+{}_{2}^{4}He$

  1. $\beta -\text{decay of }{}_{15}^{32}P$

Ans: ${}_{15}^{32}P\to {}_{16}^{32}S+{{e}^{-}}+\bar{\nu }$

  1. $\beta -\text{decay of }{}_{83}^{210}Bi$

Ans: ${}_{83}^{210}B\to {}_{84}^{210}Po+{{e}^{-}}+\bar{\nu }$

  1. ${{\beta }^{+}}-\text{decay of }{}_{6}^{11}C$

Ans: ${}_{6}^{11}C\to {}_{5}^{11}B+{{e}^{+}}+\nu $

  1. ${{\beta }^{+}}-\text{decay of }{}_{43}^{97}Tc$

Ans: ${}_{43}^{97}Tc\to {}_{42}^{97}Mo+{{e}^{+}}+\nu $

  1. Electron capture of ${}_{54}^{120}Xe$

Ans: ${}_{54}^{120}Xe+{{e}^{+}}\to {}_{53}^{120}I+\nu $


7. A radioactive isotope has a half-life of T years. How long will it take the activity to reduce to:

  1. 3.125% of Its Original Value? 

Ans: We are said that, Half-life of the radioactive isotope$=\text{ T years}$

Original amount of the radioactive isotope $={{N}_{0}}$

(a) After decay, let the amount of the radioactive isotope be N. 

It is given that only $3.125%$ of ${{N}_{0}}$remains after decay. Hence, we could write,

$\frac{N}{{{N}_{0}}}=3.125%=\frac{3.125}{100}=\frac{1}{32}$

But we know that, $\frac{N}{{{N}_{0}}}={{e}^{-\lambda t}}$

Where, $\lambda =$Decay constant and $t=$Time

$\Rightarrow -\lambda t=\frac{1}{32}$

$\Rightarrow -\lambda t=\ln l-\ln 32$

$\Rightarrow -\lambda t=0-3.4657$

$\Rightarrow t=\frac{3.4657}{\lambda }$

But, since $\lambda =\frac{0.693}{T}$

$\Rightarrow t=\frac{3.466}{\left( \frac{0.693}{T} \right)}$

$\therefore t\approx 5T\text{ years}$

Therefore, we found that the isotope will take about 5T years in order to reduce to $3.125%$ of its original value. 

  1. 1% of its original value?

Ans: After decay, let the amount of the radioactive isotope be N

It is given that only 1% of ${{N}_{0}}$ remains after decay. Hence, we could write:

$\frac{N}{{{N}_{0}}}=1%=\frac{1}{100}$

But we know, $\frac{N}{{{N}_{0}}}={{e}^{-\lambda t}}$

$\Rightarrow {{e}^{-\lambda t}}=\frac{1}{100}$

$\Rightarrow -\lambda t=\ln 1-\ln 100$

$\Rightarrow -\lambda t=0-4.602$

$\Rightarrow t=\frac{4.6052}{\lambda }$

Since we have, $\lambda =\frac{0.639}{T}$

$\Rightarrow t=\frac{4.6052}{\left( \frac{0.693}{T} \right)}$

$\therefore t=6.645T\text{ years}$

Therefore, we found that the given isotope would take about 6.645T years to reduce to 1% of its original value. 


8. The normal activity of living carbon-containing matter is found to be about 15 decays per minute for every gram of carbon. This activity arises from the small proportion of radioactive ${}_{6}^{14}C$ present with the stable carbon isotope ${}_{6}^{12}C$ . When the organism is dead, its interaction with the atmosphere (which maintains the above equilibrium activity) ceases and its activity begins to drop. From the known half-life (5730 years) of ${}_{6}^{14}C$ , and the measured activity, the age of the specimen can be approximately estimated. This is the principle of ${}_{6}^{14}C$ dating used in archaeology. Suppose a specimen from Mohenjodaro gives an activity of 9 decays per minute per gram of carbon. Estimate the approximate age of the Indus-Valley civilisation.

Ans: We are given that: 

Decay rate of living carbon-containing matter, \[R=15\text{ }decay/min\]

Let N be the number of radioactive atoms present in a normal carbon- containing matter.

Half life of ${}_{6}^{14}C,\text{ }{{\text{T}}_{\frac{1}{2}}}=5730\text{ years}$

The decay rate of the specimen obtained from the Mohenjodaro site:

\[R'=9\text{ }decays/min\]

Let N' be the number of radioactive atoms present in the specimen during the Mohenjodaro period.

Therefore, we can relate the decay constant, $\lambda $ and time, t as:

$\frac{N'}{N}=\frac{R'}{R}={{e}^{-\lambda t}}$

$\Rightarrow {{e}^{-\lambda t}}=\frac{9}{15}=\frac{3}{5}$

$\Rightarrow -\lambda t={{\log }_{e}}\frac{3}{5}=-0.5108$

$\Rightarrow t=\frac{0.5108}{\lambda }$

But we know, 

$\lambda =\frac{0.693}{{{T}_{\frac{1}{2}}}}=\frac{0.693}{5730}$

$\Rightarrow t=\frac{0.5108}{\left( \frac{0.693}{5730} \right)}=4223.5\text{ years}$

Therefore, the approximate age of the Indus-Valley civilization is found to be 4223.5 years.


9. Obtain the Amount of ${}_{27}^{60}Co$ necessary to provide a radioactive source of $8.0\text{ mCi}$strength. The half-life of ${}_{27}^{60}Co$ is 5.3 years.

Ans: We know that, 

The strength of the radioactive source could be given as, 

$\frac{dN}{dt}=8.0mCi$

$\Rightarrow \frac{dN}{dt}=8\times {{10}^{-3}}\times 3.7\times {{10}^{10}}=29.6\times {{10}^{7}}decay/s$

Where, N is the required number of atoms. 

Half life of ${}_{27}^{60}Co$, ${{T}_{\frac{1}{2}}}=5.3years$

$\Rightarrow {{T}_{\frac{1}{2}}}=5.3\times 365\times 24\times 60\times 60=1.67\times {{10}^{8}}s$

For decay constant $\lambda $, we could give the rate of decay as, 

$\frac{dN}{dt}=\lambda N$

Where, $\lambda =\frac{0.693}{{{T}_{\frac{1}{2}}}}=\frac{0.693}{1.67\times {{10}^{8}}}{{s}^{-1}}$

$\Rightarrow N=\frac{1}{\lambda }\frac{dN}{dt}=\frac{29.6\times {{10}^{7}}}{\left( \frac{0.693}{1.67\times {{10}^{8}}} \right)}=7.133\times {{10}^{16}}atoms$

Now for ${}_{27}C{{o}^{60}}$, Mass of Avogadro number of atoms $=60g$

Then, mass of $7.133\times {{10}^{16}}atoms=\frac{60\times 7.133\times {{10}^{16}}}{6.023\times {{10}^{23}}}=7.106\times {{10}^{-6}}g$

Therefore, the amount of ${}_{27}C{{o}^{60}}$that is required for the purpose is $7.106\times {{10}^{-6}}g$. 


10. The half life of ${}_{38}^{90}Sr$ is 28years. What is the disintegration rate of 15mg of this isotope?

Ans: We know that, 

Half life of ${}_{38}^{90}Sr$, ${{t}_{\frac{1}{2}}}=28years=28\times 365\times 24\times 3600=8.83\times {{10}^{8}}s$

Mass of the isotope, $m=15mg$

90g of ${}_{38}^{90}Sr$ atom contains Avogadro number of atoms. So, 15mg of ${}_{38}^{90}Sr$ contains, 

$\frac{6.023\times {{10}^{23}}\times 15\times {{10}^{-3}}}{90}=1.0038\times {{10}^{20}}\text{number of atoms}$

Rate of disintegration would be, $\frac{dN}{dt}=\lambda N$

Where, $\lambda $is the decay constant given by, $\lambda =\frac{0.693}{8.83\times {{10}^{8}}}{{s}^{-1}}$

$\therefore \frac{dN}{dt}=\frac{0.693\times 1.0038\times {{10}^{20}}}{8.83\times {{10}^{8}}}=7.878\times {{10}^{10}}atoms/s$

Therefore, we found the disintegration rate of 15mg of given isotope to be $7.878\times {{10}^{10}}atoms/s$.


11. Obtain approximately the ratio of the nuclear radii of the gold isotope ${}_{79}^{197}Au$ and the silver isotope ${}_{47}^{107}Ag$.

Ans: We know that, 

Nuclear radius of the gold isotope ${}_{79}A{{u}^{197}}={{R}_{Au}}$

Nuclear radius of the silver isotope ${}_{47}A{{g}^{107}}={{R}_{Ag}}$

Mass number of gold, ${{A}_{Au}}=197$

Mass number of silver, ${{A}_{Ag}}=107$

We also know that the ratio of the radii of the two nuclei is related with their mass numbers as:

$\frac{{{R}_{Au}}}{{{R}_{Ag}}}={{\left(\frac{{{A}_{Au}}}{{{A}_{Ag}}} \right)}^{\frac{1}{3}}}=1.2256$

Hence, the ratio of the nuclear radii of the gold and silver isotopes is found to be about 1.23.


12. Find the Q-value and the kinetic energy of the emitted α-particle in the α-decay of: [Given:$m\left( {}_{88}^{226}Ra \right)=226.02540u,\text{ m}\left( {}_{89}^{222}Rn \right)=222.01750u,\text{ }$ $\text{m}\left( {}_{86}^{220}Rn \right)=220.01137u,\text{ m}\left( {}_{84}^{216}Po \right)=216.00189u$ ]

  1. ${}_{88}^{226}Ra$

Ans: We know that, 

Alpha particle decay of ${}_{88}^{26}Ra$ emits a helium nucleus. As a result, its mass number reduces to $222=\left( 226-4 \right)$and its atomic number reduces to$86=\left( 88-2 \right)$. This is shown in the following nuclear reaction:

${}_{88}^{226}Ra\to {}_{86}^{222}Ra+{}_{2}^{4}He$

\[Q-value\text{ }of\text{ }emitted\text{ }\alpha -particle\text{ }=\text{ }\left( Sum\text{ }of\text{ }initial\text{ }mass\text{ }-\text{ }Sum\text{ }of\text{ }final\text{ }mass \right){{c}^{2}}\]Where, c = Speed of light 

It is also given that:

$m\left( {}_{88}^{226}Ra \right)=226.02540u$

$\text{m}\left( {}_{86}^{220}Rn \right)=220.01137u$

$m\left( {}_{2}^{4}He \right)=4.002603u$

On substituting these values into the above equation, 

$\text{Q value =}\left[ 226.02540-\left( 222.01750+4.002603 \right) \right]u{{c}^{2}}$

$Q\text{ value=0}\text{.005297u}{{\text{c}}^{2}}$

But we know, $1u=931.5MeV/{{c}^{2}}$

$\Rightarrow Q=0.005297\times 931.5\approx 4.94MeV$

Kinetic energy of the $\alpha $particle$=\left( \frac{\text{Mass number after decay}}{Mass\text{ number before decay}} \right)\times Q$

$\therefore K.{{E}_{\alpha }}=\frac{222}{226}\times 4.94=4.85MeV$

Hence, the Kinetic energy of the alpha particle is found to be $4.85MeV$.


  1. ${}_{86}^{220}Rn$

Ans: We know that, Alpha particle decay of ${}_{86}^{220}Rn$ could be given as, 

${}_{86}^{220}Rn\to {}_{84}^{216}Po+{}_{2}^{4}He$

We are also given, 

Mass of ${}_{86}^{220}Rn=220.01137u$

Mass of ${}_{84}^{216}Po=216.00189u$

Now, Q value could be given as, 

$Q-value=\left[ 220.01137-\left( 216.00189+4.00260 \right) \right]\times 931.5\approx 641MeV$

Now, we have the kinetic energy as, 

$K.{{E}_{\alpha }}=\left( \frac{220-4}{220} \right)\times 6.41=6.29MeV$

The kinetic energy of the alpha particle is found to be $6.29MeV$. 


13. The Radionuclide ${}_{6}^{11}C$ decays according to,

${}_{6}^{11}C\to {}_{5}^{11}B+{{e}^{+}}+\nu \text{ ; }{{\text{T}}_{\frac{1}{2}}}=20.3\min $

The maximum energy of the emitted positron is \[\mathbf{0}.\mathbf{960MeV}\].

Given the mass values: $m\left( {}_{6}^{11}C \right)=11.011434u\text{ and m}\left( {}_{6}^{11}B \right)=11.009305u$

Calculate Q and compare it with the maximum energy of the positron emitted. 

Ans: The given nuclear reaction is, 

${}_{6}^{11}C\to {}_{5}^{11}B+{{e}^{+}}+\nu $

Half life of ${}_{6}^{11}C$ nuclei, ${{T}_{\frac{1}{2}}}=20.3\min $

Atomic masses are given to be:

$m\left( {}_{6}^{11}C \right)=11.011434u$

$m\left( {}_{6}^{11}B \right)=11.009305u$

Maximum energy that is possessed by the emitted positron would be \[0.960MeV\]. The change in the Q - value \[\left( \Delta Q \right)\] of the nuclear masses of the ${}_{6}^{11}C$

$\Delta Q=\left[ m'\left( {}_{6}{{C}^{11}} \right)-\left[ m'\left( {}_{5}^{11}B \right)+{{m}_{e}} \right] \right]{{c}^{2}}$…………………….. (1)

Where,${{m}_{e}}=$Mass of an electron or positron = \[0.000548u\]

$c=$ Speed of light

$m'=$Respective nuclear masses

If atomic masses are used instead of nuclear masses, then we will have to add $6{{m}_{e}}$ in the case of ${}^{11}C$ and $5{{m}_{e}}$in case of ${}^{11}B$.

Hence, equation (1) would now reduce to,

$\Delta Q=\left[ m\left( {}_{6}{{C}^{11}} \right)-m\left( {}_{5}^{11}B \right)-2{{m}_{e}} \right]{{c}^{2}}$

Where, $m\left( {}_{6}{{C}^{11}} \right)\text{ and m}\left( {}_{5}^{11}B \right)$are the atomic masses. 

Now, we have the change in Q value as, 

$\Delta Q=\left[ 11.011434-11.009305-2\times 0.000548 \right]{{c}^{2}}=\left( 0.001033{{c}^{2}} \right)u$

But we know, $1u=931.5MeV/{{c}^{2}}$

$\therefore \Delta Q=0.001033\times 931.5\approx 0.962MeV$

We see that the Q value is almost comparable to the maximum energy of the emitted positron. 


14. The nucleus  ${}_{10}^{23}Ne$ decays by ${{\beta }^{-}}$emission. Write down the $\beta $decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:

$m\left( {}_{10}^{23}Ne \right)=22.994466u$

$m\left( {}_{11}^{23}Na \right)=22.989770u$

Ans: We know that: In ${{\beta }^{-}}$emission, the number of protons increases by 1, and one electron and an antineutrino are emitted from the parent nucleus. ${{\beta }^{-}}$emission of the nucleus could be given by, 

${}_{10}^{23}Ne\to {}_{11}^{23}Na+{{e}^{-}}+\bar{\nu }+Q$

It is also given that:

Atomic mass of ${}_{10}^{23}Ne=22.994466u$

Atomic mass of ${}_{11}^{23}Na=22.989770u$

Mass of an electron, ${{m}_{e}}=0.000548u$

Q value of the given reaction could be given as:

$Q=\left[ m\left( {}_{10}^{23}Ne \right)-\left[ m\left( {}_{11}^{23}Na \right)+{{m}_{e}} \right] \right]{{c}^{2}}$

There are 10 electrons in ${}_{10}N{{e}^{23}}$and 11 electrons in ${}_{11}^{23}Na$. Hence, the mass of the electron is cancelled in the Q-value equation.

$Q=\left[ 22.994466-22.9897770 \right]{{c}^{2}}=\left( 0.004696{{c}^{2}} \right)u$

But we have, $1u=931.5MeV/{{c}^{2}}$

$\Rightarrow Q=0.004696\times 931.5=4.374MeV$

The daughter nucleus is too heavy as compared to that of ${{e}^{-}}\text{ and }\bar{\nu }$. Hence, it carries negligible energy. The kinetic energy of the antineutrino is found to be nearly zero. 

Hence, the maximum kinetic energy of the emitted electrons is almost equal to the Q-value, i.e.,\[4.374MeV\].


15. The Q-value of a nuclear reaction $A+b\to C+d$ is defined by $Q=\left[ {{m}_{A}}+{{m}_{b}}-{{m}_{C}}-{{m}_{d}} \right]{{c}^{2}}$ where the masses refer to the respective nuclei. Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic.

Atomic masses are given to be: $m\left( {}_{1}^{2}H \right)=2.014102u,\text{ m}\left( {}_{1}^{3}H \right)=3.016049u,$

$m\left( {}_{6}^{12}C \right)=12.000000u,\text{ m}\left( {}_{10}^{20}Ne \right)=19.992439u$

  1. ${}_{1}^{1}H+{}_{1}^{3}H\to {}_{1}^{2}H+{}_{1}^{2}H$

Ans:The given nuclear reaction is:

${}_{1}^{1}H+{}_{1}^{3}H\to {}_{1}^{2}H+{}_{1}^{2}H$

Atomic mass of ${}_{1}^{1}H=1.007825u$

Atomic mass of ${}_{1}^{3}H=3.0164049u$

Atomic mass of ${}_{1}^{2}H=2.014102u$

According to the question, the Q-value of the reaction could be written as:

$Q=\left[ m\left( {}_{1}^{1}H \right)+m\left( {}_{1}^{3}H \right)-2m\left( {}_{1}^{2}H \right) \right]{{c}^{2}}$

$\Rightarrow Q=\left[ 1.007825+3.016049-2\times 2.014102 \right]{{c}^{2}}=\left( -0.00433{{c}^{2}} \right)u$

But we know, $1u=931.5MeV/{{c}^{2}}$

$\therefore Q=-0.00433\times 931.5=-4.0334MeV$

The negative Q-value of this reaction shows that the given reaction is endothermic. 


  1. ${}_{6}^{12}C+{}_{6}^{12}C\to {}_{10}^{20}Ne+{}_{2}^{4}He$

Ans: We are given that, 

Atomic mass of ${}_{6}^{12}C=12.0u$

Atomic mass of ${}_{10}^{12}Ne=19.992439u$

Atomic mass of ${}_{2}^{4}He=4.002603u$

The Q-value here could be given as, 

$Q=\left[ 2m\left( {}_{6}^{12}C \right)-m\left( {}_{10}^{20}Ne \right)-m\left( {}_{2}^{4}He \right) \right]{{c}^{2}}$

$\Rightarrow Q=\left[ 2\times 12.0-19.992439-4.002603 \right]{{c}^{2}}=\left( 0.004958{{c}^{2}} \right)u=0.004958\times 931.5$

$\therefore Q=4.618377MeV$

Since the Q-value is found to be positive, the reaction could be considered exothermic. 


16. Suppose, we think of fission of a ${}_{26}^{56}Fe$ nucleus into two equal fragments of ${}_{13}A{{l}^{28}}$. Is fission energetically possible? Argue by working out Q of the process. Given: $m\left( {}_{26}^{56}Fe \right)=55.93494u\text{ and m}\left( {}_{13}^{28}Al \right)=27.98191u$

Ans: We know that the fission of ${}_{26}^{56}Fe$ could be given as, 

${}_{26}^{56}Fe\to 2{}_{13}^{28}Al$

We are also given, atomic masses of ${}_{26}^{56}Fe\text{ and }{}_{13}^{28}Al$ as $55.93494u\text{ and 27}\text{.98191u}$ respectively. 

The Q-value here would be given as, 

$Q=\left[ m\left( {}_{26}^{56}Fe \right)-2m\left( {}_{13}^{28}Al \right) \right]{{c}^{2}}$

$\Rightarrow Q=\left[ 55.93494-2\times 27.98191 \right]{{c}^{2}}=\left( -0.02888{{c}^{2}} \right)u$

But, $1u=931.5MeV/{{c}^{2}}$

$\therefore Q=-0.02888\times 931.5=-26.902MeV$

The Q value is found to be negative and hence we could say that the fission is not possible energetically. In order for a reaction to be energetically possible, the Q-value must be positive. 


17. The fission properties of  ${}_{94}^{239}Pu$ are very similar to those of \[{}_{92}^{235}U\] .The average energy released per fission is\[\mathbf{180MeV}\]. How much energy, in MeV, is released if all the atoms in \[\mathbf{1kg}\]of pure ${}_{94}^{239}Pu$ undergo fission?

Ans: We are given that the average energy released per fission of ${}_{94}^{239}Pu$, ${{E}_{av}}=180MeV$

The amount of pure ${}_{94}P{{u}^{239}}$, $m=1kg=1000g$

Avogadro number, ${{N}_{A}}=6.023\times {{10}^{23}}$

Mass number of ${}_{94}^{239}Pu=239g$

1 mole of ${}_{94}P{{u}^{239}}$contains Avogadro number of atoms.

$1g\text{ of }{}_{94}P{{u}^{239}}\text{ contains }\left( \frac{{{N}_{A}}}{mass\text{ number}}\times m \right)\text{ atoms}$

$\Rightarrow \left( \frac{6.023\times {{10}^{23}}}{239}\times 1000 \right)=2.52\times {{10}^{24}}atoms$

Total energy released during the fission of 1kg of ${}_{94}^{239}Pu$could be calculated as:

$E={{E}_{av}}\times 2.52\times {{10}^{24}}=180\times 2.52\times {{10}^{24}}=4.536\times {{10}^{26}}MeV$

Therefore, $4.536\times {{10}^{26}}MeV$ is released if all the atoms in 1kg of pure ${}_{94}P{{u}^{239}}$undergo fission.


18. A 1000MW fission reactor consumes half of its fuel in 5.00 y. How much 

${}_{92}^{235}U$ did it contain initially? Assume that the reactor operates 80% of the time, that all the energy generated arises from the fission of ${}_{92}^{235}U$ and that this nuclide is consumed only by the fission process.

Ans: We are said that the half life of the fuel of the fission reactor, ${{t}_{\frac{1}{2}}}=5years$

$\Rightarrow {{t}_{\frac{1}{2}}}=\text{ }5\text{ }\times \text{ }365\text{ }\times \text{ }24\text{ }\times \text{ }60\text{ }\times \text{ }60\text{ }s$

We know that in the fission of 1g of ${}_{92}^{235}U$ nucleus, the energy released is equal to 200MeV.

1 mole, i.e., 235g of  ${}_{92}^{235}U$ contains \[6.023\text{ }\times \text{ }{{10}^{23}}\]atoms.

$1g\text{ of }{}_{92}^{235}U\text{ contains }\frac{6.023\times {{10}^{23}}}{234}atoms$

The total energy generated per gram of ${}_{92}^{235}U$ is calculated as:

$E=\frac{6.023\times {{10}^{23}}}{235}\times 200MeV/g=\frac{200\times 6.023\times {{10}^{23}}\times 1.6\times {{10}^{-19}}\times {{10}^{6}}}{235}=8.20\times {{10}^{10}}J/g$

The reactor operator operates only 80% of the time. Therefore, the amount of ${}_{92}^{235}U$ consumed in 5years by the 1000MW fission reactor could be calculated as, 

$\frac{5\times 80\times 60\times 60\times 365\times 24\times 1000\times {{10}^{6}}}{100\times 8.20\times {{10}^{10}}}g\approx 1538kg$

So, the initial amount of ${}_{92}^{235}U=2\times 1538=3076kg$

Hence, we found the initial amount of uranium to be $3076kg.$


19. How long can an electric lamp of 100W be kept glowing by fusion of 2.0kg of deuterium? Take the fusion reaction as

${}_{1}^{2}H+{}_{1}^{2}H\to {}_{2}^{3}He+n+3.27MeV$

Ans: The fusion reaction is given to be:

${}_{1}^{2}H+{}_{1}^{2}H\to {}_{2}^{3}He+n+3.27MeV$

Amount of deuterium, \[m=2\text{ }kg\]

1 mole, i.e., 2 g of deuterium contains \[6.023\text{ }\times \text{ }{{10}^{23}}atoms\].

2.0 kg of deuterium contains $\frac{6.023\times {{10}^{23}}}{2}\times 2000=6.023\times {{10}^{26}}atoms$ atoms

It could be inferred from the given reaction that when two atoms of deuterium fuse, \[3.27MeV\]energy is released.

Therefore, the total energy per nucleus released in the fusion reaction would be:

$E=\frac{3.27}{2}\times 6.023\times {{10}^{26}}MeV=\frac{3.27}{2}\times 6.023\times {{10}^{26}}\times 1.6\times {{10}^{-19}}\times {{10}^{6}}$

$\therefore E=1.576\times {{10}^{14}}J$

Power of the electric lamp is given to be, $P=100W=100J/s$, that is, the energy consumed by the lamp per second is 100J. 

Now, the total time for which the electric lamp glows could be calculated as, 

$t=\frac{1.576\times {{10}^{14}}}{100}=\frac{1.576\times {{10}^{14}}}{100\times 60\times 60\times 24\times 365}$

\[\therefore t\approx 4.9\times {{10}^{4}}years\]

Hence, the total time for which the electric lamp glows is found to be $4.9\times {{10}^{4}}years.$


20. Calculate the Height of the Potential Barrier for a Head-On Collision of Two Deuterons. (hint: the Height of the Potential Barrier is Given by the Coulomb Repulsion Between the Two Deuterons When They Just Touch Each Other. Assume That They Can Be Taken as Hard Spheres of Radius 2.0fm.)

Ans: When two deuterons collide head-on, the distance between their centres, d could be given as:

Radius of 1st deuteron $+$Radius of 2nd deuteron

Radius of a deuteron nucleus$=2fm=2\times {{10}^{-15}}m$

$\Rightarrow d=2\times {{10}^{-15}}+2\times {{10}^{-15}}=4\times {{10}^{-15}}m$

Also, charge on a deuteron $=$ Charge on an electron $=e=1.6\times {{10}^{-19}}C$

Potential energy of the two-deuteron system could be given by, 

$V=\frac{{{e}^{2}}}{4\pi {{\varepsilon }_{0}}d}$

Where, ${{\varepsilon }_{0}}$is the permittivity of free space. 

Also, $\frac{1}{4\pi {{\varepsilon }_{0}}}=9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}}$

$\Rightarrow V=\frac{9\times {{10}^{9}}\times {{\left( 1.6\times {{10}^{19}} \right)}^{2}}}{4\times {{10}^{15}}}J=\frac{9\times {{10}^{9}}\times {{\left( 1.6\times {{10}^{-19}} \right)}^{2}}}{4\times {{10}^{-15}}\times \left( 1.6\times {{10}^{-19}} \right)}eV$

$\therefore V=360keV$

Therefore, we found the height of the potential barrier of the two-deuteron system to be 360keV. 


21. From the relation $R={{R}_{0}}{{A}^{\frac{1}{3}}}$, where ${{R}_{0}}$ is a constant and A is the Mass Number of a Nucleus, Show That the Nuclear Matter Density Is Nearly Constant (i.e., Independent of A).

Ans: We know the expression for nuclear radius to be:

$R={{R}_{0}}{{A}^{\frac{1}{3}}}$

Where, ${{R}_{0}}$ is a Constant and $A$ is the mass number of the nucleus

Nuclear matter density would be,

$\rho =\frac{\text{Mass of the nucleus}}{\text{Volume of the nucleus}}$

Now, let m be the average mass of the nucleus, then, mass of the nucleus $=mA$

Nuclear density,

$\rho =\frac{mA}{\frac{4}{3}\pi {{R}^{3}}}=\frac{3mA}{4\pi {{\left( {{R}_{0}}{{A}^{\frac{1}{3}}} \right)}^{3}}}=\frac{3mA}{4\pi {{R}_{0}}^{3}A}$

$\therefore \rho =\frac{3m}{4\pi {{R}_{0}}^{3}}$

Therefore, we found the nuclear matter density to be independent of A and it is found to be nearly constant. 


22. For the ${{\beta }^{+}}$(positron) emission from a nucleus, there is another competing process known as electron capture (electron from an inner orbit, say, the K− shell, is captured by the nucleus and a neutrino is emitted).

${{e}^{+}}+{}_{Z}^{A}X\to {}_{Z-1}^{A}Y+\nu $

Show that if  ${{\beta }^{+}}$emission is Energetically Allowed, Electron Capture is Necessarily Allowed but Not Vice−versa.

Ans: Let the amount of energy released during the electron capture process be ${{Q}_{1}}$ . The nuclear reaction could be written as:

${{e}^{+}}+{}_{Z}^{A}X\to {}_{Z-1}^{A}Y+\nu +{{Q}_{1}}$………………………… (1)

Let the amount of energy released during the positron capture process be ${{Q}_{2}}$. The nuclear reaction could be written as:

${}_{Z}^{A}X\to {}_{Z-1}^{A}Y+{{e}^{+}}+\nu +{{Q}_{2}}$………………………… (2)

Let, ${{m}_{N}}\left( {}_{Z}^{A}X \right)\text{ be the nuclear mass of }{}_{Z}^{A}X$, ${{m}_{N}}\left( {}_{Z-1}^{A}Y \right)\text{ be the nuclear mass of }{}_{Z-1}^{A}Y$

$m\left( {}_{Z}^{A}X \right)\text{ be the atomic mass of }{}_{Z}^{A}X$

$m\left( {}_{Z-1}^{A}Y \right)\text{ be the nuclear mass of }{}_{Z-1}^{A}Y$

${{m}_{e}}\text{ be the mass of an electron}$, $c\text{ be the speed of light}$, then, the Q-value of the electron capture reaction could be given as, 

${{Q}_{1}}=\left[ {{m}_{N}}\left( {}_{Z}^{A}X \right)+{{m}_{e}}-{{m}_{N}}\left( {}_{Z-1}^{A}Y \right) \right]{{c}^{2}}$

$\Rightarrow {{Q}_{1}}=\left[ m\left( {}_{Z}^{A}X \right)-Z{{m}_{e}}+{{m}_{e}}-m\left( {}_{Z-1}^{A}Y \right)+\left( Z-1 \right){{m}_{e}} \right]{{c}^{2}}$

$\Rightarrow {{Q}_{1}}=\left[ m\left( {}_{Z}^{A}X \right)-m\left( {}_{Z-1}^{A}Y \right) \right]{{c}^{2}}$………………… (3)

The Q-value of the positron capture reaction could be given as, 

${{Q}_{2}}=\left[ {{m}_{N}}\left( {}_{Z}^{A}X \right)-{{m}_{N}}\left( {}_{Z-1}^{A}Y \right)-{{m}_{e}} \right]{{c}^{2}}$

$\Rightarrow {{Q}_{1}}=\left[ m\left( {}_{Z}^{A}X \right)-Z{{m}_{e}}-m\left( {}_{Z-1}^{A}Y \right)+\left( Z-1 \right){{m}_{e}}-{{m}_{e}} \right]{{c}^{2}}$

$\Rightarrow {{Q}_{1}}=\left[ m\left( {}_{Z}^{A}X \right)-m\left( {}_{Z-1}^{A}Y \right)-2{{m}_{e}} \right]{{c}^{2}}$………………… (4) 

It can be inferred that if \[{{Q}_{2}}>0\], then; Also, if \[Q1>0\], it does not necessarily mean that \[Q2>0\]. In other words, we could say that if ${{\beta }^{+}}$emission is energetically allowed, then the electron capture process is necessarily allowed, but not vice-versa. This is so because the Q-value must be positive for an energetically-allowed nuclear reaction.


23. In a periodic table the average atomic mass of magnesium is given as \[\mathbf{24}.\mathbf{312u}\]. The average value is based on their relative natural abundance on earth. The three isotopes and their masses are:

\[{}_{12}^{24}Mg\left( \mathbf{23}.\mathbf{98504u} \right),\text{ }{}_{12}^{25}Mg\left( \mathbf{24}.\mathbf{98584u} \right)\text{ and }{}_{12}^{26}Mg\left( \mathbf{25}.\mathbf{98259u} \right)\]

The natural abundance of ${}_{12}M{{g}^{24}}$is 78.99% by mass. Calculate the abundances of the other two isotopes.

Ans: We are given:

Average atomic mass of magnesium, \[m=24.312\text{ }u\]

Mass of magnesium ${}_{12}^{24}Mg$ isotope, \[{{m}_{1}}=23.98504\text{ }u\]

Mass of magnesium ${}_{12}^{25}Mg$ isotope, \[{{m}_{2}}=24.98584\text{ }u\]

Mass of magnesium ${}_{12}^{26}Mg$ isotope, \[{{m}_{3}}=25.98259\text{ }u\]

Abundance of ${}_{12}^{24}Mg,\text{ }{{\eta }_{1}}=78.99%$ 

Abundance of  ${}_{12}^{25}Mg,\text{ }{{\eta }_{2}}=x%$

Now, the abundance of ${}_{12}^{26}Mg$, \[{{\eta }_{3}}=100-x-78.99%=\left( 21.01-x \right)%\]

Also, we have the relation for the average atomic mass as:

$m=\frac{{{m}_{1}}{{\eta }_{1}}+{{m}_{2}}{{\eta }_{2}}+{{m}_{3}}{{\eta }_{3}}}{{{\eta }_{1}}+{{\eta }_{2}}+{{\eta }_{3}}}$

$\Rightarrow 24.312=\frac{23.98504\times 78.99+24.98584\times x+25.98259\times \left( 21.01-x \right)}{100}$

$\Rightarrow 0.99675x=9.2725255$

$\therefore x\approx 9.3%$

And, $21.01-x=11.71%$

Therefore, we found the abundance of ${}_{12}^{25}Mg$ to be 9.3% and that of ${}_{12}^{26}Mg$ to be 11.71%.


24. The neutron separation energy is defined as the energy required to remove a neutron from the nucleus. Obtain the neutron separation energies of the nuclei ${}_{20}^{41}Ca$ and ${}_{13}^{27}Al$ from the following data:

$m\left( {}_{20}^{40}Ca \right)=39.962591u,\text{ }m\left( {}_{20}^{41}Ca \right)\text{=40}\text{.962278u,  }m\left( {}_{13}^{26}Al \right)=25.986995u,\text{ }$

$m\left( {}_{13}^{27}Al \right)=26.981541u$

Ans: For a neutron removal from ${}_{20}C{{a}^{41}}$nucleus, the corresponding nuclear reaction could be written as, 

${}_{20}^{41}Ca\to {}_{20}^{40}Ca+{}_{0}^{1}n$

We are given:

$m\left( {}_{20}^{40}Ca \right)=39.962591u$

$m\left( {}_{20}^{41}Ca \right)\text{=40}\text{.962278u}$

$m\left( {}_{0}{{n}^{1}} \right)=1.008665u$

Now, the mass defect for this reaction could be given by, 

$\Delta m=m\left( {}_{20}^{40}Ca \right)+\left( {}_{0}^{1}n \right)-m\left( {}_{20}^{41}Ca \right)$

$\Rightarrow \Delta m=39.962591+1.008665-40.962278=0.008978u$

But we know, $1u=931.5MeV/{{c}^{2}}$

$\Rightarrow \Delta m=0.008978\times 931.5MeV/{{c}^{2}}$

Now, we could calculate the energy required for the neutron removal by, 

$E=\Delta m{{c}^{2}}$

$\Rightarrow E=0.008978\times 931.5=8.363007MeV$

For the case of ${}_{13}^{27}Al$, the neutron removal reaction could be written as, 

${}_{13}^{27}Al\to {}_{13}^{26}Al+{}_{0}^{1}n$

We are given,

$m\left( {}_{13}^{26}Al \right)=25.986995u$

$m\left( {}_{13}^{27}Al \right)=26.981541u$

Now, the mass defect here could be given by, 

$\Delta m=m\left( {}_{13}^{26}Al \right)+m\left( {}_{0}^{1}n \right)-m\left( {}_{13}^{27}Al \right)$

$\Rightarrow \Delta m=25.986895+1.008665-26.981541=0.014019u$

$\Rightarrow \Delta m=0.014019\times 931.5MeV/{{c}^{2}}$

Therefore, the energy that is required for the removal of neutron would be, 

$E=\Delta m{{c}^{2}}=0.014019\times 931.5$

$\therefore E=13.059MeV$


25. A source contains two phosphorous radio nuclides ${}_{15}^{32}P\text{ }\left( {{T}_{\frac{1}{2}}}=14.3d \right)$ and ${}_{15}^{33}P\text{ }\left( {{T}_{\frac{1}{2}}}=25.3d \right)$Initially, 10% of the decays come from ${}_{15}^{33}P$. How long must one wait until 90% do so?

Ans: We are given:

Half life of  ${}_{15}^{32}P\text{ }\left( {{T}_{\frac{1}{2}}}=14.3d \right)$

Half life of  ${}_{15}^{33}P\text{ }\left( {{T}_{\frac{1}{2}}}=25.3d \right)$

Now, we know that nucleus decay is 10% of the total amount of decay.

Also, the source has initially 10% of ${}_{15}^{32}P$ nucleus and 90% of ${}_{15}^{32}P$ nucleus.

Suppose after t days, the source has 10% of ${}_{15}^{32}P$ nucleus and 90% of ${}_{15}^{33}P$ 

nucleus.

Initially we have:

Number of ${}_{15}^{33}P$ nucleus $=N$

Number of ${}_{15}^{32}P$ nucleus $=9N$

Finally:

Number of ${}_{15}^{33}P$ nucleus $=9N'$

Number of ${}_{15}^{32}P$ nucleus$=N'$

For ${}_{15}^{32}P$ nucleus, we could write the number ratio as:

$\frac{N'}{9N}={{\left( \frac{1}{2} \right)}^{\frac{t}{{{T}_{\frac{1}{2}}}}}}$

$\Rightarrow N'=9N{{\left( 2 \right)}^{\frac{-t}{14.3}}}$………………………. (1)

Now, for ${}_{15}^{33}P$, we could write the number ratio as, 

$\frac{9N'}{N}={{\left( \frac{1}{2} \right)}^{\frac{1}{T{{'}_{\frac{1}{2}}}}}}$

$\Rightarrow 9N'=N{{\left( 2 \right)}^{\frac{-t}{25.3}}}$…………………………. (2)

We could now divide equation (1) by equation (2) to get, 

$\frac{1}{9}=9\times {{2}^{\left( \frac{t}{25.3}-\frac{t}{14.3} \right)}}$

$\Rightarrow \frac{1}{81}={{2}^{\left( -\frac{11t}{25.3\times 14.3} \right)}}$

$\Rightarrow \log 1-\log 81=\frac{-11t}{25.3\times 14.3}\log 1$

$\Rightarrow \frac{-11t}{25.3\times 14.3}=\frac{0-1.908}{0.301}$

$\therefore t=\frac{25.3\times 14.3\times 1.908}{11\times 0.301}\approx 208.5days$

Therefore, we found that it would take about 208.5days for 90% decay of ${}_{15}^{33}P$.


26. Under certain circumstances, a nucleus can decay by emitting a particle more massive than an $\alpha -$particle. Consider the following decay processes:

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

${}_{88}^{223}Ra\to {}_{86}^{219}Rn+{}_{2}^{4}He$

Calculate the Q-values for these decays and determine that both are energetically allowed.

Ans: Consider a ${}_{6}^{14}C$ emission nuclear reaction, 

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

We know that:

Mass of ${}_{88}^{223}Ra,\text{ }{{\text{m}}_{1}}=223.01850u$

Mass of ${}_{6}^{14}C,\text{ }{{\text{m}}_{3}}=14.00324u$

Now, the Q-value of the reaction could be given as:

$Q=\left( {{m}_{1}}-{{m}_{2}}-{{m}_{3}} \right){{c}^{2}}$

$\Rightarrow Q=\left( 223.01850-208.98107-14.00324 \right){{c}^{2}}=\left( 0.03419{{c}^{2}} \right)u$

But we have, $1u=931.5MeV/{{c}^{2}}$

$\Rightarrow Q=0.03419\times 931.5$

$\therefore Q=31.848MeV$

Hence, the Q-value of the nuclear reaction is found to be 31.848 MeV. Since the value is positive, the reaction is energetically allowed.

Now consider a ${}_{2}^{4}He$ emission nuclear reaction:

${}_{88}^{223}Ra\to {}_{86}^{229}Rn+{}_{2}^{4}He$

We know that:

Mass of ${}_{88}^{223}Ra,\text{ }{{\text{m}}_{1}}=223.01850$

Mass of ${}_{82}^{219}Rn,\text{ }{{\text{m}}_{2}}=219.00948$

Mass of ${}_{2}^{4}He,\text{ }{{\text{m}}_{3}}=4.00260$

Q-value of this nuclear reaction could be given as:

$Q=\left( {{m}_{1}}-{{m}_{2}}-{{m}_{3}} \right){{c}^{2}}$

$\Rightarrow Q=\left( 223.01850-219.00948-4.00260 \right){{c}^{2}}$

$\Rightarrow Q=\left( 0.00642{{c}^{2}} \right)u$

$\therefore Q=0.00642\times 931.5=5.98MeV$

Therefore, the Q-value of the second nuclear reaction is found to be 5.98MeV. Since the value is positive, we could say that the reaction is energetically allowed.

27. Consider the fission of ${}_{92}^{238}U$ by fast neutrons. In one fission event, no neutrons are emitted and the final end products, after the beta decay of the primary fragments, are ${}_{58}^{140}Ce\text{ and }{}_{44}^{99}Ru$. Calculate Q for this fission process. The relevant atomic and particle masses are:

$m\left( {}_{92}^{238}U \right)=238.05079u$

$m\left( {}_{58}^{140}Ce \right)=139.90543u$

$m\left( {}_{44}^{99}Ru \right)=98.90594u$

Ans: We are given:

In the fission of ${}_{92}^{238}U$, 10 $\beta -$particles decay from the parent nucleus. The nuclear reaction can be written as:

${}_{92}^{238}U+{}_{0}^{1}n\to {}_{58}^{140}Ce+{}_{44}^{99}Ru+10{}_{-1}^{0}e$

It is also given that:

Mass of a nucleus of ${}_{92}^{238}U,\text{ }{{\text{m}}_{1}}=238.05079u$

Mass of a nucleus of ${}_{58}^{140}Ce,\text{ }{{\text{m}}_{2}}=139.90543u$

Mass of nucleus of ${}_{44}^{99}Ru,\text{ }{{\text{m}}_{3}}=98.90594u$, 

Mass of a neutron ${}_{0}^{1}n,\text{ }{{\text{m}}_{4}}=1.008665u$

Q-value of the above equation would be,

$Q=\left[ m'\left( {}_{92}^{238}U \right)+m\left( {}_{0}^{1}n \right)-m'\left( {}_{58}^{140}Ce \right)-m'\left( {}_{44}^{99}Ru \right)-10{{m}_{e}} \right]{{c}^{2}}$

Where, $m'=$Represents the corresponding atomic masses of the nuclei

$m'\left( {}_{92}^{238}U \right)={{m}_{1}}-92{{m}_{e}}$

$m'\left( {}_{58}^{140}Ce \right)={{m}_{2}}-58{{m}_{e}}$

$m'\left( {}_{44}^{99}Ru \right)={{m}_{3}}-44{{m}_{e}}$

$m\left( {}_{0}^{1}n \right)={{m}_{4}}$

$Q=\left[ {{m}_{1}}-92{{m}_{e}}+{{m}_{4}}-{{m}_{2}}+58{{m}_{e}}-{{m}_{3}}+44{{m}_{e}}-10{{m}_{e}} \right]{{c}^{2}}$

$\Rightarrow Q=\left[ {{m}_{1}}+{{m}_{4}}-{{m}_{2}}-{{m}_{3}} \right]{{c}^{2}}=\left[ 238.0507+1.008665-139.90543-98.90594 \right]{{c}^{2}}$

$\Rightarrow Q=\left[ 0.247995{{c}^{2}} \right]u$

But $1u=931.5MeV/{{c}^{2}}$

\[\therefore Q=0.247995\times 931.5\text{ =}231.007MeV\]

Therefore, the Q-value of the fission process is found to be 231.007MeV.


28. Consider the D−T reaction (deuterium−tritium fusion) 

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

  1. Calculate the energy released in MeV in this reaction from the data: $m\left( {}_{1}^{2}H \right)=2.014102u$ , $m\left( {}_{1}^{3}H \right)=3.016049u$

Ans: Consider the D-T nuclear reaction, 

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

We are also given that:

Mass of \[{}_{1}^{2}H,\text{ }{{\text{m}}_{1}}=2.014102u\]

Mass of \[{}_{1}^{3}H,\text{ }{{\text{m}}_{2}}=3.016049u\]

Mass of ${}_{2}^{4}He,\text{ }{{\text{m}}_{3}}=4.002603u$

Mass of ${}_{0}^{1}n,\text{ }{{\text{m}}_{4}}=1.008665u$

Now, the Q-value of the given D-T reaction would be:

$Q=\left[ {{m}_{1}}+{{m}_{2}}-{{m}_{3}}-{{m}_{4}} \right]{{c}^{2}}$

$\Rightarrow Q=\left[ 2.014102+3.016049-4.002603-1.008665 \right]{{c}^{2}}$

$\Rightarrow Q=\left[ 0.018883{{c}^{2}} \right]u$

But $1u=931.5MeV/{{c}^{2}}$

$\therefore Q=0.018883\times 931.5=17.59MeV$


  1. Consider the radius of both deuterium and tritium to be approximately 2.0fm. 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 $2\left( \frac{3kT}{2} \right)$; k = Boltzmann’s constant, T = absolute temperature.)

Ans: We are given:

Radius of deuterium and tritium, $r\approx 2.0fm=2\times {{10}^{-15}}m$ 

Distance between the two nuclei at the moment when they touch each other,

$d=r+r=4\times {{10}^{-15}}m$

Charge on the deuterium nucleus $=e$

Charge on the tritium nucleus $=e$

Hence, the repulsive potential energy between the two nuclei could be given as:

$V=\frac{{{e}^{2}}}{4\pi {{\varepsilon }_{0}}d}$

Where, ${{\varepsilon }_{0}}=$Permittivity of free space

$\frac{1}{4\pi {{\varepsilon }_{0}}}=9\times {{10}^{9}}N{{m}^{2}}{{c}^{-2}}$

$\Rightarrow V=\frac{9\times {{10}^{9}}\times {{\left( 1.6\times {{10}^{-19}} \right)}^{2}}}{4\times {{10}^{-15}}}=5.76\times {{10}^{-14}}J=\frac{5.76\times {{10}^{-14}}}{1,6\times {{10}^{-19}}}$

$\therefore V=3.6\times {{10}^{5}}eV=360keV$

Therefore, $5.76\times {{10}^{-14}}J$ or  $360keV$of kinetic energy (KE) is needed to overcome the coulomb repulsion between the two nuclei.

However, we are also given that:

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

Where, $k=$Boltzmann constant 

$T=$Temperature required for triggering the reaction

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

Therefore, we found that the gas must be heated to a temperature of $1.39\times {{10}^{9}}K$ to initiate the reaction.


29. Obtain the maximum kinetic energy of $\beta -$particles, and the radiation frequencies of $\gamma $ decays in the decay scheme shown in figure. You are given that:

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

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

(Image will be Uploaded Soon)

Ans: It can be observed from the given $\gamma $-decay diagram that ${{\gamma }_{1}}$decays from the 1.088MeV energy level to the 0MeV energy level. Hence, the energy corresponding to ${{\gamma }_{1}}$-decay is given as:

${{E}_{1}}=1.088-8=1.088MeV$

$\Rightarrow h{{\nu }_{1}}=1.088\times 1.6\times {{10}^{-19}}\times {{10}^{6}}J$ 

Where, Planck's constant  $h=6.6\times {{10}^{-34}}Js$

${{\nu }_{1}}=$Frequency of radiation radiated by ${{\gamma }_{1}}-$decay

${{\nu }_{1}}=\frac{{{E}_{1}}}{h}$

$\Rightarrow {{\nu }_{1}}=\frac{1.088\times 1.6\times {{10}^{-19}}\times {{10}^{6}}}{6.6\times {{10}^{-34}}}=2.637\times {{10}^{20}}Hz$

It can be observed from the given $\gamma -$decay diagram that ${{\gamma }_{2}}$decays from the \[0.412MeV\]energy level to the \[0MeV\]energy level.

Now, the energy corresponding to ${{\gamma }_{2}}$-decay could be given as:

${{E}_{2}}=0.412-0=0.412MeV$

$\Rightarrow h{{\nu }_{2}}=0.412\times 1.6\times {{10}^{-19}}\times {{10}^{6}}J$

Where, ${{\nu }_{2}}=$Frequency of radiation radiated by ${{\gamma }_{2}}-$decay

${{\nu }_{2}}=\frac{{{E}_{2}}}{h}=\frac{0.412\times 1.6\times {{10}^{-19}}\times {{10}^{6}}}{6.6\times {{10}^{-34}}}=9.988\times {{10}^{19}}Hz$

It can be observed from the given $\gamma $-decay diagram that ${{\gamma }_{3}}$-decays from the \[1.088MeV\]energy level to the \[0.412MeV\]energy level.

Now, the energy corresponding to ${{\gamma }_{3}}$ -decay is given as:

${{E}_{3}}=1.088-0.412=0.676MeV$

$\Rightarrow h{{\nu }_{3}}=0.676\times {{10}^{-19}}\times {{10}^{6}}$

Where, ${{\nu }_{3}}=$Frequency of radiation radiated by ${{\gamma }_{3}}-$decay

${{\nu }_{3}}=\frac{{{E}_{3}}}{h}=\frac{0.676\times 1.6\times {{10}^{-19}}\times {{10}^{6}}}{6.6\times {{10}^{-34}}}=1.639\times {{10}^{20}}Hz$

Mass of \[m\left( {}_{78}^{198}Au \right)\text{ =}197.968233u\]

Mass of $m\left( {}_{80}^{198}Hg \right)=197.966760u$

\[1u=931.5MeV/{{c}^{2}}\]

Energy of the highest level could be given as:

$E=\left[ m\left( {}_{78}^{198}Au \right)-m\left( {}_{80}^{190}Hg \right) \right]=197.968233-197.966760=0.001473u$

$\Rightarrow E=0.001473\times 931.5=1.3720995MeV$

${{\beta }_{1}}$decays from the \[1.3720995MeV\]level to the \[1.088MeV\]level

Maximum kinetic energy of the ${{\beta }_{1}}$particle $=1.3720995\text{ }-\text{ }1.088$

$\Rightarrow K.E=0.2840995MeV$

${{\beta }_{2}}$ decays from the $1.3720995MeV$level to that of the $0.412MeV$level. Now, we find the maximum kinetic energy of the ${{\beta }_{2}}$particle to be, 

$K.{{E}_{\max }}=1.3720995-0.412=0.9600995MeV$

Therefore, we found the maximum kinetic energy of the ${{\beta }_{2}}$particle to be $0.9600995MeV$. 


30. Calculate and Compare the Energy Released By

  1. fusion of 1.0kg of hydrogen deep within Sun and

Ans: We are given:

Amount of hydrogen, \[m=1\text{ }kg=1000g\]

1 mole, i.e., 1g of hydrogen $\left( {}_{1}^{1}H \right)$ contains \[6.023\times {{10}^{23}}atoms\].

That is, 1000g of ${}_{1}^{1}H$ contains \[6.023\times {{10}^{23}}atoms\].

Within the sun, four ${}_{1}^{1}H$ nuclei combine and form one ${}_{2}^{4}He$ nucleus. In this process \[26MeV\]of energy is released.

Hence, the energy released from the fusion of 1 kg ${}_{1}^{1}H$ is:

${{E}_{1}}=\frac{6.023\times {{10}^{23}}\times 26\times {{10}^{3}}}{4}=39.1495\times {{10}^{26}}MeV$

Therefore, we found the energy released during the fusion of 1kg ${}_{1}^{1}H$ is:

${{E}_{1}}=\frac{6.023\times {{10}^{23}}\times 26\times {{10}^{3}}}{4}=39.1495\times {{10}^{26}}MeV$

Hence, the energy released during the fusion of 1kg of ${}_{1}^{1}H$ to be $39.1495\times {{10}^{26}}MeV$. 


  1. The Fission of 1.0kg of ${}^{235}U$ in a fission reactor.

Ans: We are given:

Amount of ${}_{92}{{U}^{235}}=1000gm$

1 mole, i.e., 235g of ${}_{92}^{235}U$ contains \[6.023\times {{10}^{23}}atoms\].

1000g of ${}_{92}^{235}U$ contains $\frac{6.023\times {{10}^{23}}\times 1000}{235}atoms$

We know that the amount of energy released in the fission of one atom of ${}_{92}^{235}U$ is\[200MeV\]. Therefore, energy released from the fission of 1kg of ${}_{92}^{235}U$ is:

${{E}_{2}}=\frac{6\times {{10}^{23}}\times 1000\times 200}{235}=5.106\times {{10}^{26}}MeV$

$\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{39.1495\times {{10}^{26}}}{5.106\times {{10}^{26}}}=7.67\approx 8$

Hence, we found the energy released during the fusion of 1kg of hydrogen is nearly 8 times the energy released during the fusion of 1kg of uranium. 


31. Suppose India Had a Target of Producing, by 2020 AD, 200,000 MW of Electric Power, Ten Percent of Which Was to be Obtained from Nuclear Power Plants. Suppose We are Given That, on an Average, the Efficiency of Utilization (i.e., Conversion to Electric Energy) of Thermal Energy Produced in a Reactor Was 25%. How Much Amount of Fissionable Uranium Would Our Country Need Per Year by 2020? Take the Heat Energy Per Fission of ${}^{235}U$ to be about \[\mathbf{200MeV}\].

Ans: We are given the following:

Amount of electric power to be generated, \[P=2\times {{10}^{5}}MW\], 10% of this amount has to be obtained from nuclear power plants.

Amount of nuclear power, ${{P}_{1}}=\frac{10}{100}\times 2\times {{10}^{5}}=2\times {{10}^{4}}MW$

$\Rightarrow {{P}_{1}}=2\times {{10}^{4}}\times {{10}^{6}}J/s=2\times {{10}^{10}}\times 3600\times 24\times 365J/y$

Heat energy released per fission of a ${}^{235}U$ nucleus, \[E=200\text{ }MeV\]

Efficiency of a reactor \[=25%\]

Hence, the amount of energy converted into the electrical energy per fission is calculated as:

$\frac{25}{100}\times 200=50MeV=50\times 1.6\times {{10}^{-19}}\times {{10}^{6}}=8\times {{10}^{-12}}J$

The number of atoms required for fission per year would be:

$\frac{2\times {{10}^{10}}\times 60\times 60\times 24\times 365}{8\times {{10}^{-12}}}=78840\times {{10}^{24}}atoms$

1 mole, i.e., 235g of ${{U}^{235}}$contains \[6.023\times {{10}^{23}}atoms\]

That is, the mass of \[6.023\times {{10}^{23}}atoms\]of ${{U}^{235}}=235g=235\times {{10}^{-3}}kg$

Also, the mass of:$78840\times {{10}^{24}}atoms\text{ of }{{\text{U}}^{235}}=\frac{235\times {{10}^{-3}}}{6.023\times {{10}^{23}}}\times 78840\times {{10}^{24}}=3.076\times {{10}^{4}}kg$

Hence, the mass of uranium needed per year is found to be,\[3.076\times {{10}^{4}}kg\].


NCERT Solutions for Class 12 Physics - Free PDF Download

To ace in your paper, you need to understand all the basic concepts of your subjects. For that, solutions need to be very simplified in a clear format with the help of comparatively easy methods and lesser calculation. This saves time and effort in the exam. There are live classes available for online mode of study and understanding concepts. Also, in offline mode simple free pdf are available too. Thus, learning and writing notes go simultaneously. You can have a complete series of notes by downloading these free PDFs.

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So, just grab them before heading towards the main exams. The study material is present in an enormous variety. You can choose the one which is favourable for you and quickly adaptable.

This chapter mostly concerns atomic energy and its constituents. It includes a simple introduction about atomic nuclei, atomic masses, their composition, size of nucleus, mass number and binding energy, strong relationships, nuclear force, radioactivity, and nuclear energy.

There is a greater numerical portion in the chapter and lots of numbers that you have to remember. It deals with the foundation of nuclear energy. How binding energy and mass numbers play a crucial role in the generation of nuclear power. There are many questions in the chapter and if you want a simplified solution of these questions in an easily understandable way, then download the free pdf of Vedantu consisting of all Chapter 13.


NCERT Solutions for Class 12 Physics Chapter 13 – Nuclei

Chapter 13 Nuclei

Most of the questions of Physics are in numerical format. Thus, practice and regular revision make a good command of the subject. The NCERT Books Solutions for Class 12 Physics Chapter 13 Nuclei are a moderate level, hence not too tough if you think. All you need to be a little bit conceptual, practising, and visiting a variety of questions by taking Chapter 13 as reference material.


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Chapter 13 of Class 12th Physics is not very tough and not very easy. After learning concepts and practising the maximum number of questions, you would find them of a moderate level to solve. Some benefits of Chapter 13 are listed below:

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Vedantu provides NCERT Solutions for Class 12 Physics Chapter 13 Nuclei. It is available in free PDF format. NCERT Solutions for Chapter 13 Nuclei are designed by expert Physics tutors at Vedantu. The material is designed to help students revise the complete chapter and score well in exams. NCERT Solutions for Class 12 Physics Chapter 13 are prepared as per the latest NCERT norms and exam guidelines. It provides students with a thorough knowledge of the chapter and simple explanations to the exercise problems with an emphasis on critical concepts. It is designed to help students in doubt clearance.

2. What are the benefits of NCERT Solutions for Class 12 Physics Chapter 13 Nuclei?

CBSE Class 12 Physics Chapter 13 Nuclei is an important chapter from exam perspective. NCERT Solutions explain the concepts related to Nuclei and answer questions related to the chapter. It is important to understand the chapter thoroughly to score well in exams. NCERT Solutions for Class 12 Physics Chapter 13 are the most reliable online resource. Students facing any doubt in the chapter can seek guidance from expert solutions. The tutors who have prepared the solutions have years of experience in the field of teaching. NCERT Solutions for Class 12 Chapter 13 Nuclei is a great way to be a top scorer in the Physics exam.

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NCERT Solutions for Class 12 Physics Chapter 13 Nuclei are available on its site. Following are some of the key features of NCERT Solutions for Class 12 Physics Chapter 13 offered by Vedantu:

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4.Can Vedantu Help in Achieving Maximum Marks in Board Examinations?

Vedantu is present to bring the best out of you through the best faculty of each topic they have. The solutions are appropriately created according to the latest CBSE guidelines. So, yes after practising all the questions and downloading Nuclei Class 12 Chapter 13 pdf from Vedantu, you can achieve maximum marks.

5. What is the nucleus definition?

The nucleus is the centre of an atom. It consists of positively charged particles called protons and electrically neutral particles called neutrons. Protons and neutrons form the nucleus. The atomic number is the number of protons of the nucleus.

6. What does Chapter 13 of Class 12 Physics tell us about?

Chapter 13 of Class 12 Physics is "Nuclei". The chapter mainly focuses on atoms and their constituents. It includes an Introduction About Atoms, Nuclei, Atomic Masses, Size Of Nucleus, Mass Number, Binding Energy, Nuclear Force, Nuclear Energy And Radioactivity. The chapter also consists of many numerical and examples for a better understanding of the chapter. 

7. How to make a fruitful study plan for Chapter 13 of Class 12 Physics?

The given technique will help students in making a better study plan for Chapter 13 of Class 12 Physics:

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  • Study the chapter for at least 2 hours a day.

  • Solve all the NCERT back questions and examples to understand the chapter clearly.

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8. How to score well in Chapter 13 of Class 12 Physics?

A student of Class 12 should make notes of Chapter 13 of Class 12 Physics for better understanding. He or she should go through the chapter thoroughly. Practising numerical and examples can help students to have a strong grip on that chapter. Regular revision of a particular chapter is a must. Solving question papers can also be helpful to students. They can also take the help of Vedantu as it provides NCERT Solutions of Chapter 13 of Class 12 Physics plus the question papers, available for free. Following the above steps, a student can score good marks in Class 12 Physics Exams.

9. What method should be kept in mind while downloading the PDF file of Physics Chapter 13 Class 12? 

The steps are given below to download the PDF file of Physics Chapter 13 Class 12 are; 

  • Tap Chapter 13 of Class 12 Physics. The link will open.

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In this PDF file, the solutions are written in very simple language, so that the students can understand them easily.

10. How can you express the terms "Binding Energy" and "Binding Energy per Nucleon"?

The word binding energy is described as the energy that is needed to shatter a nucleus in its constituent particles (neutrons and protons). The nucleus is broken in such a way that there is a huge separation between the constituent particles, so that they do not interact with each other. The average energy, which is used to take out one nucleon from the nucleus is known as binding energy per nucleon. 

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