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Class 12th Physics Chapter 13 Nuclei Notes

Introduction

• We will discuss about the nucleus which is the core of the atom.
• Nucleus is located in the centre of the atom and it is positively charged whereas electrons negatively charged and revolve around the nucleus in the orbits.
• We will also look at the composition of the structure of a nucleus and itsdifferent properties.
• How the present day energy crisis can be solved by using the nuclear energy.
• The interaction between different nuclei.

Structure of an atom

Composition of Nucleus

• Nucleus consists of protons and neutrons.
• Protons are positively charged particles which are present inside the nucleus and neutrons are neutral as they don’t have any charge.
• Atomic number: –
  • Atomic number constitutes the total number of protons which are present in the nucleus of that atom.
  • It is denoted by ‘Z’.
• Atomic mass:-
  • Atomic mass is the total number of neutrons and protons which are present inside the nucleus.
  • Mass of electrons is not considered while calculating the mass of the atom and only the mass of neutrons and protons are considered;
  • Since the electrons are the lightest particles their mass is not considered.
  • It is also known as Mass Number.
  • It is denoted by ‘A’.

• Nucleons  –> Protons + Neutrons
• General representation of the element: -(^A_ZX) where A = atomicmass and Z = atomic number.
  • For example:- Hydrogen ₁¹H  where atomic number=1 and mass number =1
  • Oxygen¹⁶₈O where atomic number=8 and mass number =16(8 protons and 8 neutrons).

Measurement of Atomic mass unit

• Mass of atom is very small as compared to the measurable masses which we see around us.
• Atomic Mass Unit (a.m.u) is used to measure mass of an atom
• It is denoted by u.
• Atomic mass unit is defined as (1/12^th) of the mass of the carbon.
• 1 a.m.u =(1/12) x 1.992647 x10^-26 kg
  • Where 1.992647 x10^-26 = mass of 1 carbon atom.
• 1 a.m.u =1.67 x10^-27
• To get the exact measurement of the atomic mass, an instrument known as Mass spectrometer is used.

Mass Spectrometer

Problem: –The three stable isotopes of neon: ₁₀²⁰Ne,₁₀²¹Ne and ₁₀²²Ne have respective abundances of90.51%, 0.27% and 9.22%. The atomic masses of the three isotopes are

19.99 u, 20.99u and 21.99 u, respectively. Obtain the average atomic mass of neon.

Answer:-

Atomic mass of₁₀²⁰Ne, m₁= 19.99 u

Abundance of ₁₀²⁰Ne, η₁ = 90.51%

Atomic mass of₁₀²¹Ne, m₂ = 20.99 u

Abundance of ₁₀²¹Ne, η₂ = 0.27%

Atomic mass of₁₀²²Ne, m₃ = 21.99 u

Abundance of₁₀²²Ne, η₃ = 9.22%

The average atomic mass of neon is given as:

m= (m₁ η₁+ m₂ η₂+ m₃ η₃)/ (η₁+ η₂+ η₃)

= (19.99×90.51+20.99×0.27+21.99×9.22)/ (90.51+0.27+9.22)

=20.1771u

Nuclei types

1. Isotopes: – Two nuclei with the same atomic number and different mass number are isotopes of each other.
   14. For example: – There are 3 isotopes of carbon(C) having same atomic number 6 but their mass numbersare differenti.e. 12,13 and 14.(¹²₆C), (¹³₆C), (¹⁴₆C).
2. Isobars: -The nuclei which have different atomic number but same mass number are known as isobars.
   1. For example: – Nitrogen (¹⁴₇N) and Carbon (¹⁴₆C)areboth isobars as their mass numbers aresame which is 14 but their atomic numbers are 7 and 6 respectively.
3. Isotones: -Isotones are those nuclei which have different atomic number but same number of neutrons.
   1. For example: -Boron (¹²₅B) and Carbon (¹³₆C).
   2. Boron: – Atomic number = 5 and mass number =12.
   3. Carbon: – Atomic number = 6and mass number =13.
   4. But the number of neutronsin Boron = (12-5 =7) and number of neutrons in

carbon = (13-6 = 7) are same.

4. Nuclide: – Nuclides are collection of nuclei with same atomic number having same number of neutrons.

Charge on a Nucleus

• Nucleus is positively charged and consists of protons which are (+ively) charged and neutrons are neutral.As a whole nucleus has to be positively charged.
• Charge on the nucleus is such that the entireatom is electrically neutral as a whole.
• Atom constitutes of electrons,protons and neutrons.
  • Consider an element _ZX
    • Where Z= atomic number, (the number of protons =Z and the number of electrons =Z).
  • Total charge on all the electrons e^–= -(Ze).

Total charge on the nucleus has to be equal and opposite of the charge on electron i.e. it should be =+ (Ze)for the atom to be electrically neutral.

Size of Nucleus

• Rutherfordperformed an experiment which proved that the size of the nucleus is extremely small.
• In Rutherford scattering experiment a beam of alpha particles were made to pass through a small thin gold foil.
• Very few alpha particles were deflected.
• The alpha particles got deflected because of repulsion with the nucleus as alpha particles are positively charged.They get repelled because they are both positively charged.
• Very small number of alpha particles got deflected provingthat nucleus is very small in size.
• It was found that the radius of a nucleus(R) of mass number A is given as :-
  • R=R₀A^1/3 where A = mass number and R₀=constant.

• Volume of a nucleus is ∝to the mass number.
  • V =(4/3)πR³ , Also R ∝(A)^1/3
  • => (R)³∝A
  • Therefore V ∝ (R) ³∝
• Density of nucleus is independent of mass number.

Problem:- Given the mass of iron nucleus as 55.85u and A=56,find the nuclear density?

Answer:- Given: – m_Fe = 55.85, u = 9.27 × 10^–26 kg

Nuclear density =mass/volume

= (9.27 × 10^–26)/((4 π /3)(1.2x 10^-15)³)x (1/56)

= 2.29 × 10¹⁷ kg m^–3

The density of matter in neutron stars (an astrophysical object) iscomparable to this density. This shows that matter in these objectshas been compressed to such an extent that they resemble a bignucleus.

Rutherford’s Gold Foil Experiment

Einstein’s Mass-Energy equivalence

• According to Einstein mass is also a form of energy.
• Mass – energy can also be converted into other forms of energy.
• Einstein gave mass-energy equivalence relation as: – E=mc².
• Any object which has got mass ‘m’ has mass energy associated with it and it is given as mc².
• This relation helps in understanding nuclear masses and interaction of nuclei with each other.

Problem:- Calculate the energy equivalent of 1 g of substance.

Answer:- Energy, E = 10^–3 × (3 × 10⁸)² J

E = 10^–3 × 9 × 10¹⁶ = 9 × 10¹³ J

Thus, if one gram of matter is converted to energy, there is a release of enormous amount of energy.

Nuclear binding energy

• Nuclear binding energy is the energy required to hold an atom’s protons and neutrons together in the nucleus.
• Energy required holding neutrons and protons together therefore keeps the nucleus intact.
• It can also be defined as the energyneeded to separate the nucleons from each other.

• Importance of nuclear binding energy describes how strongly nucleons are bound to each other. By determining its value we will come to know whether the neutrons and protons are tightly or loosely bound to each other.
• If nuclear binding energy is high -> high amount of energy is needed to separate the nucleons this means nucleus is very stable.

• If nuclear binding energy is low -> low amount of energy is needed to separate the nucleons this means nucleus is not very stable.
• Mass defect:-
  • Mass defect is the difference in the mass of nucleus and its constituents(neutrons and protons).
  • It is denoted by ΔM.
  • Mathematically :- ΔM = [Z m_p+ (A-Z) m_n]- M
    • Where m_p=mass of 1 proton, Z=number of protons,(A-Z)= mass of neutrons, m_N = mass of 1 neutronand M =nuclearmass of the atom.
  • For example: -(¹⁶₈O)àOxygen atom has 8 protons and 8 neutrons.
  • Mass of 8 protons à (8×1.00866) u and Mass of 8 neutrons à(8×1.00727) u.
  • Therefore Oxygen nucleus à(8p+8n) à8(1.00866 + 1.00727) = 16.12744u.
  • From spectroscopy ->Atomic mass of (¹⁶₈O) =15.9949u.
  • Mass of 8 electrons =(8×0.00055) u.
  • Therefore Nuclear mass of (¹⁶₈O) = (15.9949 – (8×0.00055)) =15.99053u.
  • Nuclear mass is less than sum of the masses of its constituents.
  • This difference in mass is known as mass defect.
  • It is also known as excess mass.
• Relation between Mass defect and Nuclear binding energy:-
  • Nuclear binding energy is denoted by E_b.
  • E_b= ΔMc²
    • Where E_b = nuclear binding energy, ΔM=mass defect.
  • As there is difference in the mass so there is energy associated with it. This energy is known as nuclear binding energy.
  • Nuclear binding energy is a measure of how well a nucleus is held together.

Problem:- Find the energy equivalent of one atomic mass unit,first in Joules and then in MeV. Using this, express the mass defectof (₈¹⁶O) in MeV/c².

Answer:-  1u = 1.6605 × 10^–27 kg

To convert it into energy units, we multiply it by c² and find thatenergy equivalent

 = 1.6605 × 10^–27 × (2.9979 × 10⁸)² kg m²/s²

= 1.4924 × 10^–10 J

= (1.4924 × 10^–10)/ (1.602×10^-19) eV

= 0.9315 × 10⁹ eV

= 931.5 MeV

Or, 1u = 931.5 MeV/c²

For (₈¹⁶O), ΔM = 0.13691 u = 0.13691×931.5 MeV/c²

= 127.5 MeV/c²

The energy needed to separate (₈¹⁶O) into its constituents is thus127.5 MeV/c².

Problem:- Find the energy equivalent of one atomic mass unit,first in Joules and then in MeV. Using this, express the mass defectof ¹⁶₈O in MeV/c².

Answer:- 1u = 1.6605 × 10^–27 kg

To convert it into energy units, we multiply it by c² and find thatenergy equivalent

= 1.6605 × 10^–27 × (2.9979 × 10⁸)² kg m²/s²

= 1.4924 × 10^–10 J

= (1.4924 × 10^–10 J)/ (1.602×10^-19) eV

= 0.9315 × 10⁹ eV

= 931.5 MeV

Or, 1u = 931.5 MeV/c²

For ¹⁶₈O, ΔM = 0.13691 u = 0.13691×931.5 MeV/c²

= 127.5 MeV/c²

The energy needed to separate ¹⁶₈O into its constituents is thus127.5 MeV/c².

Nuclear binding energy per nucleon

• Nuclear binding energy per nucleon is defined as the average energy per nucleon needed to separate a nucleus into its individual constituents.
• It is denoted by E_bn.
• Experimentally there was a graph plotted between binding energy per nucleon and the mass number(A).

• Following are the observations from the graph:-

1. Initially the graph was increasing.This implies that E_bn is very less for lesser mass number.
2. In the middle range the E_bn becomes constant.This means E_bn is independent of mass number.
3. In the end E_bnstarts decreasing.This shows E_bn is less when mass number is more.

Problem: – Obtain the binding energy (in MeV) of a nitrogen nucleus (¹⁴₇N), given m (¹⁴₇N) =14.00307u?

Answer:- Atomic mass of (¹⁴₇N ) nitrogen, m = 14.00307 u

A nucleus of (¹⁴₇N) nitrogen contains 7 protons and 7 neutrons.

Hence, the mass defect of this nucleus, ∆m = 7m_H + 7m_n − m

Where,

Mass of a proton, m_H = 1.007825 u

Mass of a neutron, m_n= 1.008665 u

Therefore, ∆m = (7 × 1.007825 + 7 × 1.008665 − 14.00307)

= 7.054775 + 7.06055 − 14.00307

= 0.11236 u

But 1 u = 931.5 MeV/c²

∆m = 0.11236 × 931.5 MeV/c²

Hence, the binding energy of the nucleus is given as:

E_b = ∆mc²

Where, c = Speed of light

E_b = 0.11236 × 931.5(MeV/c²)/c²

= 104.66334 MeV

Hence, the binding energy of a nitrogen nucleus is 104.66334 MeV.

Deriving Nuclear force from E_bn

1. Lighter nuclei: –
   1. In the initial part of the graph A(mass number) is less therefore E_bn is also less.As a result lesser energy is required to separate the nucleons.
   2. This showsnuclei areunstable.
   3. The nuclei are unstable and in order to become stable lighter nuclei combine with each other to form heavier nuclei.
   4. Let the energy of heavier nuclei formed is E’_bn and of lighter nuclei be E_bn. This implies E’_bn> E_bn.
   5. Energy is released when 2 lighter nuclei combine together to form a heavier nuclei.
   6. This process is known as Nuclear Fusion.
2. For heavier nuclei:-
   1. Mass number is very high and E_bn is very less.
   2. In order to become stable the heavier nuclei will split into 2 lighter nuclei.
   3. Energy associated with heavier nuclei =E_bn and energy associated with 2 lighter nuclei =E’_bn.
   4. This implies E’_bn> E_bn. Energy is released in this process by the heavier nuclei in order to attain stability.
   5. This process is known as Nuclear Fission.
3. Constancy of E_bnin the mid-range of A:-
   1. In this portion the mass number is increasing due to whichnumber of nucleons also increase.
   2. The force which is present between the nucleons is of short range.The strength of the force decreases as the distance increases.
   3. The nucleons are getting affected by their nearest neighbouring nucleons and not by the nucleons which are far away.
   4. As a result E_bn remains constant.
   5. But when there are too many nucleons E_bn suddenly starts decreasing.

Problem: – Obtain the binding energy of the nuclei (⁵⁶₂₆Fe) and (²⁰⁹₈₃Bi) in units of MeV from the following data: m (⁵⁶₂₆Fe) = 55.934939u and m (²⁰⁹₈₃Bi) = 208.980388 u.

Answer:- Atomic mass of (⁵⁶₂₆Fe), m_Fe = 55.934939 u

(⁵⁶₂₆Fe) nucleus has 26 protons and (56 − 26) = 30 neutrons

Hence, the mass defect of the nucleus, ∆m = ((26 × m_H) + (30 × m_n)– m_Fe)

Where,

Mass of a proton, m_H = 1.007825 u

Mass of a neutron, m_n = 1.008665 u

∆m = ((26 × 1.007825 + 30 × 1.008665) − 55.934939)

= (26.20345 + 30.25995 − 55.934939)

= 0.528461 u

But 1 u = 931.5 MeV/c²

Therefore, ∆m = 0.528461 × 931.5 MeV/c²

The binding energy of this nucleus is given as:

E_b1 = ∆mc²

Where, c =Speed of light

E_b1 = 0.528461 × 931.5(MeV/c²)/c²

= 492.26 MeV

Average binding energy per nucleon= (492.26/56)=8.76 MeV

Atomic mass of (²⁰⁹₈₃Bi), m₂ = 208.980388 u

(²⁰⁹₈₃Bi) nucleus has 83 protons and (209 − 83) = 126 neutrons.

Hence, the mass defect of this nucleus is given as:

∆m’ = (83 × m_H + 126 × m_n) − m₂

Where,

Mass of a proton, m_H = 1.007825 u

Mass of a neutron, m_n = 1.008665 u

∆m’ = (83 × 1.007825 + 126 × 1.008665 – 208.980388)

= 83.649475 + 127.091790 − 208.980388

= 1.760877 u

But 1 u = 931.5 MeV/c²

Therefore, ∆m’ = 1.760877 × 931.5 (MeV/c²) x c²

Hence, the binding energy of this nucleus is given as:

E_b2 = ∆m’c²

= 1.760877 × 931.5(MeV/c²) x c²

= 1640.26 MeV

Average binding energy per nucleon = (1640.26/209) =7.848MeV

Nuclear force

• The force with which the nucleons are bound together is known as nuclear force.
• It is the strong attractive force that binds the nucleons together.
• When the nuclear force is compared to other forces of nature like gravitational or coulomb’s force etc.it is the strongest of all the forces.
• As protons are positively charged they repel each other.This force of repulsion is given by Coulomb’s force of repulsion.
• This nuclear force is stronger than the coulomb’s force so it overcomes the force of repulsion.
• This is the reason neutrons and protons are held together inside the nucleus.

• It is independent of electric charge.Magnitude of nuclear force is same between proton-proton,proton-neutron or neutron-neutron.
• Nuclear force cannot be given mathematically.

Problem: – The neutron separation energy is defined as the energy required removing a neutronfrom the nucleus. Obtain the neutron separation energies of the nuclei (⁴¹₂₀Ca) and (²⁷₁₃Al) from the following data: m (⁴⁰₂₀Ca) = 39.962591 u, m (⁴¹₂₀Ca) = 40.962278 um (²⁶₁₃A) = 25.986895 u and m(²⁷₁₃A) = 26.981541 u.

Answer:-

For (⁴¹₂₀Ca): Separation energy=8.363007 MeV

(²⁷₁₃Al): Separation energy=13.059 MeV

(₀¹n) is removed from a (⁴¹₂₀Ca).

For a neutron nucleus. The corresponding nuclear reaction can be written as:

⁴¹₂₀Ca –> ⁴⁰₂₀Ca + ₀¹n

It is given that:

m (⁴⁰₂₀Ca) Mass = 39.962591 u,

m (⁴¹₂₀Ca) Mass = 40.962278 u

m (₀¹n)Mass = 1.008665 u

The mass defect of this reaction is given as:

Therefore, ∆m= m(⁴⁰₂₀Ca) + (₀¹n) –m(⁴¹₂₀Ca)

= (39.962591+1.008665-40.962278)u =0.008978u

But 1u=931.5 (MeV/c²)

Therefore,∆m = 0.008978 x 931.5 (MeV/c²)

Hence, the energy required for neutron removal is calculated as:

E=∆mc²

=0.008978 x931.5 =8.363007MeV

For (²⁷₁₃Al), the neutron removal reaction can be written as:

²⁷₁₃Al –> ²⁶₁₃Al + ₀¹n

It is given that:

m (²⁷₁₃Al)Mass = 26.981541 u

m (²⁶₁₃Al) Mass =25.986895u

The mass defect of this reaction is given as:

∆m =m (²⁶₁₃Al) +m (₀¹n) –m (²⁷₁₃Al)

=25.986895 + 1.008665 -26.981541

=0.0414019u

=0.0414019 x 931.5(MeV/c²)

Hence, the energy required for neutron removal is calculated as:

E=∆mc²

=(0.0414019 x 931.5) =13.059MeV

Radioactivity

• Radioactivity is a nuclear phenomenon in which an unstable nucleus undergoes decay to form stable nuclei.
• Radioactivity was discovered accidentally by a scientist Henry Becquerel (1896).

Henry Becquerel

• Only those nuclei which are unstable will show the phenomenon of radioactivity.
• Fluorescence is a property shown by fluorescent objects; when visible light falls on a fluorescent object then these objects will also start emitting light.
• Henry Becquerel performed experiments on some fluorescent objectsand he gave a hypothesis that fluorescent objects along with the visible light also emit some kind of radiation as well.
• Experimental set-up:-
  • He took a photographic plate and placed inside a box.He placed a fluorescent object on the top of the box and allowed the sun light to fall on it.
  • As the box is opaque it won’t allow the rays of light to pass through it.
  • When the photographic plate will be developed there should be no traces of image on the plate.
  • He repeated the experiment by taking different compounds and observed there was no image formed on the photographic film.
  • This showed that sunlight is the only ray which is emitted in the above case and as the plate was covered with opaque object so the sunlight is not able to reach the plate.

• He then took the compound of uranium known as uranium potassium sulphate.
• When the photographic plate was developed there was a black spot on the photographic plate.
• This shows that some radiation was emitted that reached the photographic plate.

• In order to know from where the radiation was emitted, he kept a blade on top of a box and uranium potassium sulphate compound was placed above the blade.
• After performing the experiment, when the photographic plate was developed, the photo of blade was seen on the plate.
• This shows that the radiation was emitted by the uranium potassium sulphate compound.

• Accidentally while he was performing the experiment it was not sunny that day.
• The result which he got was same i.e. when the photographic plate was developed there was the same image of the blade on the plate.

Conclusion:-

• The radiation was emitted by the uranium potassium sulphate compound and sunlight did not play any role in emitting the radiations.
• After series of experiments he proved that it was uranium alone which was responsible for emitting the radiations.
• He concluded that there are certain elements which emit radiation on their own.
• This phenomenon was named as radioactivity.

Radioactive nuclides: Nuclear stability

• Experimentallyit was observedthat the number of neutrons varies with the number of protons as we move higher in the periodic table.
• For the elements which have lesser atomic number (e.g.: O, C) the number of neutrons and the number of protons are equal.
• From the graph we can see the straight line shows the number of neutrons is equal to the number of protons.
• But as the atomic number increases gradually the number of neutrons become more than the number of protons.
• In the graph there is a specific pattern like zigzag is shownbythe stable nuclides and some which don’t show this pattern represents the unstable nuclides.
• The nuclides which are unstable emit some radiations to become;this phenomenon is known as radioactivity and the nuclides are known as radioactive nuclides.
• Nuclides to the left of the graph arestablewhich shows they have excess neutrons.
• Nuclides to the right of the graph have excess protons.Such nuclides are known as radioactive nuclides.

Radioactive Decay

• Radioactive decay is a phenomenon in which unstable nucleus decay to form stable nuclei.
• There are 3 types of radioactive decay by which a unstable nuclei becomes stable nuclei:-
  • Alpha decay:-
    • In alpha decay α – particle (helium nucleus) is emitted.
  • Beta decay:-
    • In beta decay electrons or positrons are emitted.
    • If electronsare emitted then it is known as β^– decay; and when positronsare emitted then the decay is known as β⁺
  • Gamma decay:-
    • In gamma decay high energy photons(γ rays) are emitted.
  • Law of radioactive decay:
    • Law of radioactive decay states that the number of nuclei undergoing decay per unit time is ∝to total number of nuclei in the sample.
    • Consider a sample of radioactive nuclei (Uranium). Thesample of uranium will have several uranium nuclei and they will undergo radioactive decay to convert into relatively stable lighter nuclei.
    • That means the number of uranium nuclei will gradually keep on decreasing with time.
    • For example:-
      • If in a sample there are 10 uranium nuclei and in another 100 uranium nuclei the rate of decay will be slower where there are 10 uranium nuclei as compared to 100 uranium nuclei.

Generalized expression for law of radioactive decay:-

Suppose there are ‘N’ active nuclei at any instant of time (‘t’).

In small interval of time ‘dt’ =’dN’ (number of nuclei decay).

Rate of decay (dN/dt) ∝N (number of nuclei which are present in the sample).

(dN/dt) = -(λ N)

• Whereλ = decay constant and (-ive) sign means the number of nuclei is gradually decreasing with time.
• => (dN) =-λ dt, integrating both the sides,
• =>_N0^N∫(dN)/N = -λ₀^t∫ N dt
  • Where N₀=number of nuclei at t=0 and N= number of nuclei after time t.
• [ln N]_N0^N = -λ[t]₀^t
• (ln N₀ – ln N) =-λt
• =>ln(N/N₀) = -λt
• => (N/N₀)=e^{– λt}
• =>N = N₀ e^-λt always N₀> N.
• Radioactive law is also known as exponential decay law as number of nuclei decreases exponentially which can be seen from the graph if plotted between N and t.
• It is followed only by radioactive nuclei.

Terminologies related to Radioactive decay

1. Decay Rate: –
   1. Decay rate is defined as the number of nuclei decaying per unit time.
   2. Denoted by ‘R’.
   3. Mathematically:- R =-(dN/dt)
      • Where dN=change in the number of nuclei with time and (-ive) sign shows the number of nuclei decreasing.
      • From Exponential decay law N = N₀ e^-λt, therefore
      • R=(d/dt)( N₀ e^-λt)

• =- N₀ [e^-λt(-λ)]
• R= N₀λ e^-λt ;
• At t=0 , R₀= N₀λ (equation (1))
• Therefore R=R₀ e^-λt=>decay constant also changes exponentially.

2. Activity of radioactive sample:-
   1. Activity of radioactive sample is defined as the total decay rate R of a sample of one or more radionuclides.
   2. Decay rate of whole sample is considered.
   3. I. unit: – Becquerel(Bq); 1Bq = (1 decay/sec).
   4. Other unit: – Curie(Ci); 1C i=3.7 x10¹⁰
3. Half-life of radioactive sample:-
   1. Half – life of radioactive sample is defined as the time at which number of nuclei reduces to one-half of their initial values.
   2. It is denoted by t (_1/2).
   3. Half-life tells how long radioactive nuclei can last.
   4. Mathematical expression:-
      • From radioactive law , N=N₀e^-λt (equation(1))
      • Initially number of sample = N₀.
      • Later number of samples will become=(N₀/2).
      • Therefore from equation(1)- (N₀/2) =N₀e^-λt_(1/2)
      • =>e^-λt_(1/2)= (1/2)
      • =>- λt_(1/2) = ln(1/2) (taking log on both sides)
      • =>- λt_(1/2) =ln(1) – ln(2)
      • =>λt _(1/2) = ln(2) (using ln (1) =0).
      • =>t_(1/2) = (0.693/λ)
   5. Mean-life of radioactive sample:-
      1. Mean life is defined as the average life of a nuclei in the radioactive sample.
      2. Any nuclei can decay at any interval of time.
      3. It is denoted by t_av.
      4. Mathematically:-
         • Let number of radioactive sample at t=0 =N₀.
         • From the law of radioactive decay (dN/dt) =-λN
         • => dN= -λNdt (these many nuclei will decay in time dt).
         • Therefore number of nuclei which decay between t and (t+dt) =λNdt.
         • But of the nuclei will decay fast and some will slowly decay.
         • Sum of lives of all these nuclei = tλNdt.
         • Therefore average life of the sample =t_av= (₀^∞∫(tλN)dt/N₀)
         • = (λ/N₀) ₀^∞∫ (tN) dt.
         • From exponential law of decay :- N =N₀ e^-λt
         • t_av= (λ/N₀) ∫tN₀ e^-λt dt = (λ/N₀)∫t e^-λt dt
         • t_av = λ₀^∞∫te^-λtdt
         • =λ[t∫e^-λt dt -∫(∫e^-λt.1)dt
         • =λ[(te^-λt)/(-λ)]₀^∞ -λ [(e^-λt)/(-λ) dt]₀^∞
         • =λ[0] –λ[(e^-λt)/λ²]₀^∞
         • t_av = (1/ λ)

• Using t_(1/2) = (0.693/λ)  and t_av = (1/ λ)
• t_av = (t_(1/2)/0.693)

Problem:- ( Class 12 Physics Nuclei Notes )

A radioactive isotope has a half-life of T years. How long will it take the activity to reduce to a) 3.125%, b) 1% of its original value?

Answer:-

Half-life of the radioactive isotope = T years

Original amount of the radioactive isotope = N₀

(a) After decay, the amount of the radioactive isotope = N

It is given that only 3.125% of N₀ remains after decay. Hence, we can write:

(N/N₀)= 3.125 %=( 3.125)/ (100) = (1/32)

But (N/N₀) =e^– λt

Where, λ = Decay, constant t = Time

Therefore e^{– (λt)} = (1/32)

– (λt) = ln (1) – ln (32)

– (λt) = 0 -3.4657

t= (3.4657)/ (λ)

Since λ = (0.693)/ (T)

Therefore t= (3.466)/ ((0.693)/ (T)) ≈ 5T years.

Hence, the isotope will take about 5T years to reduce to 3.125% of its original value.

(b) After decay, the amount of the radioactive isotope = N

It is given that only 1% of N₀ remains after decay. Hence, we can write:

(N/N₀)=1 %=( 1/100)

But (N/N₀) =e^– λt

Therefore e^{– (λt)} = (1/100)

– (λt) = ln (1) – ln (100)

– (λt) = 0 – 4.6052

t= (4.6052)/ (λ)

Since λ = (0.693)/ (T)

Therefore t= (4.6052)/ ((0.693)/ (T)) =6.645T years.

Hence, the isotope will take about 6.645T years to reduce to 1% of its original value.

Problem:- Obtain the amount (⁶⁰₂₇Co) of necessary to provide a radioactive source of 8.0 mCistrength. The half-life of (⁶⁰₂₇Co) is 5.3 years.

Answer:- The strength of the radioactive source is given as:

(dN/dt) =8.0 mCi

=8×10^-3x 3.7×10¹⁰

=29.6×10⁷ decay/s.

Where,

N = Required number of atoms

Half-life of (⁶⁰₂₇Co), T_1/2 = 5.3 years

= 5.3 × 365 × 24 × 60 × 60

= 1.67 × 10⁸ s

For decay constant λ, we have the rate of decay as:

(dN/dt) = (λN)

Where λ = (0.693/T_1/2) = (0.693)/ (1.67×10⁸) s^-1

Therefore N = (1/λ) (dN/dt)

= (29.6×10⁷)/ (0.693)/ (1.67×10⁸) =7.133 x10¹⁶ atoms

For (⁶⁰₂₇Co):

Mass of 6.023 × 10²³ (Avogadro’s number) atoms = 60 g

Therefore, Mass of (7.133 x 10¹⁶) atoms= (60×7.1333×10¹⁶)/ (6.023×10²³)

=7.106×10^-6g

Hence, the amount of (⁶⁰₂₇Co) necessary for the purpose is 7.106 × 10⁻⁶ g.

Alpha decay

• In alpha decay α particles are emitted.Daughter nucleus is formed from the parent nucleus.
• The atomic number decreases by 2 and Mass number increases by 4.
• It is a very spontaneous process and happens on its own.
• It occurs only in radionuclides.
• For example:-
  • ²³⁸₉₂U (unstable) –> ₉₀²³⁴Th + ₂⁴He
    • (Uranium (U) is known as parent nucleus and Thorium (Th) is known as daughter nucleus).
    • ₂⁴He is α particle.
  • General form of alpha decay: –^A_ZX (parent) à^(A-4) _(Z-2) Y + ₂⁴He
    • Where Y is daughter nucleus.

Q-value of alpha decay

• Q-value is a parameter or a characteristic of a nuclear reaction which describeswhether the reaction can take place or not.
• Q-value is defined as difference in the initial mass energy and final mass energy of decayed products.
• Consider the general equation:-
  • ^A_ZX (parent) –>^(A-4) _(Z-2) Y + ₂⁴
  • Initial rest mass energy U_i =[m (^A_ZX) – Zm_e]c²
    • Where m (^A_ZX) = rest mass and Zme = mass of electrons.
  • Final rest mass energy U_f = [m(^(A-4) _(Z-2) Y – (Z-2) m_e + m(₂⁴He) – 2m_e ] c²
  • Q =U_i – U_f
  • =[m (^A_ZX) – Zm_e – m(^(A-4) _(Z-2) Y + (Z-2) m_e – m(₂⁴He) + 2m_e]c²
  • Q = [m (^A_ZX) – m(^(A-4) _(Z-2) Y- m(₂⁴He)]c²
• If Q =(+ive) then equation is energetically allowed.
  • => Q>0, (U_i-U_f) >0=> U_i>U_f
  • This implies energy of the reactants is more than the energy of the products.
  • There is some free energy that is in the form of kinetic energy.
• If Q =(-ive) then equation is not energetically allowed.

Problem: –  Find the Q-value and the kinetic energy of the emitted α-particle in the α-decay of (a) (²²⁶₈₈Ra) and (b) (²²⁰₈₆Rn). Given m (²²⁶₈₈Ra) =226.0250 u, m (²²⁰₈₆Rn) = 220.01337u, m (²²²₈₆Rn) =222.01750u, m (²¹⁶₈₄Po) =216.00189u.

Answer:-

 (a) Alpha particle decay of (²²⁶₈₈Ra) emits a helium nucleus. As a result, its mass number reduces to (226 − 4) 222 and its atomic number reduces to (88 − 2) 86. This is shown in the following nuclear reaction.

(²²⁶₈₈Ra) –>  (²²²₈₈Ra) + (⁴₂He)

Q-value of emitted α-particle = (Sum of initial mass − Sum of final mass)

c² Where, c = Speed of light It is given that:

m (²²⁶₈₈Ra) =226.0250 u, m (²²²₈₆Rn) =222.01750u, m (⁴₂He) =4.002603u

Q-value = [226.02540 − (222.01750 + 4.002603)] u c²

= 0.005297 u c²

But 1 u = 931.5 (MeV/c²)

Therefore Q = (0.005297 × 931.5) ≈ 4.94 MeV

Kinetic energy of the α-particle =

(Mass number after decay)/ (Mass number before decay) x Q

= (222)/ (226) x 4.94 = 4.85MeV.

(b) Alpha particle decay of (²²⁰₈₆Rn)

(²²⁰₈₆Rn) –> (²¹⁶₈₄Po) + (⁴₂He)

It is given that:

Mass of (²²⁰₈₆Rn) = 220.01137 u

Mass of (²¹⁶₈₄Po) = 216.00189 u

Therefore, Q-value = [220.01137 – (216.00189+4.002603)] x931.5

≈ 641 MeV

Kinetic energy of the α-particle = (220-4)/ (220) x 6.41

= 6.29 MeV

Problem: –We are given the following atomic masses:²³⁸₉₂U = 238.05079 u

⁴₂He = 4.00260 u, ₉₀²³⁴Th = 234.04363 u,₁¹H= 1.00783 u, ²³⁷₉₁Pa = 237.05121 u

Here the symbol Pa is for the element protactinium (Z = 91).

(a) Calculate the energy released during the alpha decay of ²³⁸₉₂U .

(b) Show that ²³⁸₉₂U cannot spontaneously emit a proton.

Answer:-

(a) The alpha decay of ²³⁸₉₂U is given by equation

^A_Z X –> ^(A-4) _(Z-2Y + ⁴₂He

The energy releasedin this process is given by:

Q = (M_U – M_Th – M_He) c²

Substituting the atomic masses as given in the data, we find

Q = (238.05079 – 234.04363 – 4.00260) u × c²

= (0.00456 u) c²

= (0.00456 u) (931.5 MeV/u)

= 4.25 MeV.

(b) If ²³⁸₉₂U spontaneously emits a proton, the decay process would be

²³⁸₉₂U –> ²³⁷₉₁Pa + ¹₁H

The Q for this process to happen is

= (M_U – M_Pa – M_H) c²

= (238.05079 – 237.05121 – 1.00783) u × c²

= (– 0.00825 u) c²

= – (0.00825 u) (931.5 MeV/u)

= – 7.68 MeV

Thus, the Q of the process is negative and therefore it cannot proceedspontaneously. We will have to supply energy of 7.68 MeV to a²³⁸₉₂U nucleus to make it emit a proton.

Beta decay

• In case of beta decay either electron or positron is emitted.
• Mass number remains the same.
• β⁻decay –> Electron is emitted.
  • Atomic number increases by 1.
  • For example:- ³²₁₅P à₁₆³²S + e^– + ̅ν  where  ̅ν =anti-neutrino
  • Q-value for β⁻decay:-
    • ^A_ZXà^A _(Z+1) Y + e^–+ ̅ν
    • Initial rest mass energy U_i = [m (^A_ZX) – Zm_e]c²
      • Where m (^A_ZX) = rest mass and Zme = mass of electrons
    • Final rest mass energy U_f = [m(^A _(Z+1) Y) – (Z+1) m_e + m_e ] c²
    • =[m(^A _(Z+1) Y) -Zm_e] c²
    • Therefore Q =U_i– U_f
    • =[m(^A_ZX) -Zm_e – m(^A _(Z+1) Y) + Zm_e] c²
    • Q = [m(^A_ZX) – m(^A _(Z+1) Y)]

Example:-

• β⁺decay à Positron(same as electron but with (+ ive) charge) is emitted.
  • Atomic number decreases by 1.
  • For example:- ²³₁₁Na –>₁₀²²Ne + e⁺+ν where ν = neutrino
  • Q-value for beta (+ive) decay:-
    • ^A_ZX –>^A _(Z-1) Y + e⁺ + ν
    • Initial rest mass energy U_i = [m (^A_ZX) – Zm_e]c²
      • Where m (^A_ZX) = rest mass and Zme = mass of electrons
    • Final rest mass energy U_f = [m(^A _(Z-1) Y) – (Z-1) m_e + m_e ] c²
    • Therefore Q =U_i – U_f
    • = [m (^A_ZX) – Zm_e – m(^A _(Z-1) Y) – (Z-1)m_e – m_e] c²
    • Q = [m(^A_ZX) – m(^A _(Z-1) Y) -2 m_e] c²
  • Example to show whether nuclear reaction can take place or not:-
    • ¹₁p(proton) à₀¹n(neutron) + e⁺ + ν. The mass of neutron is greater than the mass of proton.
    • The Q-value = (-ive) as a result the above reaction is not possible.This means conversion of stable proton to neutron is not allowed.
    • Consider ₀¹n (neutron) à¹₁p (proton) + e^– + ̅ν
    • The Q-value =(+ive) as a result above reaction is possible.This means conversion of neutron to proton is allowed.

Beta-decay: neutrino and anti-neutrino

• Neutrino (ν) and anti-neutrino ( ̅ν) are neutral particles.
• They also have negligible mass.
• They have extremely high penetration power.
• They have extremely weak interaction with matter.

Problem:- The nucleus (²³₁₀Ne) decays by β^– emission. Write down the β decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:m (²³₁₀Ne) = 22.994466 u,m (²³₁₁Na) = 22.989770 u.

Answer:- In β^– emission, the number of protons increases by 1, and one electron and an antineutrino are emitted from the parent nucleus.

β^– Emission of the nucleus (²³₁₀Ne)

(²³₁₀Ne) –> (²³₁₁Na) + e^– +   ̅ν + Q

It is given that:

Atomic mass m (²³₁₀Ne) of = 22.994466 u

Atomic mass m (²³₁₁Na) of = 22.989770 u

Mass of an electron, m_e = 0.000548 u

Q-value of the given reaction is given as:

Q= [m (²³₁₀Ne) + [m (²³₁₁Na) +m_e]] c²

There are 10 electrons in and 11 electrons in (²³₁₁Na). Hence, the mass of the electron is cancelled in the Q-value equation.

Therefore Q = [22.994466 – 22.989770] c²

= (0.004696 c²) u.

But 1u = 9.31 (MeV/c²)

Therefore Q= (0.004696 x 931.5) =4.374MeV.

The daughter nucleus is too heavy as compared to e^– and   ̅ν. Hence, it carries negligible energy. The kinetic energy of the antineutrino is nearly zero. Hence, the maximum kinetic energy of the emitted electrons is almost equal to the Q-value, i.e., 4.374 MeV.

Problem:- For the β⁺(positron) emission from a nucleus, there is another competing process knownas electron capture (electron from an inner orbit, say, the K−shell, is captured by thenucleus and a neutrino is emitted).

e⁺ + ^A_ZX –> ^A_(Z-1)Y + ν

Show that β⁺if emission is energetically allowed, electron capture is necessarilyallowed but not vice−versa.

Answer:-

Let the amount of energy released during the electron capture process be Q₁. Thenuclear reaction can be written as:

e⁺ + ^A_ZX –> ^A_(Z-1)Y + ν +Q₁   (equation(1))

Let the amount of energy released during the positron capture process be Q₂. Thenuclear reaction can be written as:

^A_ZX –> ^A _(Z-1) Y + e⁺ + ν +Q₂   (equation (2))

m_N (^A_ZX) =Nuclear mass of ( ^A_ZX)

m_N(^A _(Z-1) Y) = Nuclear mass of (^A _(Z-1) Y)

m(^A_ZX) = Atomic mass of ( ^A_ZX)

m(^A _(Z-1) Y) = Atomic mass of (^A _(Z-1) Y)

m_e = Mass of an electron

c = Speed of light

Q-value of the electron capture reaction is given as:

Q₁= [m_N(^A_ZX) + m_e -m_N(^A _(Z-1) Y)] c²

= [m (^A_ZX) -Zm_e+ m_e –m (^A _(Z-1) Y) + (Z-1) m_e] c²

= [m (^A_ZX) -m^A _(Z-1) Y] c²   (equation (3))

Q-value of the positron capture reaction is given as:

Q₂=m_N(^A_ZX) -m_N(^A _(Z-1) Y) – m_e] c²

= [m (^A_ZX) -Zm_e– m (^A _(Z-1) Y) + (Z-1) m_e – m_e] c²

= [m (^A_ZX) -m^A _(Z-1) Y – 2m_e] c²   (equation (4))

It can be inferred that if Q₂> 0, then Q₁> 0; Also, if Q₁> 0, it does not necessarilymean that Q₂> 0.

In other words, this means that if β⁺emission is energetically allowed, then the electron capture process is necessarily allowed, but not vice-versa. This is because the Q-value must be positive for an energetically-allowed nuclear reaction.

Gamma decay

• In Gamma decay γ rays are emitted.γ rays are electromagnetic waves with short wavelength.
• Most of the daughter nuclei of alpha decay and beta decay are in excited state.
• As a result they are unstable.When the daughter nuclides try to transit from excited state to ground state they emit radiations.
• These radiations are known as γThis known as gamma decay.

Comparison of Alpha,Beta & Gamma Rays

Alpha Rays	Beta Rays	Gamma Rays
1.      Stream of alpha particles they are known as alpha rays.	1.      Stream of electrons they are known as beta rays.	1.      Gamma rays are Electromagnetic radiation.
2.      Alpha rays get deflected by electric field as they are positively charged.	2.Beta rays get deflected by electric field as they are negatively charged.	2. Gamma rays are undeflected by the electric field as they are neutral.
3.      Alpha rays get deflected by magnetic field as they are positively charged	3. Beta rays get deflected by magnetic field as they are negatively charged	3. Gamma rays are undeflected by the magnetic field as they are neutral
4.      Alpha rays have low penetrating power.	4.Beta rays have high penetrating power.	4.Gamma rays have largest penetrating power.
5.      Ionizing power of alpha rays is high.	5.Ionizing power of beta rays is lesser than alpha rays.	5.Ionizing power of gamma rays is least.

Nuclear Energy

• Nuclear energy is the energy that holds together the nuclei of atoms.
• Nuclear energy is obtained from nucleusby either:-
  • Breaking of heavy nucleus into 2 relatively lighter nuclei known as nuclear fissionor by
  • Combining 2 lighter nuclei to form a heavy nucleus known as nuclear fusion.
• Nuclear energy is becominga possible solutionfor the energy crisis in the world.
• Electric energy can be harnessed from nuclear energy.

Nuclear reactor

Types of Nuclear reactions

Problem:- 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 thecoin is entirely made of (₂₉⁶³Cu) atoms (of mass 62.92960 u).

Answer:– Mass of a copper coin, m’ = 3 g

Atomic mass of (₂₉⁶³Cu) atom, m = 62.92960 u

The total number of (₂₉⁶³Cu) atoms in the coin, N = (N_A x m’)/ (Mass number)

Where,

N_A = Avogadro’s number = 6.023 × 10²³ atoms /g

Mass number = 63 g

Therefore N = (6.023 × 10²³ x3)/ (63) =2.868×10²² atoms.

(₂₉⁶³Cu) nucleus has 29 protons and (63-29) = 34 neutrons.

Therefore, Mass defect of this nucleus, ∆m’ = 29 × m_H + 34 × m_n − m

Where,

Mass of a proton, m_H = 1.007825 u

Mass of a neutron, m_n = 1.008665 u

Therefore, ∆m’ = 29 × 1.007825 + 34 × 1.008665 − 62.9296

= 0.591935 u

Mass defect of all the atoms present in the coin, ∆m = 0.591935 × 2.868 × 10²²

=1.69766958×10²²u.

But 1 u = 931.5 (MeV/c²)

Therefore, ∆m = 1.69766958 × 10²² × 931.5 (MeV/c²)

Hence, the binding energy of the nuclei of the coin is given as:

E_b= ∆mc²

= 1.69766958 × 10²² × 931.5(MeV/c²)/c²

= 1.581 × 10²⁵ MeV

But 1 MeV = 1.6 × 10⁻¹³ J

E_b = 1.581 × 10²⁵ × 1.6 × 10⁻¹³

= 2.5296 × 10¹² J

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

Nuclear Fission

• Scientist named Fermi was the one who performedfirst nuclear fission reaction.
• He considered the sample of ₉₂²³⁵U and bombarded it with a neutron ¹₀n i.e.
  • ¹₀n + ₉₂²³⁵U(Target nuclei)à₉₂²³⁶U(unstable so broke down into) –>₅₆¹⁴⁴Ba + ⁸⁹₃₆kr +3¹₀n
    • Where (₅₆¹⁴⁴Ba) = Barium,(⁸⁹₃₆kr)=Krypton and (¹₀n) =neutron.
  • This reaction is known as nuclear fission reaction as heavier nuclei is broken down into lighter nuclei,thereby releasing large amounts of energy.

Types of Nuclear Fission reactions:

1. Uncontrolled chain reaction:-
   1. In uncontrolled chain reaction, the reaction is uncontrolled and rapid.The reaction keeps increasing and becomes huge.
   2. Energy of the order of mega electron volts is produced.
   3. This reaction is known as chain reaction as the product formed in first reaction initiates the second reaction and so on.
   4. One of the important applications of this reaction is in making atom bombs,hydrogen bombs and nuclear bombs.
   5. (Number of neutrons hitting next target)/ (number of neutrons emitted)>=1.
   6. Reaction that takes place :-
      1. ¹₀n + ₉₂²³⁵U –>₉₂²³⁶U(unstable so broke down into) –>₅₆¹⁴⁴Ba + ⁸⁹₃₆kr +3 ¹₀n

2. Controlled chain reaction:-
   1. In controlled chain reaction, the reaction is controlled and steady.
   2. Reaction that takes place :-
      1. ¹₀n + ₉₂²³⁵U –> ₉₂²³⁶U(unstable so broke down into) –>₅₆¹⁴⁴Ba + ⁸⁹₃₆kr +3 ¹₀n
      2. Using some methods 2 neutrons are removed and only 1 neutron is allowed to hit the next target.
   3. Energy released is less as compared to the energy released in the nuclear fission reaction.
   4. One of the most important applications is in Nuclear Reactor where electricity can be produced.
   5. (Number of neutrons hitting next target)/ (number of neutrons emitted)<1.

Nuclear Reactor

• Nuclear reactors are useful in producing electricity.
• A nuclear reactor is an arrangement to generate electricity whichmakes use of nuclear fission.

Requirements for controlled nuclear fission in reactor:

1. Neutrons to be slowed down.

• Neutrons are slowed down by using moderators which are lighter nuclei which slow down fast moving neutrons by elastic collision.
• Commonly used moderators:-

1.Water

2. Heavy Water
3. Graphite

• Consequence of use of moderators:-
  1. Multiplication factor of neutrons increases: – When a neutron hits a target nucleus along with daughter nucleus,it produces 3 neutrons.These 3 neutrons are highly energetic but they need to be slowed down, so they can hit the target nucleus.
  2. As a result high multiplication factor results in uncontrolled chain reaction.

2. Excess neutrons to be absorbed.

• As uncontrolled chain reaction is wanted therefore to absorb excess neutrons Control Rods are used.
• These control rods are inserted in the core of the nuclear reactor.
• Control rods are capable of initiating(while taking out of the reactor) and stopping(inserting in the nuclear reactor) the nuclear reaction.
• As they absorb all the excess neutronsthere are no neutrons left to start the reaction.
• Control rods are made up of neutron absorbing materials.
• They decrease the multiplication factor of neutrons to a very small value.
• Commonly used material is Cadmium.

Control Rods

Construction of Nuclear Reactor

• The core of the nuclear reactor consists of uranium (²³⁵U) in the form of cylindrical rods.These rods are dipped inside a liquid which is the moderator.
• Whenever one neutron strikes this uranium rod nuclear fission reaction starts and 3 fast moving neutrons are produced.
• Because of the moderator these 3 neutrons undergo elastic collision as a result they slow down before they strike the second rod.
• Geometry of the core is such that only one out of 3 neutrons which are emitted strike the next rod making the reaction a controlled one.
• When the control rods are inserted inside they will absorb all the extra neutrons.Since there are no neutrons nuclear fission reaction will stop.
• Large amount of energy is also released in the core.
• In order to extract the energy from the core water at very high pressure is passed through it.
• As hot water passes through it produces steam in the steam generators.
• This steam is used to run the turbines which in turn produce electricity.
• This process will keep on continuing till the uranium on the rods does not get over.Then the rods have to be replaced in the nuclear reactor.

Advantages:-

1. Energy released is extremely large.
2. Needs fuel in extremely small quantity.

Disadvantages:-

1. Spent fuel is highly radioactive and extremely hazardous to all life forms.
2. Accumulation of radioactive waste.

Problem:- A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much (²³⁵₉₂U) did it containinitially? Assume that the reactor operates 80% of the time that all the energy generatedarises from the fission of (²³⁵₉₂U) and that this nuclide is consumed only by the fissionprocess.

Answer:- Half-life of the fuel of the fission reactor, t_(1/2)= 5years

= 5 × 365 × 24 × 60 × 60 s

We know that in the fission of 1 g of (²³⁵₉₂U) nucleus, the energy released is equal to 200MeV.

1 mole, i.e., 235 g of (²³⁵₉₂U) contains (6.023 × 10²³) atoms.

1 g (²³⁵₉₂U) contains = (6.023×10²³)/ (235) atoms.

The total energy generated per gram of (²³⁵₉₂U) is calculated as:

E = ((6.023×10²³)/ (235)) x200MeV/g

= (200×6.023×10²³x1.6×10^-19x10⁶)/ (235) =8.20×10¹⁰ J/g

The reactor operates only 80% of the time.

Hence, the amount of (²³⁵₉₂U) consumed in 5 years by the 1000 MW fission reactor iscalculated as:

= (5x80x60x60x365x24x1000x10⁶) g / (100×8.20×10¹⁰)

=1538kg

Therefore, Initial amount of (²³⁵₉₂U) = 2 × 1538 = 3076 kg

Problem:- The fission properties of (²³⁹₉₄Pu) are very similar to those of (²³⁵₉₂U).The average energy released per fission is 180 MeV. How much energy, in MeV, isreleased if all the atoms in 1 kg of pure (²³⁹₉₄Pu) undergo fission?

Answer:- Average energy released per fission of (²³⁹₉₄Pu), E_av=180MeV

Amount of pure (²³⁹₉₄Pu), m = 1 kg = 1000 g

N_A= Avogadro number = 6.023 × 10²³

Mass number of (²³⁹₉₄Pu) = 239 g

1 mole of (²³⁹₉₄Pu) contains N_A atoms.

Therefore mg of (²³⁹₉₄Pu) contains ((N_A)/ (Mass number) x m) atoms

= ((6.023 × 10²³)/ (239) x1000) =2.52 x10²⁴ atoms.

Therefore Total energy released during the fission of 1 kg of (²³⁹₉₄Pu) is calculated as:

E=E_av x 2.52×10²⁴

=180×2.52×10²⁴ = 4.536×10²⁶MeV

Hence, 4.536×10²⁶MeV is released if all the atoms in 1 kg of pure (²³⁹₉₄Pu) undergo fission.

Nuclear Fusion

• In a nuclear fusion reaction two lighter nuclei combine to form a relatively heavier nucleus.
• In this process huge amount of energy is also released.
• Temperature at which protons would have enough energy to overcome the coulomb’s barrier is very high.

Thermonuclear fusion

• Increasing the temperature of the material until the particles have enough energy due to their thermal motions alone –to overcome the coulomb barrier.
• For thermonuclear fusion,extreme conditions of temperature and pressure are required.
• Example of Thermonuclear fusion is generation of energy in stars.
  • For example: From the sun we get large amount of energy and this energy generated due to the thermonuclear fusion reaction taking place in the sun.

Energy generation in sun

• In sun the energy generation is a multi- step process.There are total of 4 steps involved in the energy generation inside sun.
  • Step1:- ¹₁H(proton) + ¹₁H(proton) –>²₁H(deuteron)+ e⁺(positron) + ν(neutrino)+0.42MeV
  • Step2:- e⁺(positron) + e^–(electron) –>γ(Gamma rays) + 1.02MeV
  • Step3:- ²₁H(deuteron)+ ¹₁H(proton) –>³₂He(helium) + γ(Gamma rays)+5.49MeV
  • Step4:- ³₂H + ³₂H –>⁴₂He + ¹₁H+¹₁H+12.86MeV
  • Step 1,2,and 3 occur twice in the sun and the step 4 occurs only once.
• When all the above 4 reactions are combined together then four hydrogen atoms combine to form a⁴₂He atom with a release of 26.7MeV of energy.
• Final reaction is:- 4¹₁H +2e^{– –> 4}₂He +6 γ+2 ν+26.7MeV
• It is also known as proton-proton cycle because this process starts with protons.

Problem:- How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as  ²₁H + ²₁H –>³₂He+n+3.27MeV

Answer:- The given fusion reaction is:

²₁H + ²₁H –>³₂He+n+3.27MeV

Amount of deuterium, m = 2 kg

1 mole, i.e., 2 g of deuterium contains 6.023 × 10²³ atoms.

Therefore, 2.0 kg of deuterium contains= ((6.023×10²³)/ (2)) x (2000) =6.023×10²⁶atoms

It can be inferred from the given reaction that when two atoms of deuterium fuse, 3.27

MeV energy is released.

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

E= (3.27/2) x 6.023×10²⁶ MeV

= (3.27/2) x 6.023×10²⁶x 1.6×10^-19x 10⁶

=1.576×10¹⁴ J

Power of the electric lamp, P = 100 W = 100 J/s

Hence, the energy consumed by the lamp per second = 100 J

The total time for which the electric lamp will glow is calculated as:

= (1.576×10¹⁴)/ (100x60x60x24x365)

= (4.9 x10⁴) years.

Problem:-  From the relation R = R₀A^1/3, where R₀ is a constant and A is the mass number of anucleus, show that the nuclear matter density is nearly constant (i.e. independent of A).

Answer:- We have the expression for nuclear radius as:

R = R₀A^1/3

Where,

R₀ = Constant.

A = Mass number of the nucleus

Nuclear matter density,ρ = (Mass of the nucleus)/ (volume of nucleus)

Let m be the average mass of the nucleus.

Hence, mass of the nucleus = mA

Therefore ρ=(mA)/((4/3)πR³)=(3mA)/(4π)(R₀A^(1/3))³=(3m)/(4πR₀³)

Hence, the nuclear matter density is independent of A. It is nearly constant.

Chapter-13-Nuclei

Nucleus : Previous Year’s Questions

Q.1 Two nuclei have mass numbers in the ratio 1 : 2. What is the ratio of their nuclear densities? [2009 D]
Sol. Nuclear density is independent of mass number.

Q.2 Two nuclei have mass numbers in the ratio 1 : 8. What is the ratio of their nuclear radii? [2009]
Sol. Nuclear radius R=R0A1/3
where R0= constant, A = mass number
Therefore R1R2=(A1A2)1/3
=(18)1/3=12
R1:R2=1:2

Q.3 Assuming the nuclei to be spherical in shape, how does the surface area of a nucleus of mass number A1 compare with that of a nucleus of mass number A2? [2008 C]
Sol. Nuclear radius R=R0A1/3
where R0= constant
Therefore Surface Area S=4πR2
=4π(R0A1/3)2
=4πR20A2/3
Therefore Ratio of surface areas, S1S2=(A1A2)2/3

Q.4 How is the size of a nucleus is experimentally determined? Write the relation between the radius and mass number of the nucleus. Show that the  density of nucleus is independent of its mass number. [2011C D,2011C]
Sol. The size of the nucleus is experimentally determined using Rutherford’s α – scattering experiment and the distance of closest  approach and impact parameter.
The relation between radius and mass number of nucleus is
R=R0A1/3
where R0=1.2fm
where, A = mass number R = radius of nucleus
Nuclear density
ρ=MassofnucleusVolumeofnucleus
=mA43π(R0A1/3)3
where, m = mass of each nucleon
ρ=mA43πR30A
ρ=m43πR30
From the above formula, it is clear that r does not depend on mass number.

Q.5 Write any two characteristic properties of nuclear force. [2010]
Sol. Two characteristics of nuclear force
(i) These are short range force.
(ii) These are strong force of attractive nature.

Q.6 Draw a plot representing the law of radioactive decay. Define the activity of a sample of a radioactive nucleus. Write its SI unit. [2006 C,2007,2008 F,2010 D]
Sol. The curve representing the law of radioactive decay is shown below.


The rate or activity of a sample is defined as the rate of disintegration taking place in the sample of radioactive substance.
SI unit of activity is Becquerel (Bq).
1 Bq = 1 disintegration/s.

Q.7 State the law of radioactive decay.
Plot a graph showing the number (N) of undecayed nuclei as a function of time (t) for a given radioactive sample having half life T2√.
Depict in the plot, the number of undecayed nuclei at (a) t=3T2√ and (b) t=5T2√. [2011 D]
Sol. Law of radioactive decay : – The rate of disintegration of radioactive sample at any instant is directly proportional to the number of undisintegrated nuclei present  in the sample at that instant i.e.,
dNdt∝N
dNdt=−λN
where , N = number of undisintegrated nuclei present in the sample at any instant t and dNdt is rate of disintegration.
The required plot is shown below.

Q.8 How is the mean life of a radioactive sample related to its half life? [2011 F]
Sol. Mean life tm=1λ, where λ is decay constant.
But half life T12=In2λ
Therefore λ=In2T12
Therefore tm=T12log2=T120.693

Q.9 A nucleus undergoes β – decay. How does its                                         
(a) Mass number                                         (b) Atomic number change?
[2011 C D]
Sol. During b- decay
(a) No change in mass number.                  (b) Atomic number increases by 1.

Q.10 A nucleus 23892U undergoes alpha- decay and transforms to thorium. What is
(a) The mass number and                                              (b) Atomic number of the nucleus produced?                         [2011 C D]
Sol. In α- decay, the mass number of parent nucleus decreases by 4 units and atomic number decreases by 2 units.
Therefore 23892U→42He+23490Th
(a) Mass number of produced nucleus = 234                             (b) Atomic number of produced nucleus = 90

Q.11 What is the basic mechanism for the emission of β− and β+ particles in nuclide? Give an example by writing explicity a decay process for b– emission.
Is
(a) The energy of the emitted β – particles continuous or discrete?
(b)The daughter nucleus obtained through β – decay, an isotope or an isobar of the parent nucleus? [2010 C D]
Sol.During β− -decay from the nucleus, nuclei undergoes a change in such a way that atomic number increases by one and mass number remain same.
In β+- decay, the mass number of present radioactive nuclei remains same whereas atomic number decrease by one.
Example of β− – decay
3215P→3216S+e−+ν¯¯¯
(Half life = 26 days)
In β- decay, an electron and an antineutrino are created in following manner :
n→p+e−+ν¯¯¯
(a) The energy of emitted b-particles is continuous.
(b) As there is no change in mass number during β-decay. So, the daughter nucleus is isobar of the parent nucleus.

Q.12 A radionuclide sample has N0 nuclei at τ = 0. Its number of undecayed nuclei get reduced N0e at t = τ . What does the term t stand for?
Write, in terms of t, the time interval T in which half of the original number of nuclei, of this radionuclide would have got decayed. [2008 C D]
Sol.(i) The term t stands for mean life.
(ii) The required relation is τ = 1.44 T i.e., mean life of radioactive sample.   
= 1.44 × half life

Q.13
(a) What is meant by half life of a radioactive element?
(b) The half life of a radioactive substance is 30s. Calculate
(i) The decay constant and
(ii) Time taken for the sample to decay by 3/4th of the initial value. [2006,2009 F]
Sol.(a) Half life : – Half life of a radioactive element is the time taken by the sample to disintegrate to half of its original amount.
Half life period T12=0.693λ
where, λ is decay constant.
(b) T12=30s
(i) λ=0.693T12=0.69330
⇒0.0231s−1

(ii) Since N=N0(12)n
where, n = number of half lives
N = number of undisintegrated nuclei present in the sample.
N0 = original number of undisintegrated atom.
Here,
N=N0−34 N0
N0=14N0
N=N0(12)n
N04=N0(12)n
⇒ (12)n=(12)n
⇒ n = 2
Totaltime=n×T12
Total time taken = 60s = 1 min.

Q.14 The radioactive isotope D decays according to the sequence
D⟶β−particleD1⟶α−particleD2
If the mass number and atomic number of D2 are 176 and 71 respectively, what is
(a) The mass number
(b) Atomic number of D?                                                                                  [2007 D,2009 D,2010 D]
Sol. In β – decay, the mass number remains same and atomic number increases by 1 unit. In a- decay, the mass number decreases by 4 units and atomic number decreases by 2 units.
Let mass and atomic number of D be A and Z respectively.
Therefore AZD⟶β−particleAZ+1D1⟶α−particleA−4DZ+1−2D2
or A−4Z−1D
AZD⟶AZHD1+0−1B⟶AZHD1⟶A−4Z−1D2+42He
According to question, the mass number and atomic number of D2 are 176 and 71 respectively.
Therefore (a) Mass number of D = Z –1 = 71 ⇒ Z = 72
(b) Atomic number of D = A – 4 = 176 ⇒ 176 + 4 = 180

Q.15 In a given sample, two radioisotopes, A and B, are initially present in the ratio of 1 : 4. The half lives of A and B are respectively 100 yr and 50 yr. Find the time after  which the amounts of A and B become equal. [2012 F]
Sol. Let NA be the concentation of A, then 4 NA is the concentration of B.

Q.16 An observer, in a laboratory, starts with N0 nuclei of a radioactive sample and keeps on observing the number (N) of left over nuclei at regular intervals of 10 min each. She prepares the following table on the basis of her observation.

Use this data to plot a graph of loge(N0/N) vs time(t) and calculate the
(a) decay constant and 
(b) half life of the given sample. [2009C D]
Sol. The graph between loge(N0N) and time is shown below.

(a) Since loge(N0N)=λt
⇒ Slope of loge(N0N) vs time t graph gives decay constant (λ)
Therefore λ=3.46510=6.93020=0.3465s−1
(b) Since Half life T12=0.693λ
T12=0.6930.3465=2s
Half life T12=2s

Q.17 (a) Derive the law of radioactive decay viz.
N=N0e−λt
(b) Explain, giving necessary reactions, how energy is released during.
(i) Fission and                                                                  (ii) Fusion.   [2011 C]
Sol. Law of radioactive decay: –The rate of disintegration of radioactive sample at any instant is directly proportional to  the number of undisintegrated nuclei present in the sample at that instant i.e.,
dNdt∝N
dNdt=−λN
where,N = number of undisintegrated nuclei present in the sample at any instant t and dNdt is rate of disintegration.
The required plot is shown

Since dNdt=−λN
where λ = decay constant
N = number of undisintegrated nuclei in the sample of radioactive substance.
or dNN=−λdt
Integrating both sides, we get
∫dNN=−λ∫dt
In N=−λt+C
At t = 0, N = N0= number of undisintegrated
Therefore C = log N.
Therefore log N = −λt+logN0
or logNN0=−λt
Therefore N=N0e−λt

(b) (i) Nuclear Fission : – The phenomenon of splitting of heavy nuclei (mass number > 120) into smaller nuclei of nearly equal masses is known as nuclear fission.
In nuclear fission, the sum of the masses of the product is less than the sum of masses of the reactants. This difference of mass gets converted into energy as per E=mc2  and hence ample amount of energy is released in nuclear fission e.g.,
23592U+10n→14156Ba+9236Kr+310n+Q
Thus, energy is liberated in nuclear fission.

(ii) Nuclear Fusion : – The phenomenon of conversion of two lighter nuclei into a single heavy nucleus is called nuclear fusion.
Since, the mass of the heavier product nucleus is less than the sum of masses of reactant nuclear and therefore certain mass defect occurs which converts into energy as per Einstein’s mass energy relation. Thus, energy is released during nuclear fusion e.g.,
1H1+1H1⟶1H2+e++n+0.42MeV

Q.18 Why is it necessary to slow down the neutrons, produced through the fission of 23592U nuclei (by neutrons) to sustain a chain reaction? What type of nuclei are (preferably) needed for slowing down fast neutrons?
[2008 C]
Sol. Average kinetic energy of neutrons produced in nuclear fission of 92U235 is nearly 2 MeV and may escapse the reactor. So, there is need to  slow down the fast neutrons using appropriate substance namely moderator, into slow thermal neutrons.
Nuclei which have comparable mass of that of neutrons should be prefarable be used to slow down fast neutrons. It is due to the fact that the elastic collision between fast neutrons and slow moving protons in paraffin lead to  interchange the velocities.

Q.19 State the law of radioactive decay. Establish a mathematical relation between half life period and disintegration constant of a radioactive nucleus. [2006,2007 F]
Sol. Law of radioactive decay : – The rate of disintegration of radioactive sample at any instant is directly proportional to the number of undisintegrated nuclei present in  the sample at a instant.
i.e., dNdt∝N
dNdt=−λN , where 1 = decay constant.
N=N0e−λt … (1)
where, N = number of undisintegrated radioactive nuclei present in the sample.
N0 = original number of undisintegrated nuclei present in the sample.
For half life period
t=T12,N=N02
N02=N0e−λ/T12
where T12= half life period
⇒ eλT12=2
Taking in of both sides, we get λT12√ in e = In 2
λT12=In2
λT12=In2λ=0.693λ
T12=0.693λ

Class 12 Physics Nuclei

NCERT Solutions For Class 12 Physics Chapter 13 Nuclei

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