Chapter 13 – Nuclei Questions and Answers: NCERT Solutions for Class 12 Physics

Class 12 Physics NCERT book solutions for Chapter 13 - Nuclei Questions and Answers.

1. Questions and Answers
(a) Two stable isotopes of lithium 63Li36Liand 73Li37Li have respective abundances of 7.57.5 and 92.592.5These isotopes have masses 6.01512u6.01512uand7.01600u7.01600u, respectively. Find the atomic mass of lithium.

Ans: We are given the following information:
Mass of 63Li36Li lithium isotope, m1=6.01512um1=6.01512u
Mass of 73Li37Li lithium isotope, m2=7.01600um2=7.01600u
Abundance of63Li36Li, n1=7.5n1=7.5
Abundance of 73Li37Li, n2=92.5n2=92.5
The atomic mass of lithium atom is given by,
Substituting the given values, we get,
Therefore, we found the atomic mass of lithium atoms to be 6.940934u6.940934u.

(b) Boron has Two Stable Isotopes 105B510B and115B511B. Their respective masses are 10.01294u10.01294uand11.00931u11.00931u, and the atomic mass of boron is 10.811u10.811u. Find the abundances of 105B510B and 115B511B .

Ans: We are given:
Mass of 105B510B Boron isotope, m1=10.01294um1=10.01294u
Mass of 115B511B lithium isotope, m2=11.00931um2=11.00931u
Abundance of105B510B, n1=xn1=x
Abundance of 115B511B, n2=(100−x)n2=(100−x)
We know the atomic mass of boron to be, m=10.811um=10.811u
The atomic mass of lithium atom is given by,
Substituting the given values, we get,
And, 100−x=80.11100−x=80.11
Therefore, we found the abundance of 105B510Band 115B511Bto be 19.8919.89 and 80.1180.11respectively.

2. The three stable isotopes of neon: 2010Ne,2110Neand2210Ne1020Ne,1021Neand1022Ne have respective abundances of 90.5190.51 The atomic masses of the three isotopes are 19.99u,20.99uand21.99u19.99u,20.99uand21.99u , respectively. Obtain the average atomic mass of neon.

Ans: We are given that:
Atomic mass of 2010Ne1020Ne, m1=19.99um1=19.99u
Abundance of 2010Ne1020Ne, η1=90.51η1=90.51
Atomic mass of2110Ne1021Ne, m2=20.99um2=20.99u
Abundance of 2110Ne1021Ne, η2=0.27η2=0.27
Atomic mass of 2210Ne1022Ne, m3=21.99um3=21.99u
Abundance of 2210Ne1022Ne, η3=9.22η3=9.22
The average atomic mass of neon could be given as,
Substituting the given values, we get,
The average atomic mass of neon is thus found to be 20.177u.20.177u.

3. Obtain the binding energy (in MeV) of a nitrogen nucleus (147N)(714N), given m(147N)=14.00307um(714N)=14.00307u

Ans: We are given:
Atomic mass of nitrogen (7N14)(7N14), m=14.00307um=14.00307u
A nucleus of 7N147N14nitrogen contains 7 protons and 7 neutrons.
Hence, the mass defect of this nucleus would be, Δm=7mH+7mn−mΔm=7mH+7mn−m
Where, Mass of a proton, mH=1.007825umH=1.007825u
Mass of a neutron, mn=1.008665umn=1.008665u
Substituting these values into the above equation, we get,
But we know that, 1u=931.5MeV/c21u=931.5MeV/c2
Now, we could give the binding energy as,
Where, c=speed of light =3×108ms−2c=speed of light =3×108ms−2
Now, Eb=0.11236×931.5(MeVc2)×c2Eb=0.11236×931.5(MeVc2)×c2
Therefore, we found the binding energy of a Nitrogen nucleus to be 104.66334MeV104.66334MeV.

4. Obtain the binding energy of the nuclei 5626Fe2656Fe and 20983Bi83209Bi in units of MeV from the following data: m(5626Fe)=55.934939um(2656Fe)=55.934939u, m(20983Bi)=208.980388um(83209Bi)=208.980388u

Ans: We are given the following:
Atomic mass of 5626Fe2656Fe, m1=55.934939um1=55.934939u
5626Fe2656Fe nucleus has 26 protons and 56−26=3056−26=30neutrons
Hence, the mass defect of the nucleus would be, Δm=26×mH+30×mn−m1Δm=26×mH+30×mn−m1
Where, Mass of a proton,mH=1.007825umH=1.007825u
Mass of a neutron,mn=1.008665umn=1.008665u
Substituting these values into the above equation, we get,
But we have, 1u=931.5MeV/c21u=931.5MeV/c2
The binding energy of this nucleus could be given as,
Where, c = Speed of light
Now, we have the average binding energy per nucleon to be,
Also, atomic mass of 20983Bi83209Bi, m2=208.980388um2=208.980388u
We know that, 20983Bi83209Bi nucleus has 83 protons and 209−83=126neutrons209−83=126neutrons
Mass of a proton, mH=1.007825umH=1.007825u
Mass of a neutron, mn=1.008665umn=1.008665u
But we know, 1u=931.5MeV/c21u=931.5MeV/c2
Hence, the binding energy of this nucleus could be given as,
Average binding energy per nucleon is found to be =1640.26209=7.848MeV=1640.26209=7.848MeV
Hence, the average binding energy per nucleon is found to be 7.848MeV7.848MeV.

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 6329Cu2963Cu atoms (of mass 62.92960u62.92960u).

Ans: We are given:
Mass of a copper coin, m′=3gm′=3g
Atomic mass of 29Cu6329Cu63atom,
The total number of 6329Cu2963Cu atoms in the coin, N=NA×m′Mass numberN=NA×m′Mass number
Mass number = 63g
29Cu6329Cu63nucleus has 29 protons and
Mass defect of this nucleus would be, Δm′=29×mH+34×mn−mΔm′=29×mH+34×mn−m
Mass of a proton, mH=1.007825umH=1.007825u
Mass of a neutron, mn=1.008665umn=1.008665uΔm′=29×1.007825+34×1.008665−62.9296=0.591935uΔm′=29×1.007825+34×1.008665−62.9296=0.591935u
Mass defect of all the atoms present in the coin would be,
But we have, 1u=931.5MeV/c21u=931.5MeV/c2
Hence, the binding energy of the nuclei of the coin could be given as:
But, 1MeV=1.6×10−13J1MeV=1.6×10−13J
This much energy is needed to separate all the neutrons and protons from the given coin.

6. Write the Nuclear Reactions For:
(a) α−decay of22688Raα−decay of88226Ra

Ans: We know that, ααis basically a nucleus of Helium (2He4)(2He4)and ββis an electron (e−forβ−ande+forβ+)(e−forβ−ande+forβ+). In every αα-decay, there is a loss of 2 protons and 2 neutrons. In every β+β+-decay, there is a loss of 1 proton and a neutrino is emitted from the nucleus. In every β−β−-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,

(b) α−decay of24294Puα−decay of94242Pu

Ans: 24294Pu→23892U+42He94242Pu→92238U+24He

1. β−decay of3215Pβ−decay of1532P

Ans: 3215P→3216S+e−+ν¯1532P→1632S+e−+ν¯

2. β−decay of21083Biβ−decay of83210Bi

Ans: 21083B→21084Po+e−+ν¯83210B→84210Po+e−+ν¯

3. β+−decay of116Cβ+−decay of611C

Ans: 116C→115B+e++ν611C→511B+e++ν

4. β+−decay of9743Tcβ+−decay of4397Tc

Ans: 9743Tc→9742Mo+e++ν4397Tc→4297Mo+e++ν

5. Electron capture of 12054Xe54120Xe

Ans: 12054Xe+e+→12053I+ν54120Xe+e+→53120I+ν

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=T years=T years Original amount of the radioactive isotope =N0=N0
(a) After decay, let the amount of the radioactive isotope be N.
It is given that only 3.1253.125 of N0N0remains after decay. Hence, we could write,
But we know that, NN0=e−λtNN0=e−λt
Where, λ=λ=Decay constant and t=t=Time
But, since λ=0.693Tλ=0.693T
Therefore, we found that the isotope will take about 5T years in order to reduce to 3.1253.125 of its original value.

2. 1% of its original value?

Ans: After decay, let the amount of the radioactive isotope be N
It is given that only 1% of N0N0 remains after decay. Hence, we could write:
But we know, NN0=e−λtNN0=e−λt
Since we have, λ=0.639Tλ=0.639T
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 146C614C present with the stable carbon isotope 126C612C . 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 146C614C , and the measured activity, the age of the specimen can be approximately estimated. This is the principle of 146C614C 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,
Let N be the number of radioactive atoms present in a normal carbon- containing matter.
Half life of 146C,T12=5730years614C,T12=5730years
The decay rate of the specimen obtained from the Mohenjodaro site:
Let N’ be the number of radioactive atoms present in the specimen during the Mohenjodaro period.
Therefore, we can relate the decay constant, λλ and time, t as:
But we know,
Therefore, the approximate age of the Indus-Valley civilization is found to be 4223.5 years.

9. Obtain the Amount of 6027Co2760Co necessary to provide a radioactive source of 8.0mCi8.0mCistrength. The half-life of 6027Co2760Co is 5.3 years.

Ans: We know that,
The strength of the radioactive source could be given as,
Where, N is the required number of atoms.
Half life of 6027Co2760Co, T12=5.3yearsT12=5.3years
For decay constant λλ, we could give the rate of decay as,
Where, λ=0.693T12=0.6931.67×108s−1λ=0.693T12=0.6931.67×108s−1
Now for 27Co6027Co60, Mass of Avogadro number of atoms =60g=60g
Then, mass of 7.133×1016atoms=60×7.133×10166.023×1023=7.106×10−6g7.133×1016atoms=60×7.133×10166.023×1023=7.106×10−6g
Therefore, the amount of 27Co6027Co60that is required for the purpose is 7.106×10−6g7.106×10−6g.

10. The half life of 9038Sr3890Sr is 28years. What is the disintegration rate of 15mg of this isotope?

Ans: We know that,
Half life of 9038Sr3890Sr, t12=28years=28×365×24×3600=8.83×108st12=28years=28×365×24×3600=8.83×108s
Mass of the isotope, m=15mgm=15mg
90g of 9038Sr3890Sr atom contains Avogadro number of atoms. So, 15mg of 9038Sr3890Sr contains,
6.023×1023×15×10−390=1.0038×1020number of atoms6.023×1023×15×10−390=1.0038×1020number of atoms
Rate of disintegration would be, dNdt=λNdNdt=λN
Where, λλis the decay constant given by, λ=0.6938.83×108s−1λ=0.6938.83×108s−1
Therefore, we found the disintegration rate of 15mg of given isotope to be 7.878×1010atoms/s7.878×1010atoms/s.

11. Obtain approximately the ratio of the nuclear radii of the gold isotope 19779Au79197Au and the silver isotope 10747Ag47107Ag.

Ans: We know that,
Nuclear radius of the gold isotope 79Au197=RAu79Au197=RAu
Nuclear radius of the silver isotope 47Ag107=RAg47Ag107=RAg
Mass number of gold, AAu=197AAu=197
Mass number of silver, AAg=107AAg=107
We also know that the ratio of the radii of the two nuclei is related with their mass numbers as:
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(22688Ra)=226.02540u,m(22289Rn)=222.01750u,$$m(22086Rn)=220.01137u,m(21684Po)=216.00189u$Given:$m(88226Ra)=226.02540u,m(89222Rn)=222.01750u,$$m(86220Rn)=220.01137u,m(84216Po)=216.00189u$
1. 22688Ra88226Ra

Ans: We know that,
Alpha particle decay of 2688Ra8826Ra emits a helium nucleus. As a result, its mass number reduces to 222=(226−4)222=(226−4)and its atomic number reduces to86=(88−2)86=(88−2). This is shown in the following nuclear reaction:
Where, c = Speed of light
It is also given that:
On substituting these values into the above equation,
Q value =[226.02540−(222.01750+4.002603)]uc2Q value =[226.02540−(222.01750+4.002603)]uc2
But we know, 1u=931.5MeV/c21u=931.5MeV/c2
Kinetic energy of the ααparticle=(Mass number after decayMassnumber before decay)×Q=(Mass number after decayMassnumber before decay)×Q
Hence, the Kinetic energy of the alpha particle is found to be 4.85MeV4.85MeV.

2. 22086Rn86220Rn

Ans: We know that, Alpha particle decay of 22086Rn86220Rn could be given as,
We are also given,
Mass of 22086Rn=220.01137u86220Rn=220.01137u
Mass of 21684Po=216.00189u84216Po=216.00189u
Now, Q value could be given as,
Now, we have the kinetic energy as,
The kinetic energy of the alpha particle is found to be 6.29MeV6.29MeV.

13. The Radionuclide 116C611C decays according to,
116C→115B+e++ν;T12=20.3min611C→511B+e++ν;T12=20.3min The maximum energy of the emitted positron is 0.960MeV0.960MeV. Given the mass values: m(116C)=11.011434uand m(116B)=11.009305um(611C)=11.011434uand m(611B)=11.009305u Calculate Q and compare it with the maximum energy of the positron emitted.

Ans: The given nuclear reaction is,
Half life of 116C611C nuclei, T12=20.3minT12=20.3min
Atomic masses are given to be:
Maximum energy that is possessed by the emitted positron would be
. The change in the Q – value
of the nuclear masses of the 116C611C
ΔQ=[m′(6C11)−[m′(115B)+me]]c2ΔQ=[m′(6C11)−[m′(511B)+me]]c2…………………….. (1)
Where,me=me=Mass of an electron or positron =
c=c= Speed of light
m′=m′=Respective nuclear masses
If atomic masses are used instead of nuclear masses, then we will have to add 6me6me in the case of 11C11C and 5me5mein case of 11B11B.
Hence, equation (1) would now reduce to,
Where, m(6C11)and m(115B)m(6C11)and m(511B)are the atomic masses.
Now, we have the change in Q value as,
But we know, 1u=931.5MeV/c21u=931.5MeV/c2
We see that the Q value is almost comparable to the maximum energy of the emitted positron.

14. The nucleus 2310Ne1023Ne decays by β−β−emission. Write down the ββdecay equation and determine the maximum kinetic energy of the electrons emitted. Given that:

Ans: We know that: In β−β−emission, the number of protons increases by 1, and one electron and an antineutrino are emitted from the parent nucleus. β−β−emission of the nucleus could be given by,
It is also given that:
Atomic mass of 2310Ne=22.994466u1023Ne=22.994466u
Atomic mass of 2311Na=22.989770u1123Na=22.989770u
Mass of an electron, me=0.000548ume=0.000548u
Q value of the given reaction could be given as:
There are 10 electrons in 10Ne2310Ne23and 11 electrons in 2311Na1123Na. Hence, the mass of the electron is cancelled in the Q-value equation.
But we have, 1u=931.5MeV/c21u=931.5MeV/c2
The daughter nucleus is too heavy as compared to that of e−andν¯e−andν¯. 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.,

15. The Q-value of a nuclear reaction A+b→C+dA+b→C+d is defined by Q=[mA+mb−mC−md]c2Q=[mA+mb−mC−md]c2 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(21H)=2.014102u,m(31H)=3.016049u,m(12H)=2.014102u,m(13H)=3.016049u,

1. 11H+31H→21H+21H11H+13H→12H+12H

Ans:The given nuclear reaction is:
Atomic mass of 11H=1.007825u11H=1.007825u
Atomic mass of 31H=3.0164049u13H=3.0164049u
Atomic mass of 21H=2.014102u12H=2.014102u
According to the question, the Q-value of the reaction could be written as:
But we know, 1u=931.5MeV/c21u=931.5MeV/c2
The negative Q-value of this reaction shows that the given reaction is endothermic.

2. 126C+126C→2010Ne+42He612C+612C→1020Ne+24He

Ans: We are given that,
Atomic mass of 126C=12.0u612C=12.0u
Atomic mass of 1210Ne=19.992439u1012Ne=19.992439u
Atomic mass of 42He=4.002603u24He=4.002603u
The Q-value here could be given as,
Since the Q-value is found to be positive, the reaction could be considered exothermic.

16. Suppose, we think of fission of a 5626Fe2656Fe nucleus into two equal fragments of 13Al2813Al28. Is fission energetically possible? Argue by working out Q of the process. Given: m(5626Fe)=55.93494uand m(2813Al)=27.98191um(2656Fe)=55.93494uand m(1328Al)=27.98191u

Ans: We know that the fission of 5626Fe2656Fe could be given as,
We are also given, atomic masses of 5626Feand2813Al2656Feand1328Al as 55.93494uand 27.98191u55.93494uand 27.98191u respectively.
The Q-value here would be given as,
But, 1u=931.5MeV/c21u=931.5MeV/c2
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 23994Pu94239Pu are very similar to those of
23592U92235U.The average energy released per fission is 180MeV180MeV. How much energy, in MeV, is released if all the atoms in 1kg1kg of pure 23994Pu94239Pu undergo fission?

Ans: We are given that the average energy released per fission of 23994Pu94239Pu, Eav=180MeVEav=180MeV
The amount of pure 94Pu23994Pu239, m=1kg=1000gm=1kg=1000g
Avogadro number, NA=6.023×1023NA=6.023×1023
Mass number of 23994Pu=239g94239Pu=239g
1 mole of 94Pu23994Pu239contains Avogadro number of atoms.
Total energy released during the fission of 1kg of 23994Pu94239Pucould be calculated as:
Therefore, 4.536×1026MeV4.536×1026MeV is released if all the atoms in 1kg of pure 94Pu23994Pu239undergo fission.

18. A 1000MW fission reactor consumes half of its fuel in 5.00 y. How much
23592U92235U did it contain initially? Assume that the reactor operates 80% of the time, that all the energy generated arises from the fission of 23592U92235U 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, t12=5yearst12=5years
We know that in the fission of 1g of 23592U92235U nucleus, the energy released is equal to 200MeV.
1 mole, i.e., 235g of 23592U92235U contains
The total energy generated per gram of 23592U92235U is calculated as:
The reactor operator operates only 80% of the time. Therefore, the amount of 23592U92235U consumed in 5years by the 1000MW fission reactor could be calculated as,
So, the initial amount of 23592U=2×1538=3076kg92235U=2×1538=3076kg
Hence, we found the initial amount of uranium to be 3076kg.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 21H+21H→32He+n+3.27MeV12H+12H→23He+n+3.27MeV

Ans: The fusion reaction is given to be:
Amount of deuterium,
1 mole, i.e., 2 g of deuterium contains
2.0 kg of deuterium contains 6.023×10232×2000=6.023×1026atoms6.023×10232×2000=6.023×1026atoms atoms
It could be inferred from the given reaction that when two atoms of deuterium fuse,
energy is released.
Therefore, the total energy per nucleus released in the fusion reaction would be:
Power of the electric lamp is given to be, P=100W=100J/sP=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,
Hence, the total time for which the electric lamp glows is found to be 4.9×104years.4.9×104years.

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=2×10−15m=2fm=2×10−15m
Also, charge on a deuteron == Charge on an electron =e=1.6×10−19C=e=1.6×10−19C
Potential energy of the two-deuteron system could be given by,
Where, ε0ε0is the permittivity of free space.
Also, 14πε0=9×109Nm2C−214πε0=9×109Nm2C−2
Therefore, we found the height of the potential barrier of the two-deuteron system to be 360keV.

21. From the relation R=R0A13R=R0A13, where R0R0 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:
Where, R0R0 is a Constant and AA is the mass number of the nucleus
Nuclear matter density would be,
ρ=Mass of the nucleusVolume of the nucleusρ=Mass of the nucleusVolume of the nucleus
Now, let m be the average mass of the nucleus, then, mass of the nucleus =mA=mA
Nuclear density,
Therefore, we found the nuclear matter density to be independent of A and it is found to be nearly constant.

22. For the β+β+(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).
Show that if β+β+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 Q1Q1 . The nuclear reaction could be written as:
e++AZX→AZ−1Y+ν+Q1e++ZAX→Z−1AY+ν+Q1………………………… (1)
Let the amount of energy released during the positron capture process be Q2Q2. The nuclear reaction could be written as:
AZX→AZ−1Y+e++ν+Q2ZAX→Z−1AY+e++ν+Q2………………………… (2)
Let, mN(AZX)be the nuclear mass ofAZXmN(ZAX)be the nuclear mass ofZAX, mN(AZ−1Y)be the nuclear mass ofAZ−1YmN(Z−1AY)be the nuclear mass ofZ−1AY
m(AZX)be the atomic mass ofAZXm(ZAX)be the atomic mass ofZAX
m(AZ−1Y)be the nuclear mass ofAZ−1Ym(Z−1AY)be the nuclear mass ofZ−1AY
mebe the mass of an electronmebe the mass of an electron, cbe the speed of lightcbe the speed of light, then, the Q-value of the electron capture reaction could be given as,
⇒Q1=[m(AZX)−m(AZ−1Y)]c2⇒Q1=[m(ZAX)−m(Z−1AY)]c2………………… (3)
The Q-value of the positron capture reaction could be given as,
⇒Q1=[m(AZX)−m(AZ−1Y)−2me]c2⇒Q1=[m(ZAX)−m(Z−1AY)−2me]c2………………… (4)
It can be inferred that if
, then; Also, if
, it does not necessarily mean that
. In other words, we could say that if β+β+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
24.312u24.312u. The average value is based on their relative natural abundance on earth. The three isotopes and their masses are:
2412Mg(23.98504u),2512Mg(24.98584u)and2612Mg(25.98259u)1224Mg(23.98504u),1225Mg(24.98584u)and1226Mg(25.98259u)The natural abundance of 12Mg2412Mg24is 78.99% by mass. Calculate the abundances of the other two isotopes.

Ans: We are given:
Average atomic mass of magnesium,
Mass of magnesium 2412Mg1224Mg isotope,
Mass of magnesium 2512Mg1225Mg isotope,
Mass of magnesium 2612Mg1226Mg isotope,
Abundance of 2412Mg,η1=78.991224Mg,η1=78.99
Abundance of 2512Mg,η2=x1225Mg,η2=x
Now, the abundance of 2612Mg1226Mg,
Also, we have the relation for the average atomic mass as:
And, 21.01−x=11.7121.01−x=11.71
Therefore, we found the abundance of 2512Mg1225Mg to be 9.3% and that of 2612Mg1226Mg 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 4120Ca2041Ca and 2713Al1327Al from the following data:

Ans: For a neutron removal from 20Ca4120Ca41nucleus, the corresponding nuclear reaction could be written as,
We are given:
Now, the mass defect for this reaction could be given by,
But we know, 1u=931.5MeV/c21u=931.5MeV/c2
Now, we could calculate the energy required for the neutron removal by,
For the case of 2713Al1327Al, the neutron removal reaction could be written as,
We are given,
Now, the mass defect here could be given by,
Therefore, the energy that is required for the removal of neutron would be,

25. A source contains two phosphorous radio nuclides 3215P(T12=14.3d)1532P(T12=14.3d) and 3315P(T12=25.3d)1533P(T12=25.3d)Initially, 10% of the decays come from 3315P1533P. How long must one wait until 90% do so?

Ans: We are given:
Half life of 3215P(T12=14.3d)1532P(T12=14.3d)
Half life of 3315P(T12=25.3d)1533P(T12=25.3d)
Now, we know that nucleus decay is 10% of the total amount of decay.
Also, the source has initially 10% of 3215P1532P nucleus and 90% of 3215P1532P nucleus.
Suppose after t days, the source has 10% of 3215P1532P nucleus and 90% of 3315P1533P
Initially we have:
Number of 3315P1533P nucleus =N=N
Number of 3215P1532P nucleus =9N=9N
Number of 3315P1533P nucleus =9N′=9N′
Number of 3215P1532P nucleus=N′=N′
For 3215P1532P nucleus, we could write the number ratio as:
⇒N′=9N(2)−t14.3⇒N′=9N(2)−t14.3………………………. (1)
Now, for 3315P1533P, we could write the number ratio as,
⇒9N′=N(2)−t25.3⇒9N′=N(2)−t25.3…………………………. (2)
We could now divide equation (1) by equation (2) to get,
Therefore, we found that it would take about 208.5days for 90% decay of 3315P1533P.

26. Under certain circumstances, a nucleus can decay by emitting a particle more massive than an α−α−particle. Consider the following decay processes:
Calculate the Q-values for these decays and determine that both are energetically allowed.

Ans: Consider a 146C614C emission nuclear reaction,
We know that:
Mass of 22388Ra,m1=223.01850u88223Ra,m1=223.01850u
Mass of 146C,m3=14.00324u614C,m3=14.00324u
Now, the Q-value of the reaction could be given as:
But we have, 1u=931.5MeV/c21u=931.5MeV/c2
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 42He24He emission nuclear reaction:
We know that:
Mass of 22388Ra,m1=223.0185088223Ra,m1=223.01850
Mass of 21982Rn,m2=219.0094882219Rn,m2=219.00948
Mass of 42He,m3=4.0026024He,m3=4.00260
Q-value of this nuclear reaction could be given as:
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 23892U92238U 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 14058Ceand9944Ru58140Ceand4499Ru. Calculate Q for this fission process. The relevant atomic and particle masses are:

Ans: We are given:
In the fission of 23892U92238U, 10 β−β−particles decay from the parent nucleus. The nuclear reaction can be written as:
It is also given that:
Mass of a nucleus of 23892U,m1=238.05079u92238U,m1=238.05079u
Mass of a nucleus of 14058Ce,m2=139.90543u58140Ce,m2=139.90543u
Mass of nucleus of 9944Ru,m3=98.90594u4499Ru,m3=98.90594u,
Mass of a neutron 10n,m4=1.008665u01n,m4=1.008665u
Q-value of the above equation would be,
Where, m′=m′=Represents the corresponding atomic masses of the nuclei
But 1u=931.5MeV/c21u=931.5MeV/c2
Therefore, the Q-value of the fission process is found to be 231.007MeV.

28. Consider the D−T reaction (deuterium−tritium fusion)
1. Calculate the energy released in MeV in this reaction from the data: m(21H)=2.014102um(12H)=2.014102u , m(31H)=3.016049um(13H)=3.016049u

Ans: Consider the D-T nuclear reaction,
We are also given that:
Mass of
Mass of
Mass of 42He,m3=4.002603u24He,m3=4.002603u
Mass of 10n,m4=1.008665u01n,m4=1.008665u
Now, the Q-value of the given D-T reaction would be:
But 1u=931.5MeV/c21u=931.5MeV/c2

2. 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(3kT2)2(3kT2); k = Boltzmann’s constant, T = absolute temperature.)

Ans: We are given:
Radius of deuterium and tritium, r≈2.0fm=2×10−15mr≈2.0fm=2×10−15m
Distance between the two nuclei at the moment when they touch each other,
Charge on the deuterium nucleus =e=e
Charge on the tritium nucleus =e=e
Hence, the repulsive potential energy between the two nuclei could be given as:
Where, ε0=ε0=Permittivity of free space
Therefore, 5.76×10−14J5.76×10−14J or 360keV360keVof kinetic energy (KE) is needed to overcome the coulomb repulsion between the two nuclei.
However, we are also given that:
Where, k=k=Boltzmann constant
T=T=Temperature required for triggering the reaction
Therefore, we found that the gas must be heated to a temperature of 1.39×109K1.39×109K to initiate the reaction.

29. Obtain the maximum kinetic energy of β−β−particles, and the radiation frequencies of γγ decays in the decay scheme shown in figure. You are given that:

Ans: It can be observed from the given γγ-decay diagram that γ1γ1decays from the 1.088MeV energy level to the 0MeV energy level. Hence, the energy corresponding to γ1γ1-decay is given as:
Where, Planck’s constant h=6.6×10−34Jsh=6.6×10−34Js
ν1=ν1=Frequency of radiation radiated by γ1−γ1−decay
It can be observed from the given γ−γ−decay diagram that γ2γ2decays from the
energy level to the
energy level.
Now, the energy corresponding to γ2γ2-decay could be given as:
Where, ν2=ν2=Frequency of radiation radiated by γ2−γ2−decay
It can be observed from the given γγ-decay diagram that γ3γ3-decays from the
energy level to the
energy level.
Now, the energy corresponding to γ3γ3 -decay is given as:
Where, ν3=ν3=Frequency of radiation radiated by γ3−γ3−decay
Mass of
Mass of m(19880Hg)=197.966760um(80198Hg)=197.966760u
Energy of the highest level could be given as:
β1β1decays from the
level to the
Maximum kinetic energy of the β1β1particle =1.3720995−1.088=1.3720995−1.088
β2β2 decays from the 1.3720995MeV1.3720995MeVlevel to that of the 0.412MeV0.412MeVlevel. Now, we find the maximum kinetic energy of the β2β2particle to be,
Therefore, we found the maximum kinetic energy of the β2β2particle to be 0.9600995MeV0.9600995MeV.

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,
1 mole, i.e., 1g of hydrogen (11H)(11H) contains
That is, 1000g of 11H11H contains
Within the sun, four 11H11H nuclei combine and form one 42He24He nucleus. In this process
of energy is released.
Hence, the energy released from the fusion of 1 kg 11H11H is:
Therefore, we found the energy released during the fusion of 1kg 11H11H is:
Hence, the energy released during the fusion of 1kg of 11H11H to be 39.1495×1026MeV39.1495×1026MeV.

2. The Fission of 1.0kg of 235U235U in a fission reactor.

Ans: We are given:
Amount of 92U235=1000gm92U235=1000gm
1 mole, i.e., 235g of 23592U92235U contains
1000g of 23592U92235U contains 6.023×1023×1000235atoms6.023×1023×1000235atoms
We know that the amount of energy released in the fission of one atom of 23592U92235U is
. Therefore, energy released from the fission of 1kg of 23592U92235U is:
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 235U235U to be about

Ans: We are given the following:
Amount of electric power to be generated,
, 10% of this amount has to be obtained from nuclear power plants.
Amount of nuclear power, P1=10100×2×105=2×104MWP1=10100×2×105=2×104MW
Heat energy released per fission of a 235U235U nucleus,
Efficiency of a reactor
Hence, the amount of energy converted into the electrical energy per fission is calculated as:
The number of atoms required for fission per year would be:
1 mole, i.e., 235g of U235U235contains
That is, the mass of
of U235=235g=235×10−3kgU235=235g=235×10−3kg
Also, the mass of:78840×1024atomsofU235=235×10−36.023×1023×78840×1024=3.076×104kg78840×1024atomsofU235=235×10−36.023×1023×78840×1024=3.076×104kg
Hence, the mass of uranium needed per year is found to be,