# 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.

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

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,

m=m1n1+m2n2n1+n2m=m1n1+m2n2n1+n2

Substituting the given values, we get,

m=6.01512×7.5+7.01600×92.592.5+7.5m=6.01512×7.5+7.01600×92.592.5+7.5

∴m=6.940934u∴m=6.940934u

Therefore, we found the atomic mass of lithium atoms to be 6.940934u6.940934u.

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,

m=m1n1+m2n2n1+n2m=m1n1+m2n2n1+n2

Substituting the given values, we get,

10.811=10.01294×x+11.00931×(100−x)x+(100−x)10.811=10.01294×x+11.00931×(100−x)x+(100−x)

⇒1081.11=10.01294x+1100.931−11.00931x⇒1081.11=10.01294x+1100.931−11.00931x

∴x=19.8210.99637=19.89∴x=19.8210.99637=19.89

And, 100−x=80.11100−x=80.11

Therefore, we found the abundance of 105B510Band 115B511Bto be 19.8919.89 and 80.1180.11respectively.

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,

m=m1η1+m2η2+m3η3η1+η2+η3m=m1η1+m2η2+m3η3η1+η2+η3

Substituting the given values, we get,

m=19.99×90.51+20.99×0.27+21.99×9.2290.51+0.27+9.22m=19.99×90.51+20.99×0.27+21.99×9.2290.51+0.27+9.22

∴m=20.1771u∴m=20.1771u

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

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,

Δm=7×1.007825+7×1.008665−14.00307Δm=7×1.007825+7×1.008665−14.00307

⇒Δm=7.054775+7.06055−14.00307⇒Δm=7.054775+7.06055−14.00307

∴Δm=0.11236u∴Δm=0.11236u

But we know that, 1u=931.5MeV/c21u=931.5MeV/c2

⇒Δm=0.11236×931.5MeV/c2⇒Δm=0.11236×931.5MeV/c2

Now, we could give the binding energy as,

Eb=Δmc2Eb=Δmc2

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

∴Eb=104.66334MeV∴Eb=104.66334MeV

Therefore, we found the binding energy of a Nitrogen nucleus to be 104.66334MeV104.66334MeV.

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,

Δm=26×1.007825+30×1.008665−55.934939Δm=26×1.007825+30×1.008665−55.934939

⇒Δm=26.20345+30.25995−55.934939⇒Δm=26.20345+30.25995−55.934939

∴Δm=0.528461u∴Δm=0.528461u

But we have, 1u=931.5MeV/c21u=931.5MeV/c2

Δm=0.528461×931.5MeV/c2Δm=0.528461×931.5MeV/c2

The binding energy of this nucleus could be given as,

Eb1=Δmc2Eb1=Δmc2

Where, c = Speed of light

⇒Eb1=0.528461×931.5(MeVc2)×c2⇒Eb1=0.528461×931.5(MeVc2)×c2

∴Eb1=492.26MeV∴Eb1=492.26MeV

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

B.E=492.2656=8.79MeVB.E=492.2656=8.79MeV

Also, atomic mass of 20983Bi83209Bi, m2=208.980388um2=208.980388u

We know that, 20983Bi83209Bi nucleus has 83 protons and 209−83=126neutrons209−83=126neutrons

Where,

Mass of a proton, mH=1.007825umH=1.007825u

Mass of a neutron, mn=1.008665umn=1.008665u

Δm′=83×1.007825+126×1.008665−208.980388Δm′=83×1.007825+126×1.008665−208.980388

⇒Δm′=83.649475+127.091790−208.980388⇒Δm′=83.649475+127.091790−208.980388

∴Δm′=1.760877u∴Δm′=1.760877u

But we know, 1u=931.5MeV/c21u=931.5MeV/c2

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

Eb2=Δm′c2=1.760877×931.5(MeVc2)×c2Eb2=Δm′c2=1.760877×931.5(MeVc2)×c2

∴Eb2=1640.26MeV∴Eb2=1640.26MeV

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.

Mass of a copper coin, m′=3gm′=3g

Atomic mass of 29Cu6329Cu63atom,

m=62.92960um=62.92960u

The total number of 6329Cu2963Cu atoms in the coin, N=NA×m′Mass numberN=NA×m′Mass number

Where,

NA=Avogadronumber=6.023×1023atoms/gNA=Avogadronumber=6.023×1023atoms/g

Mass number = 63g

⇒N=6.023×1023×363=2.868×1022⇒N=6.023×1023×363=2.868×1022

29Cu6329Cu63nucleus has 29 protons and

(63−29)=34(63−29)=34

neutrons

Mass defect of this nucleus would be, Δm′=29×mH+34×mn−mΔm′=29×mH+34×mn−m

Where,

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,

Δm=0.591935×2.868×1022=1.69766958×1022uΔm=0.591935×2.868×1022=1.69766958×1022u

But we have, 1u=931.5MeV/c21u=931.5MeV/c2

⇒Δm=1.69766958×1022×931.5MeV/c2⇒Δm=1.69766958×1022×931.5MeV/c2

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

Eb=Δmc2=1.69766958×1022×931.5(MeVc2)×c2Eb=Δmc2=1.69766958×1022×931.5(MeVc2)×c2

∴Eb=1.581×1025MeV∴Eb=1.581×1025MeV

But, 1MeV=1.6×10−13J1MeV=1.6×10−13J

⇒Eb=1.581×1025×1.6×10−13⇒Eb=1.581×1025×1.6×10−13

∴Eb=2.5296×1012J∴Eb=2.5296×1012J

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

(a) α−decay of22688Raα−decay of88226Ra

For the given case, the nuclear reaction would be,

88Ra226→86Rn222+2He488Ra226→86Rn222+2He4

1. 3.125% of Its Original Value?

(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,

NN0=3.125NN0=3.125

But we know that, NN0=e−λtNN0=e−λt

Where, λ=λ=Decay constant and t=t=Time

⇒−λt=132⇒−λt=132

⇒−λt=lnl−ln32⇒−λt=lnl−ln32

⇒−λt=0−3.4657⇒−λt=0−3.4657

⇒t=3.4657λ⇒t=3.4657λ

But, since λ=0.693Tλ=0.693T

⇒t=3.466(0.693T)⇒t=3.466(0.693T)

∴t≈5Tyears∴t≈5Tyears

Therefore, we found that the isotope will take about 5T years in order to reduce to 3.1253.125 of its original value.

It is given that only 1% of N0N0 remains after decay. Hence, we could write:

NN0=1NN0=1

But we know, NN0=e−λtNN0=e−λt

⇒e−λt=1100⇒e−λt=1100

⇒−λt=ln1−ln100⇒−λt=ln1−ln100

⇒−λt=0−4.602⇒−λt=0−4.602

⇒t=4.6052λ⇒t=4.6052λ

Since we have, λ=0.639Tλ=0.639T

⇒t=4.6052(0.693T)⇒t=4.6052(0.693T)

∴t=6.645Tyears∴t=6.645Tyears

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

Decay rate of living carbon-containing matter,

R=15decay/minR=15decay/min

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:

R′=9decays/minR′=9decays/min

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:

N′N=R′R=e−λtN′N=R′R=e−λt

⇒e−λt=915=35⇒e−λt=915=35

⇒−λt=loge35=−0.5108⇒−λt=loge35=−0.5108

⇒t=0.5108λ⇒t=0.5108λ

But we know,

λ=0.693T12=0.6935730λ=0.693T12=0.6935730

⇒t=0.5108(0.6935730)=4223.5years⇒t=0.5108(0.6935730)=4223.5years

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

The strength of the radioactive source could be given as,

dNdt=8.0mCidNdt=8.0mCi

⇒dNdt=8×10−3×3.7×1010=29.6×107decay/s⇒dNdt=8×10−3×3.7×1010=29.6×107decay/s

Where, N is the required number of atoms.

Half life of 6027Co2760Co, T12=5.3yearsT12=5.3years

⇒T12=5.3×365×24×60×60=1.67×108s⇒T12=5.3×365×24×60×60=1.67×108s

For decay constant λλ, we could give the rate of decay as,

dNdt=λNdNdt=λN

Where, λ=0.693T12=0.6931.67×108s−1λ=0.693T12=0.6931.67×108s−1

⇒N=1λdNdt=29.6×107(0.6931.67×108)=7.133×1016atoms⇒N=1λdNdt=29.6×107(0.6931.67×108)=7.133×1016atoms

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.

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

∴dNdt=0.693×1.0038×10208.83×108=7.878×1010atoms/s∴dNdt=0.693×1.0038×10208.83×108=7.878×1010atoms/s

Therefore, we found the disintegration rate of 15mg of given isotope to be 7.878×1010atoms/s7.878×1010atoms/s.

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:

RAuRAg=(AAuAAg)13=1.2256RAuRAg=(AAuAAg)13=1.2256

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

1. 22688Ra88226Ra

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:

22688Ra→22286Ra+42He88226Ra→86222Ra+24He

Q−valueofemittedα−particle=(Sumofinitialmass−Sumoffinalmass)c2Q−valueofemittedα−particle=(Sumofinitialmass−Sumoffinalmass)c2

Where, c = Speed of light

It is also given that:

m(22688Ra)=226.02540um(88226Ra)=226.02540u

m(22086Rn)=220.01137um(86220Rn)=220.01137u

m(42He)=4.002603um(24He)=4.002603u

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

Qvalue=0.005297uc2Qvalue=0.005297uc2

But we know, 1u=931.5MeV/c21u=931.5MeV/c2

⇒Q=0.005297×931.5≈4.94MeV⇒Q=0.005297×931.5≈4.94MeV

Kinetic energy of the ααparticle=(Mass number after decayMassnumber before decay)×Q=(Mass number after decayMassnumber before decay)×Q

∴K.Eα=222226×4.94=4.85MeV∴K.Eα=222226×4.94=4.85MeV

Hence, the Kinetic energy of the alpha particle is found to be 4.85MeV4.85MeV.

22086Rn→21684Po+42He86220Rn→84216Po+24He

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,

Q−value=[220.01137−(216.00189+4.00260)]×931.5≈641MeVQ−value=[220.01137−(216.00189+4.00260)]×931.5≈641MeV

Now, we have the kinetic energy as,

K.Eα=(220−4220)×6.41=6.29MeVK.Eα=(220−4220)×6.41=6.29MeV

The kinetic energy of the alpha particle is found to be 6.29MeV6.29MeV.

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.

116C→115B+e++ν611C→511B+e++ν

Half life of 116C611C nuclei, T12=20.3minT12=20.3min

Atomic masses are given to be:

m(116C)=11.011434um(611C)=11.011434u

m(116B)=11.009305um(611B)=11.009305u

Maximum energy that is possessed by the emitted positron would be

0.960MeV0.960MeV

. The change in the Q – value

(ΔQ)(ΔQ)

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 =

0.000548u0.000548u

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,

ΔQ=[m(6C11)−m(115B)−2me]c2ΔQ=[m(6C11)−m(511B)−2me]c2

Where, m(6C11)and m(115B)m(6C11)and m(511B)are the atomic masses.

Now, we have the change in Q value as,

ΔQ=[11.011434−11.009305−2×0.000548]c2=(0.001033c2)uΔQ=[11.011434−11.009305−2×0.000548]c2=(0.001033c2)u

But we know, 1u=931.5MeV/c21u=931.5MeV/c2

∴ΔQ=0.001033×931.5≈0.962MeV∴ΔQ=0.001033×931.5≈0.962MeV

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

m(2310Ne)=22.994466um(1023Ne)=22.994466u

m(2311Na)=22.989770um(1123Na)=22.989770u

2310Ne→2311Na+e−+ν¯+Q1023Ne→1123Na+e−+ν¯+Q

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:

Q=[m(2310Ne)−[m(2311Na)+me]]c2Q=[m(1023Ne)−[m(1123Na)+me]]c2

There are 10 electrons in 10Ne2310Ne23and 11 electrons in 2311Na1123Na. Hence, the mass of the electron is cancelled in the Q-value equation.

Q=[22.994466−22.9897770]c2=(0.004696c2)uQ=[22.994466−22.9897770]c2=(0.004696c2)u

But we have, 1u=931.5MeV/c21u=931.5MeV/c2

⇒Q=0.004696×931.5=4.374MeV⇒Q=0.004696×931.5=4.374MeV

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.,

4.374MeV4.374MeV.

Atomic masses are given to be: m(21H)=2.014102u,m(31H)=3.016049u,m(12H)=2.014102u,m(13H)=3.016049u,

m(126C)=12.000000u,m(2010Ne)=19.992439um(612C)=12.000000u,m(1020Ne)=19.992439u

11H+31H→21H+21H11H+13H→12H+12H

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:

Q=[m(11H)+m(31H)−2m(21H)]c2Q=[m(11H)+m(13H)−2m(12H)]c2

⇒Q=[1.007825+3.016049−2×2.014102]c2=(−0.00433c2)u⇒Q=[1.007825+3.016049−2×2.014102]c2=(−0.00433c2)u

But we know, 1u=931.5MeV/c21u=931.5MeV/c2

∴Q=−0.00433×931.5=−4.0334MeV∴Q=−0.00433×931.5=−4.0334MeV

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

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,

Q=[2m(126C)−m(2010Ne)−m(42He)]c2Q=[2m(612C)−m(1020Ne)−m(24He)]c2

⇒Q=[2×12.0−19.992439−4.002603]c2=(0.004958c2)u=0.004958×931.5⇒Q=[2×12.0−19.992439−4.002603]c2=(0.004958c2)u=0.004958×931.5

∴Q=4.618377MeV∴Q=4.618377MeV

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

5626Fe→22813Al2656Fe→21328Al

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,

Q=[m(5626Fe)−2m(2813Al)]c2Q=[m(2656Fe)−2m(1328Al)]c2

⇒Q=[55.93494−2×27.98191]c2=(−0.02888c2)u⇒Q=[55.93494−2×27.98191]c2=(−0.02888c2)u

But, 1u=931.5MeV/c21u=931.5MeV/c2

∴Q=−0.02888×931.5=−26.902MeV∴Q=−0.02888×931.5=−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.

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?

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.

1gof94Pu239contains(NAmassnumber×m)atoms1gof94Pu239contains(NAmassnumber×m)atoms

⇒(6.023×1023239×1000)=2.52×1024atoms⇒(6.023×1023239×1000)=2.52×1024atoms

Total energy released during the fission of 1kg of 23994Pu94239Pucould be calculated as:

E=Eav×2.52×1024=180×2.52×1024=4.536×1026MeVE=Eav×2.52×1024=180×2.52×1024=4.536×1026MeV

Therefore, 4.536×1026MeV4.536×1026MeV is released if all the atoms in 1kg of pure 94Pu23994Pu239undergo fission.

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.

⇒t12=5×365×24×60×60s⇒t12=5×365×24×60×60s

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

6.023×10236.023×1023

atoms.

1gof23592Ucontains6.023×1023234atoms1gof92235Ucontains6.023×1023234atoms

The total energy generated per gram of 23592U92235U is calculated as:

E=6.023×1023235×200MeV/g=200×6.023×1023×1.6×10−19×106235=8.20×1010J/gE=6.023×1023235×200MeV/g=200×6.023×1023×1.6×10−19×106235=8.20×1010J/g

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,

5×80×60×60×365×24×1000×106100×8.20×1010g≈1538kg5×80×60×60×365×24×1000×106100×8.20×1010g≈1538kg

So, the initial amount of 23592U=2×1538=3076kg92235U=2×1538=3076kg

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

21H+21H→32He+n+3.27MeV12H+12H→23He+n+3.27MeV

Amount of deuterium,

m=2kgm=2kg

1 mole, i.e., 2 g of deuterium contains

6.023×1023atoms6.023×1023atoms

.

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,

3.27MeV3.27MeV

energy is released.

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

E=3.272×6.023×1026MeV=3.272×6.023×1026×1.6×10−19×106E=3.272×6.023×1026MeV=3.272×6.023×1026×1.6×10−19×106

∴E=1.576×1014J∴E=1.576×1014J

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,

t=1.576×1014100=1.576×1014100×60×60×24×365t=1.576×1014100=1.576×1014100×60×60×24×365

∴t≈4.9×104years∴t≈4.9×104years

Hence, the total time for which the electric lamp glows is found to be 4.9×104years.4.9×104years.

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

Radius of a deuteron nucleus=2fm=2×10−15m=2fm=2×10−15m

⇒d=2×10−15+2×10−15=4×10−15m⇒d=2×10−15+2×10−15=4×10−15m

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,

V=e24πε0dV=e24πε0d

Where, ε0ε0is the permittivity of free space.

Also, 14πε0=9×109Nm2C−214πε0=9×109Nm2C−2

⇒V=9×109×(1.6×1019)24×1015J=9×109×(1.6×10−19)24×10−15×(1.6×10−19)eV⇒V=9×109×(1.6×1019)24×1015J=9×109×(1.6×10−19)24×10−15×(1.6×10−19)eV

∴V=360keV∴V=360keV

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

R=R0A13R=R0A13

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,

ρ=mA43πR3=3mA4π(R0A13)3=3mA4πR03Aρ=mA43πR3=3mA4π(R0A13)3=3mA4πR03A

∴ρ=3m4πR03∴ρ=3m4πR03

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

e++AZX→AZ−1Y+νe++ZAX→Z−1AY+ν

Show that if β+β+emission is Energetically Allowed, Electron Capture is Necessarily Allowed but Not Vice−versa.

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=[mN(AZX)+me−mN(AZ−1Y)]c2Q1=[mN(ZAX)+me−mN(Z−1AY)]c2

⇒Q1=[m(AZX)−Zme+me−m(AZ−1Y)+(Z−1)me]c2⇒Q1=[m(ZAX)−Zme+me−m(Z−1AY)+(Z−1)me]c2

⇒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,

Q2=[mN(AZX)−mN(AZ−1Y)−me]c2Q2=[mN(ZAX)−mN(Z−1AY)−me]c2

⇒Q1=[m(AZX)−Zme−m(AZ−1Y)+(Z−1)me−me]c2⇒Q1=[m(ZAX)−Zme−m(Z−1AY)+(Z−1)me−me]c2

⇒Q1=[m(AZX)−m(AZ−1Y)−2me]c2⇒Q1=[m(ZAX)−m(Z−1AY)−2me]c2………………… (4)

It can be inferred that if

Q2>0Q2>0

, then; Also, if

Q1>0Q1>0

, it does not necessarily mean that

Q2>0Q2>0

. 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.

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.

Average atomic mass of magnesium,

m=24.312um=24.312u

Mass of magnesium 2412Mg1224Mg isotope,

m1=23.98504um1=23.98504u

Mass of magnesium 2512Mg1225Mg isotope,

m2=24.98584um2=24.98584u

Mass of magnesium 2612Mg1226Mg isotope,

m3=25.98259um3=25.98259u

Abundance of 2412Mg,η1=78.991224Mg,η1=78.99

Abundance of 2512Mg,η2=x1225Mg,η2=x

Now, the abundance of 2612Mg1226Mg,

η3=100−x−78.99η3=100−x−78.99

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

m=m1η1+m2η2+m3η3η1+η2+η3m=m1η1+m2η2+m3η3η1+η2+η3

⇒24.312=23.98504×78.99+24.98584×x+25.98259×(21.01−x)100⇒24.312=23.98504×78.99+24.98584×x+25.98259×(21.01−x)100

⇒0.99675x=9.2725255⇒0.99675x=9.2725255

∴x≈9.3∴x≈9.3

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%.

m(4020Ca)=39.962591u,m(4120Ca)=40.962278u,m(2613Al)=25.986995u,m(2040Ca)=39.962591u,m(2041Ca)=40.962278u,m(1326Al)=25.986995u,

m(2713Al)=26.981541um(1327Al)=26.981541u

4120Ca→4020Ca+10n2041Ca→2040Ca+01n

We are given:

m(4020Ca)=39.962591um(2040Ca)=39.962591u

m(4120Ca)=40.962278um(2041Ca)=40.962278u

m(0n1)=1.008665um(0n1)=1.008665u

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

Δm=m(4020Ca)+(10n)−m(4120Ca)Δm=m(2040Ca)+(01n)−m(2041Ca)

⇒Δm=39.962591+1.008665−40.962278=0.008978u⇒Δm=39.962591+1.008665−40.962278=0.008978u

But we know, 1u=931.5MeV/c21u=931.5MeV/c2

⇒Δm=0.008978×931.5MeV/c2⇒Δm=0.008978×931.5MeV/c2

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

E=Δmc2E=Δmc2

⇒E=0.008978×931.5=8.363007MeV⇒E=0.008978×931.5=8.363007MeV

For the case of 2713Al1327Al, the neutron removal reaction could be written as,

2713Al→2613Al+10n1327Al→1326Al+01n

We are given,

m(2613Al)=25.986995um(1326Al)=25.986995u

m(2713Al)=26.981541um(1327Al)=26.981541u

Now, the mass defect here could be given by,

Δm=m(2613Al)+m(10n)−m(2713Al)Δm=m(1326Al)+m(01n)−m(1327Al)

⇒Δm=25.986895+1.008665−26.981541=0.014019u⇒Δm=25.986895+1.008665−26.981541=0.014019u

⇒Δm=0.014019×931.5MeV/c2⇒Δm=0.014019×931.5MeV/c2

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

E=Δmc2=0.014019×931.5E=Δmc2=0.014019×931.5

∴E=13.059MeV∴E=13.059MeV

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

nucleus.

Initially we have:

Number of 3315P1533P nucleus =N=N

Number of 3215P1532P nucleus =9N=9N

Finally:

Number of 3315P1533P nucleus =9N′=9N′

Number of 3215P1532P nucleus=N′=N′

For 3215P1532P nucleus, we could write the number ratio as:

N′9N=(12)tT12N′9N=(12)tT12

⇒N′=9N(2)−t14.3⇒N′=9N(2)−t14.3………………………. (1)

Now, for 3315P1533P, we could write the number ratio as,

9N′N=(12)1T′129N′N=(12)1T′12

⇒9N′=N(2)−t25.3⇒9N′=N(2)−t25.3…………………………. (2)

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

19=9×2(t25.3−t14.3)19=9×2(t25.3−t14.3)

⇒181=2(−11t25.3×14.3)⇒181=2(−11t25.3×14.3)

⇒log1−log81=−11t25.3×14.3log1⇒log1−log81=−11t25.3×14.3log1

⇒−11t25.3×14.3=0−1.9080.301⇒−11t25.3×14.3=0−1.9080.301

∴t=25.3×14.3×1.90811×0.301≈208.5days∴t=25.3×14.3×1.90811×0.301≈208.5days

Therefore, we found that it would take about 208.5days for 90% decay of 3315P1533P.

22388Ra→20982Pb+146C88223Ra→82209Pb+614C

22388Ra→21986Rn+42He88223Ra→86219Rn+24He

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

22388Ra→20982Pb+146C88223Ra→82209Pb+614C

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:

Q=(m1−m2−m3)c2Q=(m1−m2−m3)c2

⇒Q=(223.01850−208.98107−14.00324)c2=(0.03419c2)u⇒Q=(223.01850−208.98107−14.00324)c2=(0.03419c2)u

But we have, 1u=931.5MeV/c21u=931.5MeV/c2

⇒Q=0.03419×931.5⇒Q=0.03419×931.5

∴Q=31.848MeV∴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 42He24He emission nuclear reaction:

22388Ra→22986Rn+42He88223Ra→86229Rn+24He

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:

Q=(m1−m2−m3)c2Q=(m1−m2−m3)c2

⇒Q=(223.01850−219.00948−4.00260)c2⇒Q=(223.01850−219.00948−4.00260)c2

⇒Q=(0.00642c2)u⇒Q=(0.00642c2)u

∴Q=0.00642×931.5=5.98MeV∴Q=0.00642×931.5=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.

m(23892U)=238.05079um(92238U)=238.05079u

m(14058Ce)=139.90543um(58140Ce)=139.90543u

m(9944Ru)=98.90594um(4499Ru)=98.90594u

In the fission of 23892U92238U, 10 β−β−particles decay from the parent nucleus. The nuclear reaction can be written as:

23892U+10n→14058Ce+9944Ru+100−1e92238U+01n→58140Ce+4499Ru+10−10e

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,

Q=[m′(23892U)+m(10n)−m′(14058Ce)−m′(9944Ru)−10me]c2Q=[m′(92238U)+m(01n)−m′(58140Ce)−m′(4499Ru)−10me]c2

Where, m′=m′=Represents the corresponding atomic masses of the nuclei

m′(23892U)=m1−92mem′(92238U)=m1−92me

m′(14058Ce)=m2−58mem′(58140Ce)=m2−58me

m′(9944Ru)=m3−44mem′(4499Ru)=m3−44me

m(10n)=m4m(01n)=m4

Q=[m1−92me+m4−m2+58me−m3+44me−10me]c2Q=[m1−92me+m4−m2+58me−m3+44me−10me]c2

⇒Q=[m1+m4−m2−m3]c2=[238.0507+1.008665−139.90543−98.90594]c2⇒Q=[m1+m4−m2−m3]c2=[238.0507+1.008665−139.90543−98.90594]c2

⇒Q=[0.247995c2]u⇒Q=[0.247995c2]u

But 1u=931.5MeV/c21u=931.5MeV/c2

∴Q=0.247995×931.5=231.007MeV∴Q=0.247995×931.5=231.007MeV

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

21H+31H→42He+n12H+13H→24He+n

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

21H+31H→42He+n12H+13H→24He+n

We are also given that:

Mass of

21H,m1=2.014102u12H,m1=2.014102u

Mass of

31H,m2=3.016049u13H,m2=3.016049u

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:

Q=[m1+m2−m3−m4]c2Q=[m1+m2−m3−m4]c2

⇒Q=[2.014102+3.016049−4.002603−1.008665]c2⇒Q=[2.014102+3.016049−4.002603−1.008665]c2

⇒Q=[0.018883c2]u⇒Q=[0.018883c2]u

But 1u=931.5MeV/c21u=931.5MeV/c2

∴Q=0.018883×931.5=17.59MeV∴Q=0.018883×931.5=17.59MeV

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,

d=r+r=4×10−15md=r+r=4×10−15m

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:

V=e24πε0dV=e24πε0d

Where, ε0=ε0=Permittivity of free space

14πε0=9×109Nm2c−214πε0=9×109Nm2c−2

⇒V=9×109×(1.6×10−19)24×10−15=5.76×10−14J=5.76×10−141,6×10−19⇒V=9×109×(1.6×10−19)24×10−15=5.76×10−14J=5.76×10−141,6×10−19

∴V=3.6×105eV=360keV∴V=3.6×105eV=360keV

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:

KE=2×3kT2KE=2×3kT2

Where, k=k=Boltzmann constant

T=T=Temperature required for triggering the reaction

∴T=KE3k=5.76×10−143×1.38×10−23=1.39×109K∴T=KE3k=5.76×10−143×1.38×10−23=1.39×109K

Therefore, we found that the gas must be heated to a temperature of 1.39×109K1.39×109K to initiate the reaction.

m(198Au)=197.968233um(198Au)=197.968233u

m(198Hg)=197.966760um(198Hg)=197.966760u

E1=1.088−8=1.088MeVE1=1.088−8=1.088MeV

⇒hν1=1.088×1.6×10−19×106J⇒hν1=1.088×1.6×10−19×106J

Where, Planck’s constant h=6.6×10−34Jsh=6.6×10−34Js

ν1=ν1=Frequency of radiation radiated by γ1−γ1−decay

ν1=E1hν1=E1h

⇒ν1=1.088×1.6×10−19×1066.6×10−34=2.637×1020Hz⇒ν1=1.088×1.6×10−19×1066.6×10−34=2.637×1020Hz

It can be observed from the given γ−γ−decay diagram that γ2γ2decays from the

0.412MeV0.412MeV

energy level to the

0MeV0MeV

energy level.

Now, the energy corresponding to γ2γ2-decay could be given as:

E2=0.412−0=0.412MeVE2=0.412−0=0.412MeV

⇒hν2=0.412×1.6×10−19×106J⇒hν2=0.412×1.6×10−19×106J

Where, ν2=ν2=Frequency of radiation radiated by γ2−γ2−decay

ν2=E2h=0.412×1.6×10−19×1066.6×10−34=9.988×1019Hzν2=E2h=0.412×1.6×10−19×1066.6×10−34=9.988×1019Hz

It can be observed from the given γγ-decay diagram that γ3γ3-decays from the

1.088MeV1.088MeV

energy level to the

0.412MeV0.412MeV

energy level.

Now, the energy corresponding to γ3γ3 -decay is given as:

E3=1.088−0.412=0.676MeVE3=1.088−0.412=0.676MeV

⇒hν3=0.676×10−19×106⇒hν3=0.676×10−19×106

Where, ν3=ν3=Frequency of radiation radiated by γ3−γ3−decay

ν3=E3h=0.676×1.6×10−19×1066.6×10−34=1.639×1020Hzν3=E3h=0.676×1.6×10−19×1066.6×10−34=1.639×1020Hz

Mass of

m(19878Au)=197.968233um(78198Au)=197.968233u

Mass of m(19880Hg)=197.966760um(80198Hg)=197.966760u

1u=931.5MeV/c21u=931.5MeV/c2

Energy of the highest level could be given as:

E=[m(19878Au)−m(19080Hg)]=197.968233−197.966760=0.001473uE=[m(78198Au)−m(80190Hg)]=197.968233−197.966760=0.001473u

⇒E=0.001473×931.5=1.3720995MeV⇒E=0.001473×931.5=1.3720995MeV

β1β1decays from the

1.3720995MeV1.3720995MeV

level to the

1.088MeV1.088MeV

level

Maximum kinetic energy of the β1β1particle =1.3720995−1.088=1.3720995−1.088

⇒K.E=0.2840995MeV⇒K.E=0.2840995MeV

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,

K.Emax=1.3720995−0.412=0.9600995MeVK.Emax=1.3720995−0.412=0.9600995MeV

Therefore, we found the maximum kinetic energy of the β2β2particle to be 0.9600995MeV0.9600995MeV.

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

Amount of hydrogen,

m=1kg=1000gm=1kg=1000g

1 mole, i.e., 1g of hydrogen (11H)(11H) contains

6.023×1023atoms6.023×1023atoms

.

That is, 1000g of 11H11H contains

6.023×1023atoms6.023×1023atoms

.

Within the sun, four 11H11H nuclei combine and form one 42He24He nucleus. In this process

26MeV26MeV

of energy is released.

Hence, the energy released from the fusion of 1 kg 11H11H is:

E1=6.023×1023×26×1034=39.1495×1026MeVE1=6.023×1023×26×1034=39.1495×1026MeV

Therefore, we found the energy released during the fusion of 1kg 11H11H is:

E1=6.023×1023×26×1034=39.1495×1026MeVE1=6.023×1023×26×1034=39.1495×1026MeV

Hence, the energy released during the fusion of 1kg of 11H11H to be 39.1495×1026MeV39.1495×1026MeV.

Amount of 92U235=1000gm92U235=1000gm

1 mole, i.e., 235g of 23592U92235U contains

6.023×1023atoms6.023×1023atoms

.

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

200MeV200MeV

. Therefore, energy released from the fission of 1kg of 23592U92235U is:

E2=6×1023×1000×200235=5.106×1026MeVE2=6×1023×1000×200235=5.106×1026MeV

E1E2=39.1495×10265.106×1026=7.67≈8E1E2=39.1495×10265.106×1026=7.67≈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.

200MeV200MeV

.

Amount of electric power to be generated,

P=2×105MWP=2×105MW

, 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

⇒P1=2×104×106J/s=2×1010×3600×24×365J/y⇒P1=2×104×106J/s=2×1010×3600×24×365J/y

Heat energy released per fission of a 235U235U nucleus,

E=200MeVE=200MeV

Efficiency of a reactor

=25=25

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

25100×200=50MeV=50×1.6×10−19×106=8×10−12J25100×200=50MeV=50×1.6×10−19×106=8×10−12J

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

2×1010×60×60×24×3658×10−12=78840×1024atoms2×1010×60×60×24×3658×10−12=78840×1024atoms

1 mole, i.e., 235g of U235U235contains

6.023×1023atoms6.023×1023atoms

That is, the mass of

6.023×1023atoms6.023×1023atoms

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,

3.076×104kg3.076×104kg.

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