Radioactivity

4.1 Introduction

In this chapter we discuss various nuclear processes, in which nuclei change their state and often

their identity as well by emitting or absorbing energy in the form of photons or other particles. The

process is called radioactivity.

When we go to the history, then we come to year 1896 when French physicist Becquerel

discovered that uranium salts emit radiation that could darken a photographic film, even if the film was

wrapped in black light proof paper. Unlike ordinary light, however, it could pass through black paper.

The emission of this new kind of radiation came to be called radioactivity. In the next two years, Marie

and Pierre Curie had identified three more radioactive elements: thorium, polonium, and radium. In

1903, Curies and Becquerel shared Nobel prize for physics for their work on radioactivity. Although the

discovery of nucleus was not established for another 15 years, we can date the beginning of nuclear

physics from Becquerel’s discovery in 1896.

It did not take time to realize that radiation could be divided into three types which Ernest

Rutherford called 𝛼, 𝛽 and γ-rays. Alpha-rays ionize air the most and are the most easily stopped as they

pass through matter. Beta-rays are intermediate in the ionization they cause and in their ability to

penetrate matter; gamma-rays ionize the least and penetrate farthest. Electric and magnetic deflections

show that 𝛼-rays a positive charge and 𝛽-rays a negative charge, while 𝛾-rays are neutral.

It was later established that the radiation emitted by radioactive material also originates in the

atomic nucleus. All the radioactive nuclei have excess energy which makes them unstable. In 𝛼

radioactive nuclei the excess energy is released by the emission of an alpha particle, which is a helium

nucleus ( He24 ). If the original nucleus has Z protons and N neutrons, we can write this process of alpha

decay as

(𝑍, 𝑁) → 𝛼 + (𝑍 − 2, 𝑁 − 2) ……………………… (4.1)

Thus the remaining “offspring “ proton has two fewer protons and two fewer neutrons than the

original “parent” and is therefore the nucleus of a different element. It was for establishing that

radioactivity involved this transformation of elements that Rutherford won the Nobel Prize in chemistry

in 1908. In the case of β-ray it was found that 𝑚

𝑞= −

𝑚𝑒

𝑒 , suggesting that β-rays were just electrons or

“cathode rays”. It was established that β-rays are indeed electrons, although the energy of β electrons is

much larger than that of typical cathode rays (energy of cathode rays is about 100 keV while that of β-

rays is several MeV) (John R. Taylor, 2004). The energy of β-particles is also much larger than their rest

energy 0.5 MeV. Thus the electrons are emitted by radioactive nuclei were the first particles observed

moving at relativistic speeds.

The radioactive nuclei that emit β particles do so because they have too many neutrons; that is, the

ratio, N/Z, of neutrons to protons is larger than the value that gives maximum stability. Thus the result is

that one of the excess neutrons can change into a proton by creating and ejecting an electron. The

resulting process is called beta decay can be written as

(𝑍, 𝑁) → (𝑍 + 1, 𝑁 − 1) + 𝑒 ……………………………… (4.2)

Like α decay, β decay causes a transformation of the element involved, but in this case, there is an

increase in of one in the atomic number Z.

The 𝛾-rays emitted by radioactive nuclei were found to have many properties in common with X-

rays, and it was eventually proved that, like X-rays, 𝛾-rays are electromagnetic radiation, the only

difference being their energies: X-ray photons seldom exceed 100 keV or so, because of the difficulty of

maintaining higher voltages; 𝛾 ray photons have energies up to about 10 MeV, depending on the

radioactive nucleus involved. Gamma ray is exactly analogous to that of the photons given off by atoms:

When a nucleus is in an excited state, it can emit a photon and drop to a lower level.

4.2 The Decay Law

The radiation from a radioactive sample consists of many individual particles, each of which came

from its own separate parent nucleus. The law governing the rate at which these radioactive decays

occur is called decay law.

If the unstable nuclei of a given species were identical clock-like mechanisms obeying the laws of

classical physics, we would expect all of them to decay at the same definite time after their formation

(John R. Taylor, 2004). Instead, they decay after a wide range of different times. The explanation of this

behavior lies in the probabilistic nature of quantum mechanics. We have seen , for example, that

quantum mechanics does not give the precise location of an electron in an atom; instead it gives only

the probabilities of the various possible locations. In the same way, quantum mechanics can not predict

the precise time at which a radioactive nucleus will decay, instead, it gives only the probabilities of the

various times at which the decay may occur.

Let λ be the probability that a particular nucleus will decay in any unit interval of time. This

parameter is called the decay constant of the nucleus. For example if r=0.02 per second, the probability

is 2% that the nucleus will decay in any 1-second interval. It is an experimental fact that for any given

kind of nucleus, the probability λ is a constant that is independent of how long be the age of the

nucleus.

Suppose that at time t, our sample contains a large number, 𝑁 = 𝑁(𝑡), of the radioactive nuclei, all

of the same species. Then total number of nuclei expected to decay in unit time is the rate R(t) at which

nuclei will decay, which is given by

𝑅(𝑡) = 𝜆 𝑁(𝑡) …………. (4.3)

Now, each time a nucleus decays, the number N(t) decreases by 1. Thus, R is given by rate of decrease of

N, i.e.

𝑅 = −𝑑𝑁

𝑑𝑡 ………… ….. (4.4), where negative sign indicates as time increases, N(t)

decreases.

From (4.3) and (4.4), we find that

𝑑𝑁

𝑑𝑡= −𝜆 𝑁(𝑡) ……………. (4.5)

This is first order differential equation for N(t). Rearranging this equation we get:

dN

N= −λ dt ………… (4.6)

Integrating both side

∫dN

N

N

N0= − ∫ dt

t

0 …………. (4.7)

where at time t = 0, N = N0(t) be the initial number of radioactive nuclei. Integration gives

log𝑒(𝑁/𝑁0) = −𝑡

𝑁 = 𝑁0 𝑒−𝜆𝑡 ……………………. (4.8)

The equation shows that N(t)

decreases exponentially with time.

At 𝑡 = ∞ , 𝑁 = 0, which implies that

the disintegration process never

stops. This is due to the reason that

as time increases, the rate of

disintegration decreases because

N(t) decreases according as equation

(4.5). The sketch of the equation

(4.8) is drawn in the figure 4.1.

If we take time 𝑡 =1

𝜆, then equation

(6) becomes

𝑁 = 𝑁0 𝑒−𝜆/𝜆

⇨𝑁 =𝑁0

𝑒

This time t is called life time of the radioactive sample which is defined as the time in which N drops to

the fraction 1/e of its original value (𝑁0) This is usually denoted by τ. Hence

𝜏 =1

𝜆 ……………………… (4.9)

In terms of τ, the decay law can be expressed as

𝑁 = 𝑁0𝑒−𝑡/𝜏 ………………. (4.10)

Now we define half life time of the of the radioactive sample as the time in which number of nuclei (N)

drops to half of the original value(𝑁0) . It is denoted as 𝑇1/2 .

∴ 𝑡 = 𝑇1/2 ⇨ 𝑁 =𝑁0

2 . Now equation (4.8) becomes

𝑁0

2= 𝑁0 𝑒−𝜆𝑇1/2

Figure 4.1: the number N decreases exponentially with time.

Or, 2 = eλT1/2

Or, log𝑒 2 = 𝜆𝑇1/2

Or, 𝑇1/2 =0.693

𝜆…………. (4.11)

This is the expression of the half life time of the radioactive substance.

Problem: show that decay equation 𝑁(𝑡) = 𝑁0 𝑒−𝜆𝑡 can be written as 𝑁(𝑡) =𝑁0

2𝑡/𝑇1/2

.

Solution: We have 𝑁(𝑡) = 𝑁0 𝑒−𝜆𝑡 ,putting 𝑇1/2 =log𝑒 2

𝜆 we get

𝑁(𝑡) = 𝑁0 𝑒−

log𝑒 2

𝑇1/2𝑡

Or, 𝑁0 = 𝑁(𝑡) 𝑒log𝑒 2

𝑇1/2𝑡

Or, log𝑒 𝑁0 = log𝑒 𝑁(𝑡)(𝑡/𝑇1/2) log𝑒 2

Or, log𝑒 𝑁0 = log𝑒 𝑁(𝑡) log𝑒 2𝑡/𝑇1/2

Or, 𝑁0 = 𝑁(𝑡) 2𝑡/𝑇1/2

Or, 𝑁(𝑡) =𝑁0

2𝑡/𝑇1/2

4.3 Absorption of α particles, range and stopping power

The energies of α particles like that of other charged particles can be determined from

measurements of their absorption by matter (Kaplan, 2002). Alpha particles are easily absorbed by sheet

of paper, by an aluminum foil 0.004 cm thick or by several centimeter of air. In the passage through the

matter, the α particle gets slowed down by losing energy. If the particles emitted by a source in air are

counted by counting the number of scintillations on zinc sulfide screen, it is found that their number

stays practically constant up to a certain distance R from the source, and then drops rapidly to zero. This

distance R is called the range of the particles and is related to the initial energy of the particles (Kaplan,

2002).

Range of alpha particles in air can be made with an apparatus like that shown in figure 4.2.

The source is placed on a movable block whose distance from a detector can be varied. A narrow beam

of particles emerges through a collimating slit, passes through a known thickness of air, and reaches the

detector. The detector is a thin, screened-walled ionization chamber 1 to 2 mm deep. Ions are formed

in pulses in the chamber when individual α particles pass through it.

The voltage pulses are induced on the

chamber electrode are amplified

electronically and counted. The counting rate

is then determined as a function of the

distance between source and detector. The

results of an experiment with α-particles

from polonium are shown in figure 4.3. The

ordinate is the relative number of particles,

and the abscissa is the distance from the

source. Curve A shows the fraction of

particles in the beam detected at various

distances from the source; only the results

near the end of the path are shown. All the

particles which pass through the slit are

counted until the detector is about 3.75 cm

from the source; the fraction detected falls

rapidly to 0.2 at 3.88 cm, and then decreases

somewhat more slowly to zero (Kaplan,

2002). A quantity called the extrapolated

range, 𝑅𝑒, is obtained by drawing the

tangent to the curve at is inflection point and

noting where the tangent crosses the

distance axis.

Figure 4.2: An apparatus for precise measurements of the range of α-particles

If the derivative of the number-distance curve (curve A) is computed at different distances from

the sources and then plotted against the distance, a differential range curve (curve B) is obtained. This

curve gives the relative number of particles stopping at a given distance as a function of the distance

from the source; the unit of the ordinate is so chosen that the area under the differential range curve is

unity and all the particles are accounted for. The maximum ordinate of the differential range curve

occurs at a value of the abscissa which is called the mean range, �̅�, defined so that half the particle track

lengths exceed it, while half are shorter. In the case considered, the mean range is 3.842 cm. The results

show that the track lengths of the particles in the beam are not all the same, but vary around an average

value. The effect is called straggling (Kaplan, 2002).

Alpha-particle loses a large fraction of their energy by causing ionization along their paths. The

extent of the ionization caused by an α-particle depends on the number of molecules it hits along its

path and on the way in which it hits them. Some particles hit more and others hit less than the average

number of molecules in passing through a centimeter of air. Hence, the actual distance from the source

at which their energy is completely used up is somewhat different for different particles, giving rise to

straggling. Due to straggling, the actual range of an α-particle is not definite, and to avoid this

indefiniteness either the extrapolated or the mean range is used (Kaplan, 2002).

Straggling effect is also partly due to the formation of 𝐻𝑒+ ions, by the attachment of one electron

to some of the α-particles (Tayal, 2003).These ions still possess ionizing power, and hence cause a slight

extension of the range before become neutral atoms. This method is not well suited to weak sources.

The ionization caused by a beam of α particles can be measured, and is related to the energy and

range. An electron and the positive ion which results from its removal from an atom form an ion pair,

and the intensity of the ionization caused by the particles is expressed by the specific ionization, defined

as the number of ions pairs formed per millimeter of beam path. The apparatus of figure 4.2 may be

utilized to measure the relative specific ionization produced by a beam of α-particles at different

distances from the source.

Specific ionization-distance curve

is shown in figure 4.4 for the particles

from Ra𝐶′(𝑃𝑜214). As the distance of

the α-particles from the source

increases, the relative specific

ionization increases, at first quite

slowly and then more rapidly, reaches

a maximum, and then drops sharply to

zero.

The shape of ionization curve

depends on the change in speed of an

α-particle as it transverses its path. In

producing ion pairs, the particle loses

energy and its speed decreases. When

it moves more slowly, it spends more

time in the neighborhood of the air

molecules it encounters, and the probability of producing ion pairs increases. This effect accounts for

Figure 4.4: specific ionization of a beam of α-particles as a function of distance from the source.

Ion

izat

ion

(ar

bit

rary

un

its)

the increase in specific ionization as the particle moves farther from the sources. Finally, electrons are

captured by the α-particle and a neutral helium atom is formed which can no longer cause ionization.

Range measurements can also be made with solid foils as absorbers. Thin, uniform foils are

placed over an α-emitting sample, and the thickness needed to absorb the α-particles completely is a

measure of the range. Mica, aluminum or gold-foils are used commonly because they can be prepared

easily in different thickness with reasonable uniformity. The range of α-particles in such materials is very

small.

The quantity which is related to absorption of charged particles by matter is the stopping power,

S(E), defined as the energy lost by the particle per unit path in the substance,

𝑆(𝐸) = −𝑑𝐸

𝑑𝑥 …………………. (4.12)

where E is the classical kinetic energy. The negative sign indicates that as distance 𝑥 increases, the

energy E decreases. The stopping power varies with the energy of the particle, and the range of the

particle is given by

𝑅 = ∫ 𝑑𝑥𝑅

0= ∫

𝑑𝐸

𝑆(𝐸)

𝐸0

0 …………. (4.13)

The stopping power S(E) of a substance can be determined experimentally by measuring the

energy of the particles which have gone through a certain thickness of the substance. When energy loss

is found in this way for different initial velocities, the range in the substance can be deduced from

equation (4.13) as a function of the initial energy. If the range is known as a function of energy, the

stopping power can be obtained from the relation

dR

dE=

1

𝑆(𝐸) …….. (4.14)

The energy lost by a nonrelativistic charged particle per unit length of its path in a given substance can

be expressed in the following form derived from theory,

𝑆 = −𝑑𝐸

𝑑𝑥=

4𝜋𝑧2𝑒4𝑁

𝑚𝑒v2 𝑍 log𝑒 (2𝑚𝑒v2

𝐼)……. (4.15)

where 𝑧𝑒 be the charge and the speed of the particle is v, 𝑚𝑒 is the mass of an electron, Z is the atomic

number of the substance, N is the number of atoms per cubic centimeter of the substance, and I is a

quantity, called the average excitation potential of an atom of the substance, which must be obtained

experimentally.

For an α-particle, z=2,and equation (4.15) may be rewritten as

𝑆′ = −𝑚𝑒v2

16𝜋𝑒4𝑁 𝑑𝐸

𝑑𝑥= 𝑍 log𝑒 (

2𝑚𝑒v2

𝐼) ……… (4.16)

The quantity on left side (𝑆′) is proportional to the stopping power S(E) ; it can be calculated and

measured for different values of the atomic number Z of the absorbing substance. The quantity I

depends, among other things, on Z, so that a graph of 𝑆′ against Z should give a curve which is not quite

a straight line which is also quite agree with the experiment.

4.4 Gamow’s Theory of Alpha Decay

Most of the naturally occurring radioactive nuclei are heavy nuclei, with A≥ 210, that can be

grouped into three radioactive series which are originated from the parent nuclei 𝑇ℎ90232 , 𝑈92

235 and 𝑈92238

after the alpha decay. These nuclei exist on earth today only because they have half-lives of billions of

years, so that appreciable amounts have survived the billions of years (see table 4.1). These extremely

long half-lives pose a puzzle: How can a process driven by nuclear energies of millions of electron volts

take billions of years to occur?

Long half lives of the 𝑇ℎ90232 , 𝑈92

235 and 𝑈92238 are not only puzzles posed by α decay, the another

puzzle occurs from the fact that the half lives of some α-radioactive nuclei vary over astonishing range,

from billions of years to fractions of a microsecond, while the energies released range only from about

4 to about 9 MeV. Table 4.1 shows five alpha-

emitting nuclei in order of decreasing half-life

(John R. Taylor, 2004).The second column shows

the kinetic energy released in the decay, and the

third shows the half-life.

Given these facts, the questions we must

answer are these: How do nuclei emit alpha

particles ? Given that they emit alpha particles,

why do some nuclei wait more than 10 billion

years before doing so? Why do large variation in the half-lives although small variation in the kinetic

energy ? The answers to these questions were found in 1928 by the Russian George Gamow and

independently by the American Condon and Gurneys.

Figure 4.5 shows the barrier that keeps an alpha particle from escaping promptly from nucleus.

The Coulomb potential energy U(𝑥) of the alpha particle at a distance 𝑥 from the centre of the offspring

nucleus is given by

𝑈(𝑥) =2𝑍𝑘𝑒2

𝑥 ……………… (4.17)

where 𝑥 is the seperation of alpha particle from the residual nucleus and Ze is the charge of the

residual nucleus, the charge of the alpha particle is 2e and 𝑘 ==1

4𝜋𝜀0. For 𝑥 > 𝑥0 where 𝑥0 is about

equal to the nuclear radius R; it is just the Coulomb potential , while for 𝑥 < 𝑥0, it is dominated by the

attractive nuclear force and potential energy drops abruptly. Inside the nucleus, the interaction between

the nucleons may be represented by a constant attractive potential 𝑈0, exerted over the distance 𝑥0. It

is spoken of as a potential well of depth 𝑈0 and of width 𝑥0. Hence potential energy is defined as

𝑈(𝑥) = −𝑈0 for 𝑥 < 𝑥0

Now we establish the approximate parameter of the barrier confining the alpha particle. Its right-

hand boundary is at the point 𝑥0 (figure 4.5), which is equal to the nuclear radius (i.e. the sum of the

radii of the residual nucleus and the alpha particle). Hence

𝑥0 ≈ 8 fm [ ∵ Using R = 𝑅0 𝐴1/3 ]

Therefore from equation (4.17) we get:

Table 4.1

Nucleus K.E. (MeV) 𝑇1/2

Thorium - 𝑇ℎ90232 4.1 14 billion year

Radium - 𝑅𝑎88226 4.9 1600 year

Curium- 𝐶𝑚96240 6.4 27 days

Polonium- 𝑃𝑜84194 7.0 0.7 s

Radium- 𝑅𝑎88216 9.5 0.18 μs

𝑈(𝑥) =2𝑍𝑘𝑒2

𝑥~2 × 90 ×

1.44 𝑀𝑒𝑉.fm

8 fm≈ 30 𝑀𝑒𝑉

When the alpha particle emerges from the nucleus, it moves far away, it has kinetic energy (E) and

hence total energy between 4 and

9 MeV (John R. Taylor, 2004). By

conservation of energy, this must

be the same as its energy before

the decay occurred. Therefore, its

energy inside the nucleus was also

between 4 and 9 MeV as shown

by the dashed horizontal line in

the figure 4.5. This is far below

the top of the barrier i.e. 𝑈𝑚𝑎𝑥 ≈

30 MeV. Hence the alpha particle

would be permanently trapped by the barrier if we assume it as a classical particle. However, the

observed fact is that an alpha particle does occasionally escape from the barrier. Such an escape can be

explained by the quantum mechanics and it is called quantum tunneling. Now we find out the

probability of such tunneling quantum mechanically.

Barrier penetration

Consider a beam of α particles of kinetic energy E incident from left on a potential barrier of

height V and width 𝒂 such that barrier height (V) is greater than the energy of the particle i.e.

𝑉 > 𝐸 . On both side of the barrier, potential is 𝑉 = 0. Hence potential is described as

𝑉 = 0 for 𝑥 < 𝑥0 (region I)

𝑉 = 𝑉 for 𝑥0 < x < 𝑥1 (region II)

𝑉 = 0 for 𝑥 > 𝑥1 (region III) …………………… (4.18)

Let 𝜓1, 𝜓2 and 𝜓3 are the respective wave functions in regions I, II and III as indicated in the

figure 4.6.

The corresponding Schrodinger equations are

𝑑2𝜓1

𝑑𝑥2 +2𝑚𝐸

ћ2 𝜓1 = 0 [∵ 𝑉 = 0] For region I

where 𝑚 =𝑚𝛼𝑀

𝑚𝛼+𝑀 be the reduced mass of the alpha particle and the residual nucleus.

𝑑2𝜓2

𝑑𝑥2 +2𝑚(𝐸−𝑉)

ћ2 𝜓2 = 0 For region II

Figure 4.5: Potential Energy Curve

𝑑2𝜓3

𝑑𝑥2 +2𝑚𝐸

ћ2 𝜓3 = 0 For region III

Put 2𝑚𝐸

ћ2 = 𝜆2 and

2𝑚(𝑉−𝐸)

ћ2 = 𝛾2 ………( 4.181)

Then the equations become

𝑑2𝜓1

𝑑𝑥2 + 𝜆2𝜓1 = 0 (Region I)

𝑑2𝜓2

𝑑𝑥2 − 𝛾2𝜓2 = 0 (Region II)

𝑑2𝜓3

𝑑𝑥2 + 𝜆2𝜓3 = 0 (Region III) ………. (4.19)

The solutions are

𝜓1 = 𝐴𝑒𝑖𝜆𝑥 + 𝐵𝑒−𝑖𝜆𝑥 (Region I)

𝜓2 = 𝐹𝑒𝛾𝑥 + 𝐺𝑒−𝛾𝑥 (Region II)

𝜓3 = 𝐶𝑒𝑖𝜆𝑥 + 𝐷𝑒−𝑖𝜆𝑥 (Region III)

𝒙 = 𝟎

Figure 4.6

𝜓3 = 𝐶𝑒𝑖𝜆𝑥

………(4.20)

where A is the amplitude of the wave incident on the barrier from the left; B is the amplitude of

the reflected wave in region I; F is the amplitude of the wave, penetrating the barrier in region II;

G is the amplitude of the reflected wave from the surface at 𝑥 = 𝑥1, in region II; C is the

amplitude of the transmitted wave in region III and D is the amplitude of nonexistent reflected

wave in region III, so putting D equal to zero we get the second equation in region III. For

simplicity consider 𝑥 = 𝑥0 = 0(at origin) and then 𝑥 = 𝑥1 = 𝑎.

The constants A, B, F, G and C are determined by using boundary conditions:

(i) 𝜓1 = 𝜓2 and ∂𝜓1

∂x=

𝜕𝜓2

𝜕𝑥 at 𝑥 = 0

(ii) 𝜓2 = 𝜓3 and ∂𝜓2

∂x=

𝜕𝜓3

𝜕𝑥 at 𝑥 = 𝑎 ………….. (4.21)

By substituting the values of 𝜓1 , 𝜓2 and 𝜓3 in equation (4.20) in equation (4.21) we get:

𝐴 + 𝐵 = 𝐹 + 𝐺 …………………. (4.22)

𝑖𝜆𝐴 − 𝑖𝜆𝐵 = 𝛾𝐹 − 𝛾𝐺 ……………….. (4.23)

𝐹𝑒𝛾𝑎 + 𝐺 𝑒−𝛾𝑎 = 𝐶𝑒𝑖𝜆𝑎 ……………… (4.24)

𝛾𝐹𝑒𝛾𝑎 − 𝛾𝐺 𝑒−𝛾𝑎 = 𝑖𝜆𝐶𝑒𝑖𝜆𝑎 ………………….. (4.25)

Multiplying equation (4.24) by 𝛾 and then adding and then subtracting with equation (4.25) we get:

2𝛾𝐹𝑒𝛾𝑎 = 𝛾𝐶𝑒𝑖𝜆𝑎 + 𝑖𝜆𝐶𝑒𝑖𝜆𝑎

⇨ 𝐹 =1

2𝐶(1 + 𝑖𝜆/𝛾)𝑒(𝑖𝜆−𝛾)𝑎 …………… (4.26)

and 2𝛾𝐺𝑒−𝛾𝑎 = 𝛾𝐶𝑒𝑖𝜆𝑎 − 𝑖𝜆𝐶𝑒𝑖𝜆𝑎

⇨ 𝐺 =1

2𝐶 (1 −

𝑖𝜆

𝛾) 𝑒(𝑖𝜆+𝛾)𝑎 ……………………………… (4.27)

From equation (4.22) and (4.23) we get:

2𝑖𝜆𝐴 = 𝑖𝜆𝐹 + 𝑖𝜆𝐺 + 𝛾𝐹 − 𝛾𝐺

⇨ 𝐴 =1

2(1 +

𝛾

𝑖𝜆) 𝐹 +

1

2(1 −

𝛾

𝑖𝜆) 𝐺 …………………. ……….. (4.28)

Substituting the value of F and G in equation (4.28) we get:

𝐴 =𝐶

4(1 +

𝛾

𝑖𝜆) (1 + 𝑖𝜆/𝛾)𝑒(𝑖𝜆−𝛾)𝑎 +

𝐶

4(1 −

𝛾

𝑖𝜆) (1 −

𝑖𝜆

𝛾) 𝑒(𝑖𝜆+𝛾)𝑎 ……… (4.29)

Since the velocity of alpha particle in I region is same as in III region, transmission probability of

incident α-particle is given by (Tayal, 2003)

𝑇 =Transimitted flux

incident flux=

|𝐶|2×v

|𝐴|2×v=

|𝐶|2

|𝐴|2 …………. (4.30)

In practice 𝛾𝑎 ≫ 1, hence first term of equation (4.29) can be neglected in comparison to the second.

Hence, we have now

|𝐴|2

|𝐶|2 = (𝐴

𝐶) (

𝐴

𝐶)

∗

=1

16(1 −

𝛾

𝑖𝜆) (1 −

𝑖𝜆

𝛾) 𝑒(𝑖𝜆+𝛾)𝑎 (1 +

𝛾

𝑖𝜆) (1 +

𝑖𝜆

𝛾) 𝑒(−𝑖𝜆+𝛾)𝑎

=1

16(1 +

𝛾2

𝜆2) (1 +

𝜆2

𝛾2) 𝑒2𝛾𝑎

∴ Transmissivity of the barrier 𝑇 =|𝐶|2

|𝐴|2 =16𝜆2𝛾2

(𝜆2+𝛾2)2 𝑒−2𝛾𝑎 …………. (4.31)

For 2𝛾𝑎 ≫ 1, the most important factor in this equation is the exponential, which then will be

extremely small. The factor in front of the exponential part is usually of the order of magnitude of unit,

maximum value is four. Thus, for order of magnitude calculations we can, write

𝑇 = 𝑒−2𝛾𝑎 ……………. (4.32)

The transmission or escape probability depends on two variables: 𝛾, barrier thickness 𝑎 . If the

barrier is high and wide like that the case in α decay, both 𝛾 and 𝑎 are large and the escape probability is

very small. Furthermore, if 𝛾 is large, 𝑇 = 𝑒−2𝛾𝑎 is very sensitive to the barrier thickness 𝑎 . This is

the main reason that the observed half-lives vary over

such an enormous range.

Derivation of Geiger-Nuttal Law

Equation (4.32) gives escape probability for any particle

confined by a rectangular barrier of figure 4.6. Here our

interest is in an alpha particle confined by the

nonrectangular barrier (of figure 4.5) i.e. for a barrier

where potential is not constant in the region 𝑥0 < 𝑥 <

𝑥1. To find the corresponding escape probability, we can

approximate the actual barrier by a succession of 𝑛

rectangular barriers as in figure 4.7. The total escape probability is the product of the 𝑛 individual

probabilities (John R. Taylor, 2004)

𝑇 = 𝑇1𝑇2 … … . 𝑇𝑛

= e−2γ1∆x1e−2γ2 ∆x1e−2γ3∆x3 … … . . e−2γn∆xn = exp(−2 ∑ 𝛾𝑖 ∆𝑥𝑖) ….. (4.33)

where each 𝛾𝑖 is given by equation 4.181 with the appropriate barrier height 𝑉𝑖. In the limit that all

∆𝑥𝑖 → 0, the sum in (4.33) becomes an integral, and we obtain the desired probability,

𝑇 = exp [−2 ∫ 𝛾(𝑥)𝑑𝑥𝑥1

𝑥0] ……….. (4.34)

where integral has taken through whole region between 𝑥 = 𝑥0 and 𝑥 = 𝑥1 where Coulomb repulsion

V(r) is greater than the energy E of an alpha particle. Substituting the value of 𝛾 from equation 4.181 we

get

Figure 4.7

𝑇 = exp [−2√2m

ħ∫ {𝑉(𝑥) − 𝐸}1/2𝑑𝑥

𝑥1

𝑥0] ………….. (4.35)

Let us assume that an alpha particle moves inside the potential well with a certain velocity 𝑣0 and

hence it hits the wall with frequency 𝑣0

2𝑥0 where 𝑥0 is the radius of the nucleus. Alpha particles has the

probability T of leaking out at each hit. Hence multiplication of frequency, with which the α particle

strikes the barrier, with escape probability T will give us decay constant λ. Hence

𝜆 = frequency × Probability

⇨ 𝜆 =𝑣0

2𝑥0exp [−

2√2m

ħ∫ (𝑉(𝑥) − 𝐸)1/2𝑑𝑥

𝑥1

𝑥0] ………. (4.36)

Taking logarithm of equation (4.36)

log𝑒 𝜆 = log𝑒𝑣0

2𝑥0−

2√2m

ħ ∫ {𝑉(𝑥) − 𝐸}1/2𝑑𝑥

𝑥1

𝑥0

= log𝑒𝑣0

2𝑥0−

2√2m

ħ ∫ {

2𝑍𝑒2

4𝜋𝜀0𝑥− 𝐸}

1/2

𝑑𝑥𝑥1

𝑥0 …………… (4.37)

where 𝑉(𝑥) =2𝑍𝑒2

4𝜋𝜀0𝑥

is the electrostatic P.E. of an alpha particle at a distance 𝑥 from the centre of the residual(i.e.

daughter) nucleus , 𝑍𝑒 is the charge of that nucleus (i.e. of daughter nucleus which is equal to the

nuclear charge minus the alpha particle charge of 2e), E is the kinetic energy of the alpha particle at

distance 𝑥. The region from 𝑥 = 𝑥0 to 𝑥 = 𝑥1 is called the thickness of the barrier.

When 𝑥 = 𝑥1, 𝐸 =V (figure 4.5)

∴ 𝐸 =2𝑍𝑒2

4𝜋𝜀0𝑥1 ⇨ 𝑥1 =

2𝑍𝑒2

4𝜋𝜀0𝐸

Therefore equation (4.37) becomes

log𝑒 𝜆 = log𝑒𝑣0

2𝑥0−

2√2mE

ħ ∫ {

𝑥1

𝑥− 1}

1/2𝑑𝑥

𝑥1

𝑥0 ……. (4.38)

Put 𝑥 = 𝑥1cos2𝜙 and 𝑥0 = 𝑥1𝑐𝑜𝑠2𝜙0

. ……… (4.39)

∴ 𝑑𝑥 = x12 cos ϕ (−) sin ϕ dϕ = −2x1 cos ϕ sin ϕ dϕ

{𝑥1

𝑥− 1}

1/2𝑑𝑥 = (

1

cos2𝜙− 1)

1/2 (−2𝑥1 cos 𝜙 sin 𝜙 𝑑𝜙)

= −2 sin 𝜙

cos 𝜙 𝑥1 cos 𝜙 sin 𝜙 𝑑𝜙

= −2𝑥1sin2𝜙 𝑑𝜙

= −𝑥1(1 − cos 2𝜙)𝑑𝜙

⇨∫ {𝑥1

𝑥− 1}

1/2𝑑𝑥

𝑥1

𝑥0= −𝑥1 ∫ (1 − cos 2𝜙)𝑑𝜙

𝑥1

𝑥0

= −𝑥1 [𝜙 −sin 2𝜙

2]

𝑥0

𝑥1

= −𝑥1[𝜙 − sin 𝜙 cos 𝜙]𝑥0

𝑥1

= −𝑥1 {𝑐𝑜𝑠−1 (𝑥

𝑥1)

1/2

− (1 −𝑥

𝑥1)

1/2(

𝑥

𝑥1)

1/2}

𝑥0

𝑥1

[∵ from eqution 4.39]

= 𝑥1 {𝑐𝑜𝑠−1 (𝑥0

𝑥1)

1/2

− (1 −𝑥0

𝑥1)

1/2(

𝑥0

𝑥1)

1/2} ………..…….. (4.40)

The width of the potential barrier is very large compared with the nuclear radius i.e. 𝑥1 ≫ 𝑥0

(Murugeshan, 2014). Therefore

𝑐𝑜𝑠−1 (𝑥0

𝑥1)

1/2

≈𝜋

2− (

𝑥0

𝑥1)

1/2

[∵ 𝑥1 ≫ 𝑥0 ⇨ 𝑥0

𝑥1≈ 0 ≈ sin {(

𝑥0

𝑥1)

1/2} = cos {

𝜋

2− (

𝑥0

𝑥1)

1

2}]

and (1 −𝑥0

𝑥1)

1/2

≈ 1

Then equation (4.40) becomes

∫ {𝑥1

𝑥− 1}

1/2𝑑𝑥

𝑥1

𝑥0= 𝑥1 {

𝜋

2− (

𝑥0

𝑥1)

1

2− (

𝑥0

𝑥1)

1/2}

= 𝑥1 {𝜋

2− 2 (

𝑥0

𝑥1)

1

2}

Putting 𝑥1 =2𝑍𝑒2

4𝜋𝜀0𝐸

∫ {𝑥1

𝑥− 1}

1/2𝑑𝑥 =

2𝑍𝑒2

4𝜋𝜀0𝐸{

𝜋

2− 2 (

𝑥0 4𝜋𝜀0𝐸

2𝑒2𝑍)

1

2}

𝑥1

𝑥0

= {Ze2

4ε0E− 2 (

2e2Z x0

4πε0E)

1/2

}

Putting this value in equation 4.38, we get:

log𝑒 𝜆 = log𝑒𝑣0

2𝑥0−

2√2mE

ħ [{

Ze2

4ε0E− 2 (

2e2Z x0

4πε0E)

1/2

} ]

= log𝑒𝑣0

2𝑥0− (

2√2𝑚 𝑒2

𝜀0 4ħ) 𝑍𝐸−

1

2 + 𝑍1/2𝑥01/2

4√2m

ħ (

2e2

4πε0)

1

2

= log𝑒𝑣0

2𝑥0+ 𝑍1/2𝑥0

1/2 4𝑒

ħ (

𝑚

𝜋𝜀0)

1/2

−𝑒2

ħ𝜀0 (

𝑚

2)

1/2

𝑍 𝐸−1/2 ……… (4.41)

After substituting the value of constants 𝑒, 𝑚, ħ 𝑎𝑛𝑑 𝜀0 ,we get:

log𝑒 𝜆 = log𝑒𝑣0

2𝑥0+ 2.97𝑍1/2𝑥0

1/2− 3.95 𝑍 𝐸−1/2 ……………….. (4.42)

where 𝑥0 is in fermi, and E is in MeV.

From equation 4.36,

𝜆 =𝑣0

2𝑥0 𝑇

⇨ log𝑒 𝜆 = log𝑒𝑣0

2𝑥0+ log𝑒 𝑇

Or, log𝑒 𝜆 − log𝑒𝑣0

2𝑥0 = log𝑒 𝑇

From equation 4.42,

log𝑒 𝑇 = 2.97𝑍1/2𝑥01/2

− 3.95 𝑍 𝐸−1/2 ………… (4.43)

Changing the base of log from e to 10 in equation (4.42) we get:

log10 𝜆 = log10 (𝑣0

2𝑥0) + 0.4343 (2.97𝑍1/2𝑥0

1/2− 3.95 𝑍 𝐸−1/2)

Or, log10 𝜆 = log10 (𝑣0

2𝑥0) + 1.29 𝑍1/2𝑥0

1/2− 1.72 𝑍 𝐸−1/2 ………. (4.431)

For the second term, change in atomic number and nuclear radius are negligible when compared to the

changes in energy and the first term is almost same for heavier nuclei (Murugeshan, 2014). So equation

(4.431) reduces to

log10 𝜆 = 𝑐 + 𝑑𝐸−1/2 ………… (4.44)

According to equation (4.44), the emitters having lesser decay constants emit alpha particles of greater

energy (E) which is called Geiger and Nuttal law. In this way we can able to determine the Geiger-Nuttal

law from the idea generated by Gamow.

If we substitute the typical values 𝑍 = 90, 𝐸 = 6 𝑀𝑒𝑉, and radius 𝑥0 = 8 fm into equation (4.43),

we get

𝑇 ≈ exp (2.97 × 901

2 × 81

2 − 3.95 × 90 ×1

√6) = exp(79.7 − 145.13) ≈ 3.8 × 10−29

As our anticipation, the probability T that an alpha particle striking the surface of a nucleus will escape is

extremely small.

Half-Lives

Since experimental data are generally given in terms of half-lives, let us assemble our results to give an

expression for the half-life of an alpha-emitting nucleus. The half-life is ln 2

𝜆 and the decay constant λ=

𝑓𝑇, where 𝑓 is the frequency with which alpha particles strike the nuclear surface. Therefore,

𝑡1/2 =ln 2

𝜆=

0.693

𝑓𝑇

Taking natural logarithms to give

ln 𝑡1/2 = − ln 𝑇 + ln0.693

𝑓

From equation (4.43)

log𝑒 𝑡1/2 = 3.95 𝑍 𝐸−1/2 − 2.97𝑍1

2𝑥0

1

2 + 𝑐 ……….. (4.45)

Figure 4.9: Relationship between half-life and α-disintegration energy for some even-even α-emitters (Kaplan, 2002)

Alpha-disintegration energy(MeV)

𝐥𝐨

𝐠𝟏

𝟎𝒕 𝟏

/𝟐

where constant c is ln0.693

𝑓 .

An alpha particle is formed from nucleons , each with potential energy about −50 𝑀𝑒𝑉 and kinetic

energy about 25 MeV. Hence the particle should have a kinetic energy of roughly 100 MeV,

corresponding speed 𝑣~𝑐/4. For a nuclear diameter d of about 15 fm, the frequency with which an

alpha particle appear at the surface of the nucleus would be roughly

𝑓~𝑣

𝑑 ~

𝑐/4

15 fm~5 × 1021𝑠−1

Hence

𝑐 = ln0.693

𝑓 ≈ −50

To test our theory , consider a set of alpha-emitting nuclei that all have the same value of Z. For

example, we could consider the isotopes 𝑃𝑜84192 , 𝑃𝑜84

194 , 𝑃𝑜84196 , 𝑃𝑜 ,84

208 𝑃𝑜84210 , 𝑃𝑜84

212 , 𝑃𝑜84214 , 𝑃𝑜84

216 of

polonium , all of which decay exclusively by alpha emission. Since these nuclei all have the same Z and

very nearly the same radius R, the second term in equation (4.45) is approximately constant, and hence

(4.45) reduces to

log𝑒 𝑡1/2 = 3.95 𝑍 𝐸−1/2 + constant ………… (4.46)

That is, our theory of alpha decay predicts that a plot of ln 𝑡1/2 against 𝐸−1/2 for a set of isotopes should

be a straight line. The remarkable success of this prediction can be seen in figure 4.8, which shows that

the data for the eight isotopes of polonium mentioned above. The line shown is the least-square fit to

the data. Its slope is 328, in good agreement with the value predicted b equation (4.46):

slope= 3.95 𝑍 = 3.95 × 82 ≈ 324 ,where Z=82 is that of daughter nuclei of polonium after α

decay.

The line’s intercept with the vertical axis (not shown in the figure 4.8) is at −125 while our predicted

value from equation (4.46) is

“constant”= −2.97 (𝑍𝑅)1

2 + 𝑐 = −2.97 × (82 × 7)1

2 − 50 ≈ 121

where we have taken 𝑅 = 7 fm.

Given the rough approximation that went into this value, this agreement is entirely satisfactory. The

agreement between theory and experiment is even better if we allow for the small variations of R in the

“constant” of equation 4.46 (John R. Taylor, 2004).

If we use the result (4.45) to compute half-lives of various different elements, we must recognize that

the three variables E, Z and R can all vary. However, neither Z nor R varies very much. For example, 81 ≤

𝑍 ≤ 90 for all residual nuclei in the natural radioactive series. Thus, the single most important effect is

still the energy dependence of the term 3.95 𝑍 𝐸−1/2 in log𝑒 𝑡1/2 . Small difference in E correspond to

appreciable difference in this term and hence enormous difference in 𝑡1/2. In particular , as E ranges

from 4 to 9 MeV, this term is the main reason that 𝑡1/2 drops by some 24 orders of magnitude as noted

in the beginning of this section 4.4. ( In this way theory agrees with the experimental observed ideas.)

The relationships between the logarithm of the half-life and the α-disintegration energy of the elements

with even values of Z are shown in figure (4.9). These are analogous curves fore even-Z-odd-A nuclides,

and for odd-Z nuclides.

Figure 4.8 Plot of 𝒍𝒏 𝒕𝟏/𝟐 against 𝑬−𝟏/𝟐 for eight alpha emitting isotopes of polonium. (Half-lives, 𝒕𝟏/𝟐 in

seconds and energy release, K, in MeV.) The number beside each point is the mass number A of the isotope. The line is the least-squares fit to the data and the axis on the right shown 𝒕𝟏/𝟐 (John R. Taylor, 2004).

4.5 Theory of Beta Decay and discovery of the Neutrino

Beta decay is the most interesting radioactive decay process. There was a serious puzzle over the

number of particles ejected in β decay in the past. The solution of this puzzle lead to the discovery of

new particle called neutrino.

According to the Pauli’s principle, nuclei with atomic number close to neutron number (i.e. 𝑍 ≈ 𝑁)

tend to have lower rest energy than their isobars with Z much different from N. However in heavier

nuclei, the Coulomb repulsion of the proton implies

that the minimum occurs with Z appreciably less

than N. Figure 4.51 shows the masses of the 𝐴 =

231 isobars, with the minimum at 𝑃𝑎91231 for which

𝑍 = 91 and 𝑁 = 140. Any nucleus with too many

neutrons, such as 𝑇ℎ90231 or with too many protons,

such as 𝑈92231 , has a rest mass energy larger than

the minimum energy at 𝑃𝑎91231 . Higher energy

means instability, so the nucleus say 𝑇ℎ90231

somehow transfer itself into the lower energy

𝑃𝑎91231 ; and indeed, experiment shows that there is

such a mechanism. The excess energy in 𝑇ℎ90231

creates and ejects an electron; and one neutron is

simultaneously converted into a proton. The

resulting transformation is

𝑇ℎ14190231 → 𝑃𝑎14091

231 + 𝑒−

is the process that Rutherford called β decay. Two such β decay are indicated on the figure 4.51, where

they are labeled 𝛽− to emphasize that they produce negative electrons.

A nucleus such as 𝑁𝑝93231 with too many protons, is also unstable. In fact, for most such nuclei there are

two possible modes of decay. The excess energy in 𝑁𝑝93231 can create and eject a positron (the positively

charged version of the electron) changing a proton into a neutron. The resulting transformation,

𝑁𝑝13893231 → 𝑈13992

231 + 𝑒+

is called 𝛽+ decay. Alternately, the 𝑁𝑝93231 nucleus can convert a proton into a neutron by capturing one

of the atomic electrons

Figure 4.51: The rest mass of the isobars with A= 231 have their minimum at protactinium with Z=91 and N=140. The possible decays of the other species are indicated as 𝛽− , 𝛽+, and EC.

𝑁𝑝13893231 + 𝑒− → 𝑈13992

231

This is a process called electron capture. Figure 4.51 indicates both of these processes for 𝑁𝑝93231 .

Now we will discuss the origin of the new particle neutrino. Let us consider the case of 𝛽− decay, for

which the observed process can be written as

(𝑍, 𝑁) → (𝑍 + 1, 𝑁 − 1) + 𝑒− …… (4.47)

If the parent nucleus (𝑍, 𝑁) is at rest and if the process really involves only two final bodies indicated in

equation (4.47), then the energies of those two bodies are uniquely determined by conservation of

energy and momentum. With the parent at rest, the total kinetic energy of the final particles is

𝐾 = ∆𝑀𝑐2 = [𝑚𝑛𝑢(𝑍, 𝑁) − {𝑚𝑛𝑢(𝑍 + 1, 𝑁 − 1) + 𝑚𝑒}]𝑐2

= [𝑚𝑛𝑢(𝑍, 𝑁) + 𝑍𝑚𝑒 − {𝑚𝑛𝑢(𝑍 + 1, 𝑁 − 1) + 𝑚𝑒 + 𝑍𝑚𝑒}]

= {𝑚𝑎𝑡𝑜𝑚(𝑍, 𝑁) − 𝑚𝑎𝑡𝑜𝑚(𝑍 + 1, 𝑁 − 1)}𝑐2 …… (4.48)

Since the electron is so much lighter than the recoiling nucleus, the nucleus takes very little kinetic

energy. So we can approximate all the kinetic energy must go to the electron given by equation (4.48).

However, experiment shows that our prediction i.e. all electrons should be ejected with the same

energy is completely wrong. The energies range continuously from 𝐾 = 0 up to the predicted value of

1.16 𝑀𝑒𝑉 (or very close to it). While a few electrons have nearly the predicted energy, the great

majority have much less. This apparent failure of conservation of energy was beginning to be recognized

in the late 1920s, and efforts were made to find the missing energy.

In the early 1930s, Pauli suggested that there might be another particle emitted in β decay, which

interacted only very weakly with matter and hence could carry away the missing energy undetected.

With two ejected particle (in addition to the residual nucleus), the released energy could be shared

between the two particles in many different ways. If the electron is at rest, then the another particle

took all the available energy or vice versa. Thus the electron’s kinetic energy could range anywhere from

zero to some 𝐾𝑚𝑎𝑥 , as observed. This gives the reason why the β-ray spectrum is continuous while other

spectrum like optical, X-ray, 𝛾- ray spectrum are the line spectra, which are not continuous.

From the observed facts about β decay, we can immediately infer the main properties of Pauli’s

proposed particle− its charge, mass and interaction with matter. First, its charge must be zero because

the equation(4.47) already conserves charge. Since some of the electrons have kinetic energy close to

the predicted value(i.e. 1.16 MeV) such that no additional particle was emitted during β decay. This

infers that new particle have mass very close to zero. Thus new particle has zero charge and very small

mass and Fermi named it the neutrino (little neutral one), denoted by ν.

The great difficulty of detecting the neutrino implies that it interacts very weakly with all matter.

Like electron, neutrinos are completely immune to the strong nuclear forces; in addition, they are

immune to all electromagnetic forces. It eventually became clear that there are two closely related

particles, the neutrino ν and the antineutrino �̅� , and it is actually that is emitted in 𝛽− decay:

(𝑍, 𝑁) → (𝑍 + 1, 𝑁 − 1) + 𝑒− + �̅� ………… (4.49)

On the other hand, it is antineutrino that emerges in 𝛽+ decay,

(𝑍, 𝑁) → (𝑍 + 1, 𝑁 − 1) + 𝑒+ + 𝜈 …………. (4.50).