Radii of the nuclei,
Where A is the mass number and
Mass defect:
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.
Another important relation which helps in CBSE NCERT solutions for class 12 physics chapter 13 nuclei is Einstein's mass-energy relation.
Where c is the speed of light.
Another main concept of NCERT grade 12 chapter 13 nuclei is the law of radioactive decay. This is given by
Where N is the number of nuclei at any time t, is the number of nuclei at any time and lambda is disintegration constant.
The half-life of a radionuclide is given by
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.
Mass of the two stable isotopes and their respective abundances are and and and .
m=6.940934 u
The atomic mass of boron is 10.811 u
Mass of the two stable isotopes are and respectively
Let the two isotopes have abundances x% and (100-x)%
Therefore the abundance of is 19.89% and that of 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})
The average atomic mass of neon is 20.1771 u.
Q. 13.3 Obtain the binding energy( in MeV ) of a nitrogen nucleus , given m
m_{n} = 1.00866 u
m_{p}= 1.00727 u
Atomic mass of Nitrogen m= 14.00307 u
Mass defect m=7m_{n}+7m_{p }- m
m=71.00866+71.00727 - 14.00307
m=0.10844
Now 1u is equivalent to 931.5 MeV
E_{b}=0.10844931.5
E_{b}=101.01186 MeV
Therefore binding energy of a Nitrogen nucleus is 101.01186 MeV.
Q. 13.4 (i) Obtain the binding energy of the nuclei and in units of MeV from the following data:
m_{H} = 1.007825 u
m_{n} = 1.008665 u
The atomic mass of is m=55.934939 u
Mass defect
m=0.528461
Now 1u is equivalent to 931.5 MeV
E_{b}=0.528461931.5
E_{b}=492.2614215 MeV
Therefore the binding energy of a nucleus is 492.2614215 MeV.
Average binding energy
m_{H} = 1.007825 u
m_{n} = 1.008665 u
The atomic mass of is m=208.980388 u
Mass defect
m=1261.008665+831.007825 - 208.980388
m=1.760877 u
Now 1u is equivalent to 931.5 MeV
E_{b}=1.760877 931.5
E_{b}=1640.2569255 MeV
Therefore the binding energy of a nucleus is 1640.2569255 MeV.
Mass of the coin is w = 3g
Total number of Cu atoms in the coin is n
n=2.87110^{22}
m_{H} = 1.007825 u
m_{n} = 1.008665 u
Atomic mass of is m=62.92960 u
Mass defect m=(63-29)m_{n}+29m_{H }- m
m=341.008665+291.007825 - 62.92960
m=0.591935 u
Now 1u is equivalent to 931.5 MeV
E_{b}=0.591935931.5
E_{b}=551.38745 MeV
Therefore binding energy of a nucleus is 551.38745 MeV.
The nuclear energy that would be required to separate all the neutrons and protons from each other is
nE_{b}=2.87110^{22}551.38745
=1.583210^{25} MeV
=1.583210^{25}1.610^{-19}10^{6}J
=2.533110^{9 }kJ
Q.13.6 (i) Write nuclear reaction equations for
The nuclear reaction equations for the given alpha decay
Q.13.6 (ii) Write nuclear reaction equations for
The nuclear reaction equations for the given alpha decay is
Q.13.6 (iii) Write nuclear reaction equations for
The nuclear reaction equations for the given beta minus decay is
Q.13.6 (iv) Write nuclear reaction equations for
The nuclear reaction equation for the given beta minus decay is
Q.13.6 (v) Write nuclear reaction equations for
The nuclear reaction for the given beta plus decay will be
Q.13.6 (vii) Write nuclear reaction equations for
The nuclear reaction for electron capture of is
(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.
In this case
It will take 5T years to reach 3.125% of the original activity.
(b) In this case
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.
Also
but
Therefore
The age of the Indus-Valley civilisation calculated using the given specimen is approximately 4217 years.
Required activity=8.0 mCi
1 Ci=3.710^{10} decay s^{-1}
8.0 mCi=810^{-3}3.710^{10} =2.9610^{8} decay s^{-1}
T_{1/2}=5.3 years
Mass of those many atoms of Cu will be
7.1210^{-6}^{ }^{ }g of is necessary to provide a radioactive source of 8.0 mCi strength.
Q. 13.10 The half-life of is 28 years. What is the disintegration rate of 15 mg of this isotope?
T_{1/2}=28 years
The number of atoms in 15 mg of is
N=1.003810^{20}
The disintegration rate will be
=-1.003810^{20}7.8510^{-10}
=-7.8810^{10} s^{-1}
The disintegration rate is therefore 7.8810^{10 }decay s^{-1}.
Q.13.11 Obtain approximately the ratio of the nuclear radii of the gold isotope and the silver isotope
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
Q.13.12 Find the Q-value and the kinetic energy of the emitted -particle in the -decay of
Mass defect is m
m=226.02540-222.0175-4.002603
m=0.005297 u
1 u = 931.5 MeV/c^{2}
Q-value=m931.5
=4.934515 MeV
By using Linear Momentum Conservation and Energy Conservation
The kinetic energy of alpha particle =
=
=4.847 MeV
Mass defect is m
m=220.01137-216.00189-4.002603
m=0.006877 u
1 u = 931.5 MeV/c^{2}
Q-value=m931.5
=6.406 MeV
By using Linear Momentum Conservation and Energy Conservation
The kinetic energy of alpha particle =
=
=6.289 MeV
Q.13.13 The radionuclide decays according to
The maximum energy of the emitted positron is .
and
calculate Q and compare it with the maximum energy of the positron emitted.
If we use atomic masses
Q-value= 0.001033931.5=0.9622 MeV which is comparable with a maximum energy of the emitted positron.
The decay equation is
(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=0.004696931.5
Q=4.3743 eV
The emitted nucleus is way heavier than the 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.
The above negative value of mass defect implies there will be a negative Q value and therefore the reaction is endothermic
The above positive value of mass defect implies Q value would be positive and therefore the reaction is exothermic
The reaction will be
The mass defect of the reaction will be
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 =n
Energy per fission (E)=180 MeV
Total Energy released if all the atoms in 1 kg undergo fission = E n
=180 2.5210^{24}
=4.53610^{26 }MeV
The amount of energy liberated on fission of 1 atom is 200 MeV.
The amount of energy liberated on fission of 1g
Total Energy produced in the reactor in 5 years
Mass of which underwent fission, m
=1537.8 kg
The amount present initially in the reactor = 2m
=21537.8
=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}500
The energy liberated by fusion of 2.0 kg of deuterium atoms E
Power of lamp (P)= 100 W
Time the lamp would glow using E amount of energy is T=
=4.9910^{4} years
For a head-on collision of two deuterons, the closest distances between their centres will be d=2r
d=22.0
d=4.0 fm
d=410^{-15} m
charge on each deuteron = charge of one proton=q =1.610^{-19} C
The maximum electrostatic potential energy of the system during the head-on collision will be E
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
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
For the electron capture, the reaction would be
The mass defect and q value of the above reaction would be
where m_{N }and m_{N} are the nuclear masses of elements X and Y respectively
For positron emission, the reaction would be
The mass defect and q value for the above reaction would be
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 exercises
Let the abundances of and be x and y respectively.
x+y+78.99=100
y=21.01-x
The average atomic mass of Mg is 24.312 u
The abundances of and are 9.3% and 11.71% respectively
The reaction showing the neutron separation is
But 1u=931.5 MeV/c^{2}
Therefore E=(0.008978)931.5
E=8.363007 MeV
Therefore to remove a neutron from the nucleus 8.363007 MeV of energy is required
The reaction showing the neutron separation is
But 1u=931.5 MeV/c^{2}
Therefore E=(0.014019)931.5
E=13.059 MeV
Therefore to remove a neutron from the nucleus 13.059 MeV of energy is required
Let initially there be N_{1} atoms of and N_{2} atoms of and let their decay constants be and respectively
Since initially the activity of is 1/9 times that of we have
(i)
Let after time t the activity of be 9 times that of
(ii)
Dividing equation (ii) by (i) and taking the natural log of both sides we get
where and
t comes out to be 208.5 days
Q.13.26 Under certain circumstances, a nucleus can decay by emitting a particle more massive than an -particle. Consider the following decay processes:
Calculate the Q-values for these decays and determine that both are energetically allowed.
1 u = 931.5 MeV/c^{2}
Q=0.03419931.5
=31.848 MeV
As the Q value is positive the reaction is energetically allowed
1 u = 931.5 MeV/c^{2}
Q=0.00642931.5
=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
The mass defect for the above reaction would be
In the above equation, m_{N} represents nuclear masses
but 1u =931.5 MeV/c^{2}
Q=0.247995931.5
Q=231.007 MeV
Q value of the fission process is 231.007 MeV
Q.13.28 (i) Consider the D–T reaction (deuterium-tritium fusion)
(a) Calculate the energy released in MeV in this reaction from the data:
The mass defect of the reaction is
1u = 931.5 MeV/c^{2}
Q=0.018883931.5=17.59 MeV
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
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
Therefore the temperature required to initiate the reaction is
decays from 1.088 MeV to 0 V
Frequency of is
Plank's constant, h=6.6210^{-34} Js
Similarly, we can calculate frequencies of and
The energy of the highest level would be equal to the energy released after the decay
Mass defect is
We know 1u = 931.5 MeV/c^{2}
Q value= 0.001473931.5=1.3721 MeV
The maximum Kinetic energy of would be 1.3721-1.088=0.2841 MeV
The maximum Kinetic energy of would be 1.3721-0.412=0.9601 MeV
(a)
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}
(b) The energy released in fission of one atom is 200 MeV
Number of atoms present in 1 kg of is N
The energy released on fission of N atoms is E_{2}
Let the amount of energy to be produced using nuclear power per year in 2020 is E
(Only 10% of the required electrical energy is to be produced by Nuclear power and only 25% of thermo-nuclear is successfully converted into electrical energy)
Amount of Uranium required to produce this much energy is M
(N_{A}=6.02310^{23}, Atomic mass of Uranium is 235 g)
=3.07610^{4} kg