NUCLEAR PHYSICS BSc. 6th Semester Class notes, Summaries of Physics

Download NUCLEAR PHYSICS BSc. 6th Semester Class notes and more Summaries Physics in PDF only on Docsity!

Dr. Upakul Mahanta, Department of Physics, Bhattadev University

NUCLEAR PHYSICS

BY

DR. UPAKUL MAHANTA,

Asst.Prof., Department of Physics, Bhattadev University

BSc. 6th Semester Class notes

Dr. Upakul Mahanta, Department of Physics, Bhattadev University

Dr. Upakul Mahanta, Department of Physics, Bhattadev University

Dr. Upakul Mahanta, Department of Physics, Bhattadev University

Chapter 1

Nuclear Forces and Stability of Nuclei

Teacher: Any co-curricular activities?

Student: Sir, ”Dramatics”

Teacher: Really...! What did you do?

Student: Sir, acted like I understand NUCLEAR PHYSICS.....

1.1 Introduction

Everything we can see in the night time sky is made of nuclear matter. Nuclear physics describes how the Sun generates the energy we need for life on Earth, how all the atoms in your body were made in stars and what happens in stars when they die. Nuclear physics research tries to answer the fundamental questions: Where Do We Come From? What Are We? Where Are We Going? Nuclear physics is the field of physics that studies atomic nuclei and their constituents and interactions. Other forms of nuclear matter are also studied. Nuclear physics should not be confused with atomic physics, which studies the atom as a whole, including its electrons. The history of nuclear physics as a discipline starts with the discovery of radioactivity by Henri Becquerel in 1896 while investigating phosphorescence in uranium salts. In the years that followed, radioactivity was extensively investigated, notably by Marie and Pierre Curie as well as by Ernest Rutherford and his collaborators. By the turn of the century physicists had also discovered three types of radiation emanating from atoms, which they named alpha, beta, and gamma radiation. The 1903 Nobel Prize in Physics was awarded jointly to Becquerel for his discovery and to Marie and Pierre Curie for their subsequent research into radioactivity. Rutherford was awarded the Nobel Prize in Chemistry in 1908 for his ”investigations into the disintegration of the elements and the chemistry of radioactive substances”. The key experiment behind this announcement was performed in 1910 at the University of Manchester: Ernest Rutherford’s team performed a remarkable experiment in which Geiger and Marsden under Rutherford’s supervision fired alpha particles (helium nuclei) at a thin film of gold foil. Rutherford’s analysis of the data in 1911, led to the Rutherford model of the atom, in which the atom had a very small, very dense nucleus containing most of its mass, and consisting of heavy positively charged particles with embedded electrons in order to balance out the charge. That’s where the entire things start developing out.

1.2 Fundamental Forces of Nature:

As you sit on your chair, reading this article, with your laptop or desktop and your android phone nearby, you may be unaware of the many forces acting upon you. A force is defined as a push or pull that changes an object’s state of motion or causes the object to deform. Newton defined a force as anything that caused an object to accelerate according to F = ma, where F is force, m is mass and a is acceleration. The familiar force of gravity pulls you down into your seat, toward the Earth’s center. You feel it as your weight. Why don’t you fall through your seat? Well, another force, electromagnetism, holds the atoms of your seat together, preventing your atoms from intruding on those of your seat. The remaining two forces work at the atomic level, which we never feel, despite being made of atoms. The strong force holds the nucleus together. Lastly, the weak force is responsible for radioactive decay, specifically, beta decay where a neutron within the nucleus changes into a proton and an electron, which is ejected from the nucleus. Thus there are four fundamental forces present in nature. Let’s now become a bit more technical about these forces

1. Strong force (also known as strong nuclear force:) The strong interaction is very strong, but very short-ranged (order is 10−^15 m). It is responsible for holding the nuclei of atoms together. It is basically attractive, but can be effectively repulsive in some circumstances. Mesons are the force carrier for it in case of nucleon (ie protons and neutrons) and Gluons are the force carrier in the case of quarks.

Dr. Upakul Mahanta, Department of Physics, Bhattadev University

• Mass Defect: Mass defect refers to the difference in mass between of a nucleus and the sum of the masses of the protons, neutrons that is its constituent particles. Another alternative way of saying the same thing is the difference between the mass of an isotope and its mass number. Well to be honest and in simple terms what you will found is that the actual mass of the nucleus is always going to be different than the sum of the indivisual mass of the total number of neutrons and protons. To put in a mathematical way

∆m = (nprotmprot + nneutmneut) − M

where ’n’s are numbers, ’m’s are masses and M is mass of the formed nucleus.

• Binding energy: Nuclear binding energy is the minimum energy that would be required to disassemble the nucleus into its component parts. These component parts are ofcourse neutrons and protons. The binding energy is always a positive number, and thus we need to spend energy in moving the nucleons away from each other (attracted by nuclear force). Other way of saying the same thing is that it’s the amount of energy that has got utilised while holding the nucleons together inside the nucleus. Now where does this energy come from? Our authority Albert Einstein had the last laugh. His simple innocent looking formula E = mc^2 , describing the equivalence of energy and mass says that adding energy increases the mass (both weight and inertia), whereas removing energy decreases the mass. It is not advisable to talk about mass being converted to energy or similar expressions. It is better to say that, in measuring an objects mass, we are determining its energy. Suddenly people pointed their fingers towards the mass defect. They said the mass defect times c squared is the energy that holds the nucleus which is nothing but the binding energy. Mathematically ∆E = ∆m × c^2

Let’s now find how much 1 amu does in MeV corresponds to. This oftens comes in your exam with 2 marks. Here is how we do it. Well obviously using Einstein’s formula E = mc^2

E = mc^2 = 1. 6605 × 10 −^27 kg × (2. 9979 × 108 )^2 (m/sec)^2 = 15. 0639 × 10 −^11 J

=

15. 0639 × 10 −^11 J

1. 6022 × 10 −^13 M eVJ = 931. 5 M eV

As we know that 1. 6022 × 10 −^19 J = 1eV. Now paste this in your memory chip.

1 amu = 931.5 MeV

Let us take an example for the α-particle which is the 2 He^4 nucleus. This nucleus contains 2 protons and 2 neutrons. Now sum of the masses of the indivisual constituents can be calculated as follows

M ′^ = 2 × 1. 007276 amu + 2 × 1. 008665 amu = 4. 031882 amu

And the mass (M) of the α-particle is 4.001506 amu. That means we are in a position to calculate the mass defect of the nucleus which is

∆m = 4. 031882 amu − 4. 001506 amu = 0. 030376 amu

Hence the binding energy will be

∆E = ∆m × 931. 5 M eV /amu = 0. 030376 amu × 931. 5 M eV /amu = 28. 29 M eV

Well this is the amount of energy gets released from one single He^4 nucleus. Now if I take 4 gms of He^4 it will contain Avogadro’s number of nuclei. Then the release of energy will be 28. 29 M eV × 6. 023 × 1023 M ol−^1 which is almost, after doing a little bit of algebra, 2. 56 × 106 mega joules of energy. How tremendous this energy is? Well, it will heat up about 3 million gallons of water from room temperature to boiling point. Can you imagine just about 4 gms of He^4 has the ability to heat up about 2 million gallons of water? Pretty impressive, right?

Dr. Upakul Mahanta, Department of Physics, Bhattadev University

1.4 Nuclear stability:

Nuclear Stability is a concept that helps to identify the stability of a nuclear species. For example two isotope can have different abundance means their stability is different in different ways. But before we move on let us think about the following question. Why does every nucleus wants to get stability? Think of yourself when you are tired and ready for sleep. In this case you will most likely just stay put and not do anything as if you don’t have anything to spare. The major underlying reason is: ”Nature seeks the lowest energy stat”. In the lowest energy state, things are most stable and less likely to change. One way to view this is that energy makes things happen. If a nucleus is at its lowest energy state, it has no energy to spare to make a change occur. The following information that talks about stability is all based on the nucleus tending towards the lowest energy state. Unstable nuclei will try and become stable by getting to a lower energy state. They will typically do this by emitting some form of radioactivity and change in the process. The main factors that determine nuclear stability are

1. the neutron-proton ratio 2. the packing fraction of the nucleus. 3. the binding energy per unit nucleon Let us have some basic ideas about what these are actually.
   • The neutron-proton ratio: The neutron-proton ratio (N/Z ratio or nuclear ratio) of an atomic nucleus is the ratio of its number of neutrons to its number of protons which is a principal factor for determining whether a nucleus is stable. Elements with (Z< 20) are lighter and these elements’ nuclei and have a ratio of 1:1 and prefer to have the same amount of protons and neutrons amongst stable and naturally occurring nu- clei. But for heavier nuclei this ratio generally increases as the atomic number increases. This is because electrical repulsive forces between protons scale with distance dif- ferently than strong nuclear force attractions. In particu- lar, most pairs of protons in large nuclei are far enough, then the electrical repulsion dominates over the strong nu- clear force, and thus instability increases. This means if the nucleus has to be still holding up then more number of neutrons will be needed just to give more number of attractive forces in the nuclear core as the neutrons are chargeless. Thus N/Z ratio will become more than 1 for heavier nuclei. The graph in the right side is what I am saying
   • Packing Fraction: It is defined as mass defect per unit nucleon. The value of packing fraction depends upon the manner of packing of the nucleons with in the nucleus. It’s value can be neg- ative, positive or even zero. A positive packing fraction describes a tendency towards instability. A negative pack- ing fraction means isotopic mass is less than actual mass number indicates stability of the nucleus. From the figure it is clear that the packing fraction beyond mass number 200 becomes positive and increases with increase in mass number. In general, lower the packing fraction, greater is the binding energy per nucleon and hence greater is the stability. Mathematically it is defined as

pf = IsotopicM ass − M assN umber M assN umber

× 104

• Binding Energy per unit nucleon: Well the binding energy curve is obtained by dividing the total nuclear binding energy by the number of nucleons. As

Dr. Upakul Mahanta, Department of Physics, Bhattadev University

the simplest ways of predicting the nuclear stability is based on whether nucleus contains odd/even number of protons and neutrons:

Table 1.2: Odd-Even rule of nuclear stability

Protons Neutrons No. of Stable Nuclides Stability Odd Odd 4 least stable Odd Even 50 ↓ Even Odd 57 ↓ Even Even 168 most stable

catch of this table:

• Nuclides containing odd numbers of both protons and neutrons are the least stable means more radioactive.
• Nuclides containing even numbers of both protons and neutrons are most stable means less radioactive.
• Nuclides contain odd numbers of protons and even numbers of neutrons are less stable than nuclides containing even numbers of protons and odd numbers of neutrons.

Q. Based on the even-odd rule presented above, predict which one would you expect to be radioactive in each pair? (a) 8 O^16 & 8 O^17 (b) 17 Cl^35 & 17 Cl^36 (c) 10 N e^20 & 10 N e^17 (d) 20 Ca^40 & 20 Ca^45 (e) 80 Hg^195 & 80 Hg^196 Answer: (a) The 8 O^16 contains 8 protons and 8 neutrons (even-even) and the 8 O^17 contains 8 protons and 9 neutrons (even- odd). Therefore, 8 O^17 is radioactive.

(b) The 17 Cl^35 has 17 protons and 18 neutrons (odd-even) and the 17 Cl^36 has 17 protons and 19 neutrons (odd- odd). Hence, 17 Cl^36 is radioactive.

(c) The 10 N e^20 contains 10 protons and 10 neutrons (even-even) and the 10 N e^17 contains 10 protons and 7 neu- trons (even-odd). Therefore, 10 N e^17 is radioactive.

(d) The 20 Ca^40 has even-even situation and 20 Ca^45 has even-odd situation. Thus, 20 Ca^45 is radioactive.

(e) The 80 Hg^195 has even number of protons and odd number of neutrons and the 80 Hg^196 has even number of protons and even number of neutrons. Therefore, 80 Hg^195 is radioactive.

1.5 Nuclear Structure and Dimensions:

The radius of a nucleus is not well defined, since we cannot describe a nucleus as a rigid sphere with a given radius. However, we can still have a practical definition for the range at which the density of the nucleons inside a nucleus approximate our simple model of a sphere for many experimental situations (e.g. in scattering experiments). A simple formula that links the nucleus radius to the mass number is the empirical radius formula

R = R 0 A

(^13)

where R 0 = 1.12 fm and 1 fm = 10−^15 m. But from this we actually arrive at a very fundamental conclusion which may also come in your exam with 2 marks. Q. Show that nuclear density is constant for all nuclei? (Also comes in your final exam) Answer: We know that

R = R 0 A

(^13)

4 3

π R^3 =

π R^30 A

V ol. =

π R^30 A

Therefore density ρ is

ρ =

A

4 3 π R

3 0 A^

4 3 π R

3 0

which is constant term. Thus it can be shown that the nuclear density is constant for all nuclei.

Dr. Upakul Mahanta, Department of Physics, Bhattadev University

1.6 Yukawa’s meson theory:

The window of my little world opened out only to the garden of science, but from that window, enough light streamed in.” Heidiki Yukawa.

During 1934, Yukawa often lay awake at night thinking about this problem. He had a notebook at the side of his bed, so that he could record any thoughts that he might have. Sometimes he believed that he was close to a solution, but when he thought through his ideas in the morning, they proved to be worthless. One night, however, an insight came to him -there must be a relationship between the intensity of the force and the mass of the binding particle. That makes Yukawa to put forward the following proposals

• The nuclear binding force is transmitted by the exchange of massive charged particles (the heavy quanta).
• The range of the force is inversely proportional to the mass of the quantum.
• The heavy quanta are unstable, decaying via the weak interaction.

On the basis of this idea, Yukawa calculated that this binding particle would have a mass 200 times that of an electron. Shortly after Yukawa’s prediction a particle with almost precisely this mass was discovered in cosmic ray phenomena. It looked at first that Yukawa had been uncannily accurate, but there were problems with the particle found in the cosmic ray records. Although its mass was 207 times that of an electron, it was a fermion with half- integral spin rather than a boson of integral spin as Yukawa predicted for the carrier of the strong nuclear force. It turned out that the cosmic ray particle was not the particle Yukawa was talking about. Later three particles with masses approximately 270 times that of an electron were found. These did have the properties that Yukawa had predicted. One was of positive charge, one of negative charge and one was neutral. He called this particle a meson. He was the first person to theorize that the strong nuclear force between protons and neutrons was me- diated by mesons, specifically the pion. The discovery of the pion in 1947 resulted in a Nobel Prize for Yukawa in 1949.

According to the Yukawa’s theory every nucleon emits and reabsorbs pions continuously. There are three types of pions. Namely charged pions (π+, π−) and neutral pion (π^0 ). Since these are massive quanta they carry momentum with them. The associated transfer of momentum is equivalent to an action of a force. Left figure below shows a proton sends a neutral pion to another proton and thus the repulsive information gets carried out. Middle figure shows neutron sends a neutral pion to another neutron and thus the attractive information gets carried out and the right figure a proton sends a +ve pion and itself becomes a neutron. If a nearby neutron absorbs that pion then it becomes a proton. Then that proton can do the same thing. Then that proton can do the same thing. An exchange of -ve pion is not possible. This exchange can be likened to constantly hitting a ping-pong ball or a tennis ball back and forth between two people. As long as this meson exchange can happen, the strong force is able to hold the participating nucleons. But the nucleons must be extremely close together in order for this exchange to happen.

Q. Write a short note on Yukawa’s meson theory. 3 marks

Dr. Upakul Mahanta, Department of Physics, Bhattadev University

Chapter 2

Radioactivity: α, β and γ-decay

In Japan, radiation created monsters (Godzilla), in America radiation created superheros (Super-

man), here nothing, we are cool..!

2.1 Introduction

In the last chapter we spoke about stability of nuclei. But not all the nuclei are stable. What does that lead to? This chapter will focus on what the unstable nuclei will do? Well it’s just simple. They will decay.

Because the nucleus experiences the intense conflict between the two strongest forces in nature, it should not be surprising that there are many nuclear isotopes which are unstable and emit some kind of radiation. Radioactive decay (also known as nuclear decay or radioactivity) is the process by which an unstable atomic nucleus loses energy by emitting radiation, such as an α particle, β particle or γ particle. To put it in another way the atomic nuclei that dont have enough binding energy to hold the nucleus together due to an excess of either protons or neutrons are going to disintegrate. Let’s have some history. Radioactivity was discovered in 1896 by the French scientist Henri Becquerel, while working with phosphorescent materials. These materials glow in the dark after exposure to light, and he suspected that the glow produced in cathode ray tubes by X-rays might be associated with phosphorescence. He wrapped a photographic plate in black paper and placed various phosphorescent salts on it. All results were negative until he used uranium salts. The uranium salts caused a blackening of the plate in spite of the plate being wrapped in black paper. It soon became clear that the blackening was also produced by non-phosphorescent salts of uranium and metallic uranium. These radiations were given the name ”Becquerel Rays”.

2.2 Properties of Radioactivity:

The modes and characteristic energies that comprise the decay scheme for each radioisotope are specific. If instru- mentation is sufficiently sensitive, it is possible to identify which isotopes are present in a sample. But that will cost lot of your money. Radioactive decay will change one nucleus to another if the product nucleus has a greater nuclear binding energy than the initial decaying nucleus. The difference in binding energy (comparing the before and after states) determines which decays are energetically possible and which are not. But let me put all the information about radioactivity in a straight forward form.

• It is entirely a nuclear phenomenon is due to the instability of the nucleus.(Remember the N/Z ratio)
• It is a spontaneous and irreversible process. (Well that’s obviously the thing going to be, once it’s emitted it’s emitted, you can never rerun it back.)
• It is independent of external factors such as pressure, temperature, state of substance, electrical field, magnetic field, catalyst etc.(Take a sample, push it, shake it do anything you want to do! The nucleus will show extreme disrespect to your activity)
• A radioactive element emitsα, β or γ radiations which probabilistic in nature and does not depend on the age of the nucleus or how it was created. (You never can predict when a certain nuclei is going to emit a particle.)
• A radioactive element does not emit α and β particles simultaneously.(Two bullets are fired from a gun simultane- ously right! Well that’s impossible. Common sense.)
• The original radioactive nucleus or element is called a parent element and the new element formed is known as daughter element. (That’s the terminolgy we use)
• It’s a first order reaction.( means it will need infinite time to finish)
• The physical and chemical properties of daughter element are different than that of the parent element.(I will tell you later why this is so).

Dr. Upakul Mahanta, Department of Physics, Bhattadev University

2.2.1 Artificial and Natural Radioactivity

The harsh reality is that radioactivity has not been invented by man; it has been there, existing in the universe since time immemorial. The of nuclei which takes place in nature, is called natural radioactivity. However there are elements beyond uranium which have been artificially made. They are called the transuranium elements which can be made to disintegrate into other nuclei by colliding with slow moving neutrons. This is called artificial radioactivity. Thus it is customery to check the difference between these two types.

Table 2.1: Difference between Artificial and Natural Radioactivity

Natural Radioactivity Artificial Radioactivity

1. Radioactivity that takes place on its own in nature 1. It is induced by man in laboratories
2. Occurs in elements with atomic number greater than 82 2. Can be induced in elements with low atomic numbers.
3. It usually have long half life. 3. This usually have short half life.
4. Decay partcles are α, β & γ 4. Decay partcles are α, β, −β & γ

Well there hasn’t been a question in your exam from this table. But I request you to remember these as this is also probabilistic in nature...may or may not come in your exam.

2.2.2 Properties of α, β & γ-decay

In radioactive processes, particles or electromagnetic radiation are emitted from the nucleus. The most common forms of radiation emitted have been traditionally classified as α, β, and γ radiation. Let’s now inspect the characteristics of them.

• Characteristics of α-decay 1 These particles are helium nuclei 2 He^4 .(Alpha rays consist of stream of positively charged particles carrying charge of +2 units and a mass almost equal to 4 amu) 2 They affect photographic plate 3 They are deflected only slightly towards the negative plate in electric field. They are also deflected by magnetic field. (see they are charged and hence Lorentz force is in action) 4 These particles can ionize gases. Alpha rays have maximum ionizing power. (Again because these particles will interact with the medium as they are charged ) 5 They have a velocity of the order of 1 × 107 ms−^1. 6 They have very little penetrating power.(mass is the culprit in this case)
• Characteristics of β-decay 1 Beta rays are electrons (^) − 1 e^0 .( these rays are made up of streams of negatively charged particles with a negligible mass. 2 They affect photographic plate. 3 They get deflected to the maximum extent towards the positive plate in electric field. They are also deflected by magnetic field.( Again Lorentz force) 4 Their ionising power is less than that of α- rays. ( It is about one hundredth of α- particles). 5 Their velocity varies with the source sometimes reaches 2. 7 × 108 ms−^1. 6 Their penetration power is about 100 times more than that of α- particles. (Since mass is too small ).
• Characteristics of γ-decay 1 They are electromagnetic radiations (photons) like X-rays having very short wavelength, in the range of 10−^10 m to 10 −^13 m. 2 They affect photographic plate. 3 They are unaffected by electric and magnetic fields.(No charge no Lorentz force) 4 Their ionizing power is low, and is about one hundredth of β- particles.(No charge no ionisation) 5 Their velocity is same as that of light. 6 Their penetrating power is very high, about 100 times more than that of β - particles.(since they donot interact they keep on moving moving... and moving)

In addition, there are a couple of less common types of radioactive decay, these are as follows:

• Positron emission Although positron emission doesnt occur with naturally occurring radioactive isotopes, it does occur naturally in a few man-made ones. A positron is essentially an electron that has a positive charge instead of a negative charge. A positron is formed when a proton in the nucleus decays into a neutron and a positively charged electron. The positron is then emitted from the nucleus.

Dr. Upakul Mahanta, Department of Physics, Bhattadev University

1. The number of nuclei of the radioactive isotope in a sample to halve.
2. The count rate from a sample containing the radioactive isotope to fall to half its starting level. Either way you are correct if write in your exam. Since half-life is kind of stability in the time frame that indicates we must have some connection between the decay constant and half-life. Let’s now investigate this on a mthematical ground. (go back to your HS 2nd year book.) Now with a bit reorientation of the last mathematical expression we obtain

ln

N

N 0

= −λt

2. 303 log

N

N 0

= −λt

using the definition of half-life at t= T 12 the initial number N 0 will be N 20 which is the number nuclei present at that instant. On substitution we get

2. 303 log

N 0 2 N 0

= −λT (^12)

2. 303 log

= −λT (^12)

− 2. 303 × 0 .3010 = −λT (^12)

T 12 =

λ

The longer the half-life of a nucleus, the lower the radioactive activity. A nucleus with a half-life that is a million times greater than another will be a million times less radioactive. Thus half-life is a convenient way to assess the rapidity of a decay, but it should not be confused with the average life span of a radioactive nucleus.

2.3.2 Average life of a radioactive nucleus

Radioactive atoms disintegrate spontaneously and it is not possible to predict which atom is going to disintegrate next which I have told you n number of times. The practical way is you take a sample of the radioactive atoms and wait for all of them to decay away, and keep track of how long each atom lasts. The atom which disintegrates at first is said to have zero (0) life and the atom which disintegrate last is said to have infinite life. That means what is the mean life of all that nuclei becomes a legitimate question. Thus the sum of all the lifetimes of the atoms, divided by the original number of nuclei, is the mean lifetime. In other words, the mean lifetime is simply the arithmetic average of the lifetimes of the individual nuclei. Thus we can put it in a mathematical way Tavg will be

Tavg =

∫^0 t N dt ∞ 0 N dt

You can think this as follows. Say in sample of 100 nuclei 5 lived for 1 sec, 15 lived for 2 secs, 25 lived 10 sec like that. So the event 1 sec occured 5 times, the event 2 sec occured 15 times and like that. Thus the number times the life-time will give the numerator and ofcourse integrated over 0 to ∞ will cover all possible values of time. Which then is getting divided by the number of nuclei present at that time. Thus we can rewrite it as

0 t N^0 e

−λt (^) dt ∫ (^) ∞ 0 N^0 e −λt (^) dt

0 t e

−λt (^) dt ∫ (^) ∞ 0 e −λt (^) dt

Now method of substitution we give if we replace λ t = x then λ dt = dx and t = x λ without effecting the limits of

the integral.

Tavg =

0

x λ e

−x dx ∫ λ ∞ 0 e −x dx λ

=

λ

0 x e

−x (^) dx ∫ (^) ∞ 0 e −x (^) dx

Do you remember Γ-function which we learned in our 2nd semester. That was Γ (n) =

0 x

n− 1 e−xdx. Now

compare the last equation with Γ-function. Immediately you will realise that the numerator is Γ (1) and denominator is Γ (0) which is simply 1. Thus we get

Tavg =

λ

Dr. Upakul Mahanta, Department of Physics, Bhattadev University

• Relationship between Half-life & Average life of a radioactive nucleus

Well it’s straight forward. We have

T 12 =

λ

and Tavg =

λ

Now divide on by the other you will have your answer as T 12 = 0. 693 × Tavg. Thus the average life time is more than

the half-life. Now you may think why is the difference in terms of physics. Let me put in this way. Imagine that you put 100 radioactive atoms in a box and observe them decay, one by one, until they are all gone. If you add up the lifetimes of the 100 individual atoms and divide the sum by 100, you get the average lifetime. On the other hand, the half-life is a kind of median lifetime. Half of the atoms live longer and half live shorter than this time. This is the time at which 50 of the nuclei have decayed and 50 remain in the box. Well, there are always a few Methuselah nuclei that live much longer than usual, purely by chance. Methuselah....! Don’t worry just a fancy word! (Methushelah is the man reported to have lived the longest according to the Hebrew Bible. It’s also a name of a wine bottle of eight times the standard size. A full wine bottle is 750 ml I guess. Thus all I wanted say it’s mnemonic for larger, bigger, longer etc....). These nuclei pull up the average, but they have no effect on the median. Meanwhile, the shortest time that a nuclei can live is zero. There are no nuclei with negative lifetimes that could pull the average down. So the average lifetime is somewhat longer than the half life.

Q. Derive an expression for radioactive decay law. Q. Derive an expression for half life of a radioactive substance/nucleus/sample etc. Q. Derive an expression for average life of a radioactive substance/nucleus/sample etc. Q. Derive a relationship between half life and average life of a radioactive substance/nucleus/sample etc. (In this first you have derive the expression for these half and average life and then you divide. Remember that.! )

2.4 Gamow’s theory of α-decay:

In the middle of 1928 a Russian theoretical physicist, Gamow, published a reinterpretation of the Geiger-Nuttall law.(I will tell you later on in details what this law is all about. In brief this law, proposed in 1912 , relates the range of an alpha-particle to the half-life of the radioactive nucleus which emits it. The shorter the half-life was, the longer the range of the emitted particle.) Alpha-particles are held in the nucleus by an energy barrier and physicists had thought this barrier had to be overcome for an alpha-particle to be emitted. Gamow’s interpretation involved wave mechanics to suggest the alpha-particle could ’tunnel’ through the energy barrier, and that this chance of escaping was greater for higher energy particles. Hence the reason alpha decay occurs is because the nucleus has too many protons which cause excessive repulsion. In an attempt to reduce the repulsion, a Helium nucleus is emitted. The way it works is that the Helium nuclei are in constant collision with the walls of the nucleus and because of its energy and mass, there exists a nonzero probability of transmission. That is, an α-particle will tunnel out of

The basic assumptions made in deriving the theory is:

1. The -particle may exist as an entity within the nucleus.
2. Inside the nucleus the energy of the α-particle is entirely kinetic and held in the nucleus by a potential barrier.
3. Hence a finite likelihood is always there of making a tunnel through the nucleus each time a collision occurs. The word in the last assumption ”likelihood” has made the entire event a probabilistic one. What can be interpreted is that the α-particle constantly keeps on colliding the nuclear wall and in the it gets a chance to leak through it. To be more mathematical

Dr. Upakul Mahanta, Department of Physics, Bhattadev University

number of nuclei in a star generates a steady fusion reaction over millions or even billions of years. Now taking the logarithm in the last expression we get

ln T = − 2 k 2 L

Thus we get the 2nd part of the equation (2.1). But the L appearing in the expression is for a particle in box with dimension L. However in the case of nucleus L is not going to be. It’s obvious. See the α-particle coming out of the nucleus will go forever upto where nobody knows. So I can’t put it as limit like L. Hence what I will do is to integrate over a range. Thus the last expression gets modified to

ln T = − 2

∫ R 0

R

k 2 (r) dr (2.2)

Now remember my k 2 was

2 m(V −E)

ℏ more accurately^ k^2 (r)^ =

2 m[V (r)−E] ℏ.^ So all our problem lies now for figuring out the expression of V(r). Now this V(r) can be written down as the following

V (r) = 0 r < R 0

=

2 Ze^2 4 πǫr

r > R 0

where R 0 is the nuclear dimension and one thing you note that I have written here is ǫ, not ǫ 0. Well nucleus is not empty space right! Again from diagram it is seen that at r = R , V(r) = E. See that wavy line is cutting the exponential

line. That distance is my r = R means we can safely write E = 2 Ze

2

4 πǫR or^ ER^ =^

2 Ze^2 4 πǫ. Now we are in a position to rewrite my k 2 (r) as follows

k 2 (r) =

2 m ℏ

2 Ze^2 4 πǫr

− E

2 m ℏ

ER

r

− E

2 mE ℏ

R

r

With this equation(2.2) will take shape as

ln T = − 2

∫ R 0

R

2 mE ℏ

R

r

− 1 dr (2.3)

Here now we will use the method of substitution. So what Gamow did was the following. He substituted r = R cos^2 θ.

And hence the other things like the variables becomes as

r = R cos^2 θ dr = − 2 R cosθ sinθ dθ

and limits of integration becomes at r = R, θ = cos−^1 1(c−^1 1) and at r = R 0 , θ = cos−^1

R 0 R (c

− 1

R 0 R ).^ On substituting all these in equation (2.3) we get

ln T = − 2

∫ (^) c−^1

√ (^) R R^0 c−^11

2 mE ℏ

sec^2 θ − 1 − 2 R cosθ sinθ dθ

2 mE ℏ

R

∫ (^) c− 1 √ (^) R R^0 c−^11

tanθ 2 cosθ sinθ dθ

2 mE ℏ

R

∫ (^) c−^1

√ (^) R R^0 c−^11

2 sin^2 θ dθ

2 mE ℏ

R

∫ (^) c− 1 √ (^) R R^0 c−^11

(1 − cos 2 θ) dθ

2 mE ℏ

R

∫ (^) c−^1

√ (^) R R^0 c−^11

(dθ − cos 2 θ dθ)

2 mE ℏ R (θ − cosθ sinθ)

c−^1

√ (^) R R^0

c−^11

= 2

2 mE ℏ

R

[

cos−^1

R 0

R

− cos

cos−^1

R 0

R

1 − cos^2 θ

]

2 mE ℏ

(0 − 0 sinθ)

Dr. Upakul Mahanta, Department of Physics, Bhattadev University

ln T = 2

2 mE ℏ

R

cos−^1

R 0

R

− cos

cos−^1

R 0

R

√ 1 − cos

cos−^1

R 0

R

cos

cos−^1

R 0

R

2 mE ℏ

R

[

cos−^1

R 0

R

R 0

R

R 0

R

]

As the potential barrier is very very large as compared to the nuclear dimension ie RR 0 the last expression becomes

drastically simplified. First

1 − R R^0 ≈ 1 and second cos−^1

R 0

R ≈^

π

2 −^

R 0 R. The second one is bit tricky. Let’s see how that has been achieved.

cos

π 2

R 0

R

= cos

( (^) π 2

cos

R 0

R

• sin

( (^) π 2

sin

R 0

R

= 0 + 1 × sin

R 0

R

R 0

R

π 2

R 0

R

= cos−^1

R 0

R

Now replacing all these we get

ln T = 2

2 mE ℏ

R

[

π 2

R 0

R

R 0

R

× 1

]

2 mE ℏ

R

[

π 2

R 0

R

]

2 mE ℏ

2 Ze^2 4 πǫE

[

π 2

R 0

2 Ze^2 4 πǫE

]

2 m ℏ

Ze^2 πǫ

E

[

π 2

2 πǫE R 0 Ze^2

]

2 m ℏ

Ze^2 πǫ

E

π 2

2 m ℏ

Ze^2 πǫ

E

2 πǫE R 0 Ze^2

=

m √ 2 ℏǫ

Ze^2

E − 4

e ℏ

mZR 0 πǫ

Now putting the last expression in equation (2.1) we get

ln λ = ln

v 2 R 0

m √ 2 ℏǫ

Ze^2

E − 4

e ℏ

mZR 0 πǫ

And that’s the Gamow theory for us. Well that’s hell lot of a mathematics. But it’s simple if you able to keep a track on it. In your exam I doubt that this will come. All I am engaged in doing my duty.

2.5 Geiger-Nuttall Law:

In Gamow’s theory of α-decay we have considered an alpha particle in a nucleus as a particle in a box. The particle is in a bound state because of the presence of the strong interaction potential. It will constantly bounce from one side to the other, and due to the possibility of quantum tunneling by the wave through the potential barrier, each time it bounces, there will be a small likelihood for it to escape. But once it comes out of the nucleus how far will it travel before getting detected. Or in a way where should we place our detector so that we can have an α-particle detection. See in the Gamow’s theory the disintegration constant depends on the energy of the α-particle meaning it’s the energy content of α-particle because of which it will travel. Geiger and Nuttall made experimental study between the decay constant (λ) and the range of the α-particle (Rα) for different α-emitters. What they have found is the following. For an α-emitting radioactive substance the logarithm of the decay constant (λ) and the logarithm of