B.Sc. Physics · 2020. 12. 28. · CHAPTER # 1: LIGHT AND QUANTUM PHYSICS CHAPTER # 2: WAVE NATURE OF MATTER CHAPTER # 3: NUCLEAR PHYSICS CHAPTER # 4: ENERGY FROM NUCLEUS B.Sc. Physics

B.Sc. Physics

Complete Notes of

MODERN PHYSICS

CHAPTER # 1: LIGHT AND QUANTUM PHYSICS

CHAPTER # 2: WAVE NATURE OF MATTER

CHAPTER # 3: NUCLEAR PHYSICS

CHAPTER # 4: ENERGY FROM NUCLEUS

B.Sc. Physics MODERN PHYSICS in Past Papers of University of

Sargodha

B.Sc. Physics, Paper C, Annual 2017






B. Sc. Physics (H.R.K) Chapter 49: Light and Quantum Physics (Edition 2015-16)

1 Muhammad Ali Malik, Whatsapp # +923016775811, [email protected], www.facebook.com/HomeOfPhysics

LIGHT AND QUANTUM PHYSICS \

The evidences in support of wave behavior of radiation are overwhelming. The radiations (including not only light but

all of electromagnetic radiations) show the phenomenon of reflection, refraction, interference, diffraction and

polarization, which can be understood by considering radiation as a wave.

In this chapter, we now move off in a new direction and consider experiments that can be understood only by making

quite a different assumption about electromagnetic radiation, namely, that it behave like a stream of particles.

49.1 Thermal Radiations

The radiations emitted by a body due to its temperature are called thermal radiations.

All bodies not only emit the thermal radiations, but also absorb these radiations from surroundings. If

the rate of emission of radiation is equal to the rate of absorption for a body, then the body is said to be in

thermal equilibrium.

The radiations emitted by a hot body depend not only on the temperature but also on the material of

which the body is made, the shape and the nature of the surface.

49.1.1 Cavity Radiator or the Black Body

A black body or cavity radiator is that which absorbs

approximately all radiations falling on it. So a black body has maximum

rate of emission and absorption of radiations.

A perfect black body does not exist. However a small hole in a

cavity whose inner wall are lamped black is the nearest approach to a

perfect black body.

For an ideal radiator (black body), the spectrum of the emitted

thermal radiation depends only on the temperature of radiating body and

not on the material, nature of the surface, size or shape of the body.

49.1.2 Radiant Intensity

Energy emitted per unit area per unit time over all the wavelengths, is called radiant intensity. It is

denoted by . Or

Power radiated per unit area over all the wavelengths is called radiant intensity.

49.1.3 Stephen-Boltzmann Law

The radiant intensity is directly proportional to the forth power of absolute temperature.

Mathematically,

where is a universal constant, called Stephen-Boltzmann constant. Its value is

.



49.1.4 Spectral Radiancy

Spectral radiancy of the black body is defined as the radiant intensity per unit wavelength at given

temperature. it is denoted by . Mathematically, it is described as:

This shows that the product of gives energy emitted per unit area per unit time over all the

wavelength lies in the range from to .

The energy emitted per unit area per unit time over all the wavelengths at a particular temperature is

obtained by integrating the equation (1), i.e.,

∫

49.1.5 The Wien Displacement Law

Statement. The wavelength for which the spectral radiancy becomes is inversely

proportional to the absolute temperature of the black body.

If is the wavelength corresponding to maximum spectral radiancy

for temperature , then the Wien displacement is described mathematically as:

The constant has the value of

Problem 1: What are the surface temperatures, radiant intensity and total radiated power of stars.

Sirius

Sun

Betelgeuse

Solution:

Surface Temperatures According to Wein’s Displacement Law:

Sirius

Sun



Betelgeuse

Radiant Intensity

By Stephan Boltzmann’s law:

Sirius

Sun

Betelgeuse

Total Radiated Power (L)

The total radiated power can be find out by expression:

Sun

Betelgeuse

The total radiated power of Betelgeuse is about 38000 times larger

49.1.6 Failure of Classical Physics to Solve the Problem of Energy Distribution along the Curve of Black

Body Radiation

The derivation of theoretical formula for distribution of spectral radiancy among various wavelengths has

remained an unsolved problem over a long period of time. On the basis classical physics, following two

formulas were derived to solve the problem of the energy-distribution along the curve of the black body

radiation.

Rayleigh-Jeans formula

Wien’s formula

49.1.7 Rayleigh-Jeans Formula

Rayleigh-Jeans derived a theoretical formula for the distribution of spectral radiancy among various

wavelengths on the basis of classical physics, which is given by:

Where k is the Boltzmann constant.

This formula was excellent for longer wavelengths but not good for shorter wavelengths.



48.1.8 Wien’s Formula

On the basis of analogy between the spectral radiancy curve and Maxwell speed distribution curve,

the Wien’s formula about spectral radiancy is described as

Where a and b are constant. Wien’s formula was in good agreement with the experimental curve for shorter

wavelength but not good for longer wavelengths.

Conclusion

Rayleigh-Jeans and Wien’s formulae were failed to solve the problem of distribution of spectral

radiancy among various wavelengths in cavity radiations because these formulae were based upon the

classical theory.

49.1.9 Success of Quantum Physics to Solve the Problem of Energy Distribution along the Curve of

Black Body Radiation

49.1.10 Max-Plank’s Quantum theory of Radiation

In 1900, Max-Planks gave the new concept about the nature of radiation, called Quantum theory of

radiation in which Planks assumed the discrete nature of radiation. He assumed the atoms of the cavity emit

and absorb radiation in the form of packet of energy, called quanta. The energy of each quanta is directly

proportional to the frequency:

Where h is the plank’s constant having the numerical value of .

In addition to this concept Max-Planks made the following assumption to derive his radiation law:

The atoms of the cavity behave like tiny harmonic oscillators.

The oscillators radiate and absorb energy only in the form of packets or bundles of electromagnetic

waves.

An oscillator can emit or absorb any amount of energy which is the integral multiple of .

Mathematically,

Where n is an integer.

Using these assumptions, Plank’s derived his radiation law given by:

Where and are the constant whose values can be chosen from the best fit of experimental curve. Within

two months, Plank’s succeeded to reform his law as:



49.1.11 Derivation of Rayleigh-Jeans Formula from Max-Plank’s Radiation Law

Plank’s radiation law approaches to Rayleigh-Jeans law at very long wavelengths. The Plank’s

radiation law is given by:

-------------- (1)

Putting

For a very long wavelengths i.e.,

And

Putting these values in equation (1), we get:

Back substituting the value of , we have

This is the Rayleigh-Jeans formula.

49.1.12 Derivation of Wien’s Formula from Max-Plank’s Law

Plank’s radiation law approaches to Wien’s formula at very short wavelength. The Plank’s radiation

law is given by:

Putting and

For very short wavelengths, i.e.,

So

Putting these values in equation (2), we have:



This is the Wien’s formula for spectral radiancy.

ءںیمہ اس اکر ریخ وک رتہب انبےن ےکےئل آپ یک دمد یک رضورت ےہ۔ ور ونسٹ اینپ یتمیق آرا رکںی ای سیف کب اڈیسی رپ جیسیم رکںی۔ا ای لیم

ہ رہش ٹنمن ڈرگی اکجل ون دمحم یلع کلم، وگر

[email protected], www.facebook.com/HomeOfPhysics

49.2 Photo Electric Effect

The interaction between radiation and atoms of the

cavity lead to the idea of quantization of energy. It means that

energy can be emitted and radiated in the form of packets.

Photo electric effect is another example of interaction between

radiation and matter.

Definition: When a light of suitable frequency falls on a metal

surface, the electrons are emitted out. These electrons are called

photo electrons and this phenomenon is called photo electric

effect.

49.2.1 Experimental Set-up

The apparatus used to study the photo electric effect is shown in the figure. The light of a suitable

frequency falls on the metal surface, which is connected to the negative terminal of variable voltage source. If

the frequency is high enough, the electrons are emitted out from metal plate and accelerate towards anode.

These electrons are called photo electrons and current flows through circuit due to photo electrons is called

photo electric current. This phenomenon is called photo electric effect.

49.2.2 Maximum K.E of Photo Electrons

The maximum K.E of photo electrons can be measured by reversing the polarity of the battery. Now

the photo electric current will be reduced. The photo electric current does not drop to zero immediately

because the photo electrons emit from metal plate with different speeds. Some will reach the cathode even

though the potential difference opposes their motion. However if we make the reversed potential difference

large enough (called stopping potential) at which the photo electric current drops to zero. This potential

difference multiplied by the electronic charge gives the maximum kinetic energy of photo electrons.

Mathematically, the maximum kinetic energy of photo electrons is described as:

where is the stopping potential and is the charge of an electron.

mailto:[email protected]://www.facebook.com/HomeOfPhysics



49.2.2 Experimental Results

Photo electrons are emitted out from the given metal surface when the frequency of incident light is

equal to or greater than a critical value , called threshold frequency,

whatever the intensity of light may be.

Photo electric emission will not take place from a given metal surface if the

frequency of the incident light is less than the

threshold frequency whatever the intensity

of light may be.

Threshold frequency depends upon

the nature of the metal surface.

The energy of photo electrons

depends upon the frequency of incident

light and independent of the intensity of light.

The number of photo electrons emitted per second is directly

proportional to the intensity of light provided that the frequency of light

is equal to or greater than the threshold frequency.

49.2.4 Threshold Frequency

The frequency of the incident light required to remove least tightly bound electron from the metal

surface, is called threshold frequency.

49.2.5 Photo Electric Effect on the Basis of Classical Wave Theory

According to classical wave theory, the light consists of electromagnetic waves and their function is

to transfer energy from one place to another.

When light falls on metal surface, it transfer energy to the electrons continuously. When an electron

acquires sufficient energy, it escapes out the metal surface.

This theory successfully explains the emission electrons apparently, but this theory can’t explain the

three major features of photo electric effect.

The Intensity Problem

The Frequency Problem

Time Delay Problem



49.2.5.1 The Intensity Problem

According to classical wave theory, the light consists of oscillating electric and magnetic vector,

which increases in amplitude as the intensity of light beam is increased. Since the force applied to the

electrons is ‘ ’, therefore the kinetic energy of the electrons should also increased with the intensity of

light. However, the experimental results suggest that the K.E of electrons is independent of intensity of light.

49.2.5.2 The Frequency Problem

According to classical wave theory, the photo electric effect should occur for any frequency of light

provided that the incident light is intense enough to supply the energy needed to eject the photo electrons.

However, the experimental results show that there exists a critical frequency for each material. If the

frequency of the incident light is less than the photo electric effect does not occur, mo matter how much the

intensity of light is.

49.2.5.3 Time Delay Problem

The classical wave theory predicts that there must be a time interval between the incidence of light on

the metal surface and the emission of photo electrons. During this time, the electron should be absorbing

energy from the beam until it has accumulated enough energy to escape the metal surface. However the

experimental results show that there is no detectable time interval between the incidence of light and emission

of photo electrons provided that frequency of light is equal to or greater than the threshold frequency.

Conclusion

These three major features could not be explained on the basis of wave theory of light. However

Quantum light theory has successfully explained the photo electric effect which was proposed by the Einstein

in 1905.

49.2.6 Photons

In 1905, Einstein proposed quantum theory to explain the photo electric effect, according to which the

light consist of bundles or packets of energy, called photons. The energy E of a single photon is

where is the frequency of light and is the planks constant .

According to quantum theory, on photon of energy is absorbed by a single electron. If this energy is

greater than or equal to a specific amount of energy (called work function), then the electrons will be ejected,

otherwise not.

49.2.7 Work Function

The minimum amount of energy required to eject the electrons out of metal surface, is called work

function , which can be described as:

Here is the threshold or cur off frequency and is known as cut off wavelength.



49.2.8 Threshold Frequency or Cut Off Frequency

It is the minimum frequency of incident light at which the photoelectric effect takes place.

49.2.9 Einstein’s Photo Theory or Quantum Theory of Photoelectric Effect

When a incident light of suitable energy is exposed to the metal surface, then a part of this energy is

utilized in ejecting electron from metal surface and the excess energy ( ) becomes the kinetic energy of

photoelectrons. If the electrons does not lose any energy by the internal collision as it escapes from the metal,

then its energy will be maximum which can be described by the

formula:

As and

------------ (1)

(

)

From equation (1), it is clear that the stopping potential and

the frequency of light has a linear relationship and graph between these

quantities is a straight line.

Conclusion

In 1916, Millikan showed that Einstein’s equation agreed with experiments in every detail. Hence the

photon theory explains that, if the frequency of light is less than the threshold frequency , then no

photoelectrons will be emitted, how much the intensity of light may be.

If the frequency of light is equal to greater than cut off value , then weakest possible beam of light

can produce photo electric effect. Because the energy of one photon depends upon the frequency of light i.e.,

and not on the intensity of light.

Sample problem 6. Find the work function of sodium for which the threshold frequency is

.

Solution: Threshold frequency is

Work Function

As

Problem 29. An atom absorbs a photon having a wavelength 375 nm and immediately emits another

photon having wavelength 580 nm. What was the energy absorbed in this process?

Solution

Wavelength of incident photon

Wavelength of emitted photon



Energy Absorbed

(

) (

)

(

)

ءانبےن ےکےئل آپ یک دمد یک رضورت ےہ۔ںیمہ اس اکر ریخ وک رتہب ور ونسٹ اینپ یتمیق آرا رکںی ای سیف کب اڈیسی رپ جیسیم رکںی۔ا ای لیم


[email protected],

www.facebook.com/HomeOfPhysics

Problem 35. (a) The energy needed to remove an electron from metallic sodium is 2.28 eV. Does the

sodium show photoelectric effect for red light with ? (b) What is the cur off wavelength for

photoelectric emission from sodium and to what color this wavelength corresponds?

Solution:

Wavelength of incident Photon

Work function

Energy of incident photon

As , so no photoelectric effect will take place.

(b) Cut off Wavelength

Work function

The cut off wavelength corresponds to green color.

Problem 36: Find maximum kinetic energy in eV of photoelectrons if the work function of the material

is 2.33 eV and frequency of radiation is

Solution:

Work function

Frequency of incident photon

Maximum kinetic energy of Photoelectrons

The maximum kinetic energy of electrons can be find out by expression:




Now

Therefore,

Problem 38: light of wavelength 200 nm falls on an Aluminum surface. In Aluminum 4.2 eV is required

to remove an electron. What is:

a) Kinetic energy of fastest electron

b) Kinetic energy of slowest electrons

c) Find stopping potential

Solution:

(a)

(b)

The minimum K.E of photoelectrons is zero as it consumes all of its K.E in colliding with the atoms

of the metal.

(c)

As

Also

Therefore:

Problem 39: If the work function for the metal is . (a) What would be the stopping potential for

light having a wavelength of . (b) what would be the maximum speed of the emitted

photoelectrons at the metal surface?

Solution: (a)

(b)

As



√

√

ءںیمہ اس اکر ریخ وک رتہب انبےن ےکےئل آپ یک دمد یک رضورت ےہ۔ ور ونسٹ اینپ یتمیق آرا رکںی ای سیف کب اڈیسی رپ جیسیم رکںی۔ا ای لیم



49.3 The Compton Effect

In 1923, Compton performed an experiment and he observed that the wavelength of X-rays changes

after scattering from a graphite target. Compton explained his experimental results postulating that the

incident X-rays beam consists of photons, and these photons experienced Billiard-Ball like collision with the

free electrons in the scattering target.

The experimental set-up for observing the Compton effect is shown in the figure below.

Compton effect can be explained by considering the elastic collision between X-ray photon and

electrons.

Consider an incident photon having energy and momentum

and

respectively. The

photon is scattered by a stationary electron along an angle with its regional direction, as shown in the figure.

The energy and momentum of the scattered photon are

and

respectively. The

rest mass energy of electron is , which is recoiled making an angle θ with the original direction of

incident photon.

During the elastic collision, both the energy and momentum remains conserved. So the energy

equation in this case is:




Dividing equation by ‘ ’, we get:

Squaring both sides of the equation,

(

)

(

)

--------- (1)

The conservation of momentum along the original direction of incident photon (along x-axis) is

Squaring both sides, we get:

--------- (2)

The conservation of momentum along the direction perpendicular to the original direction of photon

(y-axis) is:

Squaring both sides, we get:

--------- (3)

Adding equation (2) and (3)

--------- (4)

Subtracting equation (4) from equation (1), we get:

(

)



√

(

)

(

)

(

)

(

)

(

)

This is the expression of Compton shift, which shows consistency with experimental results. As collision is a

particle phenomenon, hence the particle nature of radiations was confirmed.

Sample Problem 8: X-rays with are scattered from a carbon target. The scattered

radiation is viewed at to the incident beam. (a) What is Compton shift ? What kinetic energy is

imparted to the recoiling electron?

Solution: (a)

(b) K.E imparted to electron

(

) (

)

Problem 51: A particular X-ray photon has a wavelength . Calculate the photon’s (a) energy

(b) frequency (c) the momentum

Solution: (a)



(b)

(c)

Problem 55: Photon of wavelength are incident on free electrons. Find the wavelength of

photon that is scattered at from the incident direction? Do the same if the if the scattering angle is

.

Solution:

a) for

b) for

(a)

(b)

This result shows that as the angle of scattering increases, the wavelength of scattered photon also

increases.

ءےہ۔ںیمہ اس اکر ریخ وک رتہب انبےن ےکےئل آپ یک دمد یک رضورت ور ونسٹ اینپ یتمیق آرا رکںی ای سیف کب اڈیسی رپ جیسیم رکںی۔ا ای لیم




B. Sc. Physics (H.R.K) Chapter 50: The Wave Nature of Matter (Edition 2015-16)

16 Muhammad Ali Malik, Whatsapp # +923016775811, [email protected] www.facebook.com/HomeOfPhysics

In this chapter, we present experimental evidence supporting the claim that matter, long regarded as made up

of particles, has an equally convincing wave aspect. Heisenberg’s uncertainty principle will show us the limit

to which we can extend the concept of “particle” to into quantum mechanics.

We then introduce the Schrodinger’s equation, the fundamental equation to quantum mechanics, which deals

with the wave behavior of the particles.

THE WAVE NATURE OF MATTER

50.1 Dual Nature of Radiation

The electromagnetic radiations like light, X-rays etc., can produce the phenomenon of interference,

diffraction and polarization due to their wave nature. But under certain circumstances they can produce

photoelectric effect and Compton Effect which is the evidence of their particle nature. It means that

electromagnetic radiations have dual nature; wave as well as particle nature.

50.2 Dual Nature of Matter

In a similar way the particles like electrons, neutrons and protons etc. must have dual nature. If the

beam of electrons accelerated through a known potential difference ‘V’, is made to fall on a double slit and

after passing through the double slit, they are allowed to strike on a fluorescent screen. It has been observed

that pattern obtained on the screen is similar to the pattern of interference of light.

50.2.1 Double Slit Experiment

In Double slit experiment, a filament produces a spray of electrons which are accelerated through a

potential difference of about 50 kV. After passing through double slit, the electrons produce a visible

interference pattern on fluorescent screen, which can be photographed.

50.3 De Broglie Hypothesis

A wave is associated with every moving particle. Such a wave is called matter wave whose

wavelength can be find out by expression:

Where h is Plank’s constant, m is mass and v is velocity of moving object.

As the value of Plank’s constant is very small and is of the order of , the wavelength

associated with ordinary object (e.g., A moving tennis ball) is so small and is difficult to observe. But for the

small objects like electrons and neutron etc., the wave behavior of particles is dominant.

رکںی ای سیف کب اڈیسی رپ جیسیم رکںی۔ ور ونسٹ ای لیم ءا ںیمہ اس اکر ریخ وک رتہب انبےن ےکےئل آپ یک دمد یک رضورت ےہ۔اینپ یتمیق آرا






50.4 The Davison-Germer Experiment

The De Broglie hypothesis was confirmed by Davison and Germer. The schematic diagram of

Davison and Germer experimental setup is shown in the figure.

The electrons from a heated filament F are accelerated

by an adjustable potential difference V. The beam of electrons is

allowed to fall on a nickel crystal. The diffracted beam of

electrons is detected by a movable detector at different values of

angles.

It is observed that there is a strong diffracted beam

obtained for and . This situation is

similar to the diffraction of light produced by the diffraction

grating.

In nickel crystal, the atoms are arranged in definite order; hence the crystal surface acts like a

diffraction grating. The first order maxima is obtained at an angle . The wavelength associated with

electrons can be determined by using the equation:

Here for the first order diffraction peak

Inter-planner spacing for nickel

Angle of diffraction

The wavelength of the matter wave, can be find out by using de Broglie hypothesis:

The kinetic energy of the electron is:

√

√

By putting the value of constants and , we have:

So, the value of de Broglie wavelength is in good agreement with the experimentally observed

wavelength associated with electrons.

50.5 G. P. Thomson Experiment

In 1927, G. P. Thomson (son of J. J. Thomson) performed an experiment and confirmed the de

Broglie equation of matter waves.

He obtained a fine beam of electrons accelerated through a potential difference of 15 kV and made it

to fall on a target which was not a single crystal, but it was made up of a large number of tiny, randomly

oriented crystallites (powdered aluminum).



A photographic plate was placed parallel to the target on which the diffraction pattern was obtained. It

was observed that the diffraction pattern of electrons was very similar to the diffraction of X-rays. As the

diffraction is a wave property, so the wave nature of electrons was confirmed experimentally.

The wavelength of the matter waves associated with the electrons can be determined by using the

Bragg’s equation:

------------ (1)

Where d is the inter-planner spacing, is the glancing angle and m is the order of diffraction.

If V is the potential difference through which the electrons are accelerated, then the kinetic energy of

electrons is:

√

According to de Broglie hypothesis

√

√

G. P. Thomson observed that the de Broglie wavelength associated with moving electrons was good in

agreement with experimentally observed value, that was find out by using Bragg’s law.

The atomic structure of solids are studied the diffraction beam of electrons. The electrons are less

penetrating than X-rays, so the electrons are used to study the surface morphology of solids.

Sample problem 2: Calculate the de Broglie wavelength of (a) a virus particle of mass

moving at the speed of , and (b) an electron whose kinetic energy is .

Solution.

Mass of virus

Velocity

De Broglie Wavelength

( )

(b) Kinetic Energy of electron

√


( )



Problem 1. A bullet of mass 41 g travels at 960 m/s. what wavelength can we associate of the bullet not

reveal itself though diffraction effects?

Solution

Mass of bullet

Velocity of bullet


( )

For diffraction condition:

Wavelength Size of diffracting aperture

As the wavelength of associated with bullet is very small, so diffraction effects of bullets can’t be revealed.

Problem 3: Calculate the wavelength of a 1 keV (a) electron (b) photon (c) neutron.

Solution: Kinetic Energy


We know

( )

√ ( )

(a) For an electron

So

√ ( )

√

(b) For photon

(c) For neutron

So

√ ( )

√

Problem 4: The wavelength of yellow spectral emission line of sodium is . At what K.E would

an electron have same de Broglie wavelength?

Solution: Given wavelength

(a) For an electron

As we know:

√ ( )

( )

( )



Problem 5: If the de Broglie wavelength of proton is 0.113 pm, (a) What is the speed of the proton and

(b) what electric potential would the proton have to be accelerated from rest to acquire this speed?

Solution

(a) Wavelength associated with proton

Velocity of proton

Mass of proton

As

(b) Electric Potential

For present case:

Problem 15: A neutron crystal spectrometer utilizes crystal planes of spacing in a

Beryllium crystal. What must be the Bragg’s angle so that only neutrons of energy are

reflected? Consider only fist order spectrum.

Solution:

Given Inter-planner spacing

Bragg’s Angle

Order of diffraction

Kinetic Energy

According to Bragg’s Law:

√

√

( )

Sample Problem 4. Thermal neutrons (293 K) coming out of a nuclear reactor fall on a crystal. The

spacing between the Bragg’s planes is . Find the most probable de Broglie wavelength in

the beam of these neutrons. (b) Find the Bragg’s scattering angle for the first order Bragg’s Diffraction.

Solution: (a) Boltzmann’s constant

Temperature

Order of diffraction

Inter-planner Spacing

√

√



Now according to Bragg’s Law:

(

)

50.6 Waves and Particles

The evidence of the matter is wave like is very strong. On the other hand, the evidence that a matter is

particle like is equally as strong. In these situations, the description of wave as well as the particle nature of

matter has remained a challenge for the scientists.

One property that we like for the particles (even particles with the wave like nature) to have the ability

to be localized. For example, an electron in an atom of 0.1 nm diameter is localized in certain region of space.

On the other hand, a wave cannot be localized in space and time like a particle.

The amplitude of a matter wave carries information about the location of the particle. The wave has

the large amplitude where the particle is likely to be found, and it has the small amplitude where the particle is

unlikely to be found. If the wave has the constant amplitude throughout a given region of space, the particle is

equally likely to be found anywhere in that region. If the amplitude of the matter wave is zero in a specific

region, then the particle never found there.

50.7 Localizing Wave in Space

A wave packet is associated with a moving particle. Many waves adds up to make a wave packet of

length and adds to zero everywhere else. Thus for the present case, the particle is localized in space. The

particle is likely to be found in the region of size and unlikely to found outside that region.

This wave packet no longer contains a single wave number but rather a spread of wave numbers

centered about . Let is the rough measure of the spread of wave numbers. The product of and is

proves to be of the order of unity:

This expression tells that the smaller the value of , the larger must be the range of wave numbers .

Conversely, the narrower the spread in , the less localized the particle will be.

50.8 Localizing Wave in Time

A particle is localized in space as well as in time. So, the space variable x must be replaced by time

variable t (as the wavelength by the time period ). And the wave number must be replaced by the

angular frequency .

Similarly, the spread of wave number must be replaced by and the displacement by the

interval of time . So we have

It means that the product is of the order of unity.



50.9 Derivation of Heisenberg’s Uncertainty Relationship by Localization of Wave in Space

Consider a particle of mass m is moving along x-axis with velocity . By Localization of Wave in space, we

know:

( )

Where is the region in which the particle is likely to be found and is the rough measure of the spread of

wave numbers.

According to the de Broglie Hypothesis, the wavelength of the matter wave associated with moving

particle is described as:

Where h is the Plank’s constant and is the linear momentum of moving particle.

The angular wave number associated with this particle is described as:

(

)

The spread of wavenumber will be:

(

)

( )

Putting value of in equation (1), we get:

(

)

This is the expression of Heisenberg Uncertainty Relationship for the particle moving along x-axis. If

the motion of the particle depends upon the three coordinates x, y, z, the generalizing above relation, we have:

According to these relationships:

It is not possible to determine both the position and

the momentum of a particle with ultimate precision.

The width of the wave packet indicates the probable location of the particle, and is the range

in momentum.

It means that due to the wave nature, the exact position x of a particle cannot be determine, but it will

be in the range . Similarly, the true or exact momentum of the particle cannot be determined, but it will

be in the range of .



50.10 Derivation of Uncertainty Principle by Single-Slit Diffraction of Electron Experiment

Consider the experiment of diffraction of electrons by single slit. Let a beam of electrons moving with

speed passes through a single slit of width . The diffraction pattern is obtained on the screen B as shown

in the figure.

Due to the wave nature, the electron

beam bends on the either side of the central

point producing the diffraction pattern. Let

is the uncertainty in the component of

velocity along y- axis for the first minimum,

then we have

Consider the angle is very small, then

:

( )

In case of location of the first minimum of diffraction, we have:

Here m=1 and if is very small, then

( )

Comparing equation (1) and (2), we have:

According to the de Broglie hypothesis:

(

)

It means that, if the position of the particle is made more and more precise, the uncertainty in the

momentum of the particle increases and vice versa. On other words,

It is not possible to determine both the position and the momentum

of a particle with unlimited precision.

رکںی ای سیف کب اڈیسی رپ جیسیم رکںی۔ ور ونسٹ ای لیم ءا ںیمہ اس اکر ریخ وک رتہب انبےن ےکےئل آپ یک دمد یک رضورت ےہ۔اینپ یتمیق آرا






50.11 The Energy-Time Uncertainty Relationship

The wave nature of a particle can be represented by the wave packet having angular frequency . So

the spread of angular frequency and the time interval are related as:

( )

According to Einstein’s photon equation is

where is frequency of photon.

The uncertainty in the frequency of the matter waves will be:

As

, put in equation (1):

(

)

This is another form of Uncertainty principle. It may be stated as:

It is not possible to determine both the energy and time co-ordinates with ultimate precision.

Problem 23: A nucleus in an excited state will return to its ground state, emitting a gamma ray in the

process. If its mean life time is 8.7 ps in a particular excited state of energy 13.2 MeV, find the

uncertainty in the energy of the corresponding emitted gamma-ray photon.

Solution:

By Heisenberg Uncertainty Principle:

Problem 25: A microscope using photons is employed to locate an electron in an atom within a distance

of 12 pm. What is the minimum uncertainty in the momentum of the electron located in this way?

Solution:

Now as



Problem 26. Imagine playing a base ball in a universe where the planks constant is 0.60 J-s. what

would be the uncertainty in the position of a 0.5 kg base ball moving at 20 m/s with an uncertainty in

velocity is 1.2 m/s? Why would be hard to catch such a ball?

Solution:

Planks constant

Mass of the ball

Uncertainty in velocity

Now

The base ball will be in a region anywhere in a distance of 0.159 m=15.9 cm. so it will be hard to catch such a

ball.

Sample Problem 7: The rest mass energy of a massive particle is . The uncertainty of

measurement is only . What is the mean life interval between production and decay of this

particle?

Solution:

Sample Problem 6: (a) A free 10 eV electron moves in the x-direction with a speed of .

Assume that the speed can be measured to a precession of 1%. With what precision its position can be

simultaneously measured? (b) A golf ball has a mass of 45 g and a speed of 40 m/s, what you can

measure with a precision of 1%. What would be the uncertainty in its position?

Solution: Velocity of electron


Now

(a) Mass of golf ball

Velocity of electron


Now



50.12 The Wave Function

The wave nature of a particle can be represented by wave function , which is the function of space

and time co-ordinates. The behavior of the particle in terms of wave can be determined by knowing the wave

function for every point in the space and for every instant of time:

( )

Explanation

Consider a matter wave associated with a particle of mass travelling in the direction of increasing x

and on which no force acts, so called free particle. To describe the displacement associated with such a wave,

the American physicist Erwin Schrodinger introduced a quantity ( ) for such a free particle, called wave

function. The wave function for a free particle moving in the direction of increasing x is given by:

( ) ( )

Here is the amplitude of the wave, (

) is the wave number and ( ) is the angular

frequency. As this wave function contains the imaginary number ( √ ), so it a complex quantity.

Physical Interpretation of Wave Function

The physical interpretation of the wave function was given by the German Physicist Max Born. He

asserted that physical meaning should not be given to itself, but to the product of and its complex

conjugate . Specifically, Born postulated:

The product gives the probability that the particle

in question will be found between position and .

50.13 Schrodinger’s Equation

The Schrodinger equation is used to find out the expression of wave function of moving particle in a specific

direction. As it is described earlier, that a wave function is the function of both space and time variables. So

we can write the wave function for the particle moving along axis as:

( ) ( ) ( )

That is, the wave function of a particle can be described as the product of space dependent wave

function ( ) and time dependent wave function ( ). In the rest of the chapter we will focus our attention to

the space dependent portion of wave function.

Now the Schrodinger’s equation for a particle travelling in the direction is:

( )

( ) ( ) ( )

where is the total energy of the particle and ( ) is its potential energy.

50.14 Schrodinger’s Equation for a Free Particle

If the particle is a free particle, its potential energy ( ) is a constant which we can take to be zero

for all values of . The total energy E of the moving particle must be taken entirely kinetic. That is, in which

we must have

, in which is the momentum of the particle. With this assumption, the

Schrodinger’s equation becomes:



( )

( )

( )

( )

Put

--------------- (1), where is the wave number.

( )

( )

( )

( )

Putting

, we have:

( ) ( )

( ) ( )

The characteristic equation is described as:

√

The characteristic solution of this equation will be: ban

( ) ------------ (2)

As the particle is moving along positive x-direction, so the equation (2) will become:

( ) ------------ (3)

where A is the amplitude of the wave. The expression in equation (3) is wave function for the free particle

moving along direction.

Probability Density

For the free particle, the probability density P(x) is described as:

( ) [ ][ ]

Thus the probability density of a free particle is constant and is independent of or . Thus we conclude

that the particle can be find with equal probability, at any point along the x-directin from to .

Wave Number

From equation (1), we have

For the free particle, the total energy

, in which is the momentum of the particle.

( ⁄ )

Using de Broglie hypothesis, we have



50.15 Particle in a Well or One Dimensional Box

Let a particle of mass m is moving in a one-dimensional box of length . Consider that the particle

moves back and forth along x-axis between the perfectly hard and infinite high walls of the box, from

to , and no force acts on it during its motion. The particle suffers elastic collisions and its total energy

remains constant.

As there is no force acting on the particle, therefore

For convenience, we take the potential energy of the particle as zero, i.e., , inside the box. Since

the walls of the box are infinitely high, therefore the potential energy of the particle outside the box is infinite.

Since the particle cannot have an infinite amount of energy, so it cannot exist outside the box. Hence the wave

function of the particle ( ) is zero for and .

( )

( )

( )

( )

( )

( )

Put

--------------- (1)

( )

( )

Putting

, we have

( ) ( )

( ) ( )

For characteristic solution, we have:

√

The characteristic solution of this equation will be:

( )

( ) ( ) ( )

( ) ( ) ( ) --------- (2)

Let

Thus equation (2) will become:

( ) ---------- (3)

Where the constants A and B can be evaluated from the boundary conditions, which are:



(i) at

And (ii) at

Applying the 1st boundary condition, the equation (3):

Applying the 2nd

boundary condition:

, where n is an integer

----------- (4)

Thus equation (3) will become:

( )

----------- (5)

The solution of the wave function for the particle in a box, since we have not yet determined the

constant A. for this purpose, we make use of the normalization condition:

∫ | ( )|

For the present case:

∫| ( )|

∫ (

)

∫[ (

) ]

√

Thus the equation (4) will become:

( ) √

Energy of the particle in the Box

From equation (1) and (4), we have:

----------------- (6)

For ,

. Thus the equation (6) will become:

This shows that the particle in the box have discrete values. Hence, the energy of the particle in the box is

quantum box is quantized.



50.16 The Potential Step and Barrier Tunneling

When a particle is moving in a region of a constant potential suddenly comes across another region of

different constant potential, the common boundary of the two regions is

called potential step. In the figure, the height of the potential step is ,

say at . According to the classical physics, the particle coming from

region I, approach the potential barrier of potential step with energy,

and are slow down by the force

, so that the kinetic

energy is converted into potential energy. If the particles have sufficient

energy to overcome the barrier, then there will be total transmission. And

if , then the particles are stopped by the barrier and their motion will be reversed. In this case, there is

the total reflection of the beam.

Quantum mechanically, for this potential step, we have the potential function as:

( )

( )

For the present case, the Schrodinger wave equation will be:

( )

( ) ( ) ( )

( )

[ ( )] ( )

( )

[ ( )] ( ) ---------------- (1)

Case-1.

Region-1

Let the particles with total energy E moves from region-I to region-II along x-axis. In region-I,

( ) . Therefore, the equation (1) will become:

( )

( )

Put

--------------- (2)

where is the wave number.

( )

( )


( )

------------ (3)

The first and the second term of equation (3) represent the incident and reflected particles respectively.

Region-II

For region-II, ( ) . Therefore, the eqution (1) for the case of region-II will be:

( )

[ ] ( )

Put ( )

--------------- (4)




( )

( )


( )

------------ (5)

In equation (5), the first term represents the transmitted wave. And the second term represents a wave coming

from in the negative direction. Clearly, for , no particle can flow to region-I and D must be zero.

Therefore, the equation (5) becomes:

( )

------------ (6)

Case-2.

Region-I.

When is less than , then solution of Schrodinger wave equation for region-I is:

( )

( )

Put

------------ (7)


( )

( )


( )

------------ (8)

The first and the second terms corresponds to the incident and reflected beams respectively.

Region-II.

When is less than , then solution of Schrodinger wave equation for region-II is:

( )

[ ] ( )

As , therefore the Schrodinger wave equation will become:

( )

[ ] ( )

Put ( )

--------------- (9)


( )

( )


( )

------------ (10)

The first and the second terms corresponds to the incident and reflected beams respectively. The

equation (10) describes that there is always be a probability for a particle to move through a barrier.

ور ونسٹ ای لیم ءا ہںیمہ اس اکر ریخ وک رتہب انبےن ےکےئل آپ یک دمد یک رضورت ےہ۔اینپ یتمیق آرا رہش ٹنمن ڈرگی اکجل ون رکںی ای سیف کب اڈیسی رپ جیسیم رکںی۔دمحم یلع کلم، وگر



B. Sc. Physics (H.R.K) Chapter 54: Nuclear Physics (Edition 2015-16)


NUCLEAR PHYSICS

Deep within the atom lies its nucleus, occupying only of the volume of the atom but providing most of its mass as well as

the force that holds it together.

In order to understand the structure of nucleus, our task becomes easier by many similarities between the study of atoms and

the study of nuclei. Both systems are governed by the laws of quantum mechanics. Like atoms, nuclei have excited states that

can decay to ground state through the emission of photons (gamma rays).

In this chapter we study the structure of nuclei and their constituents. We consider some experimental techniques for studying

their properties, and we conclude with a description of the theoretical basis for understanding the structure of nucleus.

54.1 Rutherford Experiment for the Discovery of the Nucleus

In 1897 J. J. Thompson discovered electrons. He also suggested a model of atom. According to his

model, an atom consisted equal amount of positive and negative charge. The positive charge of the atom was

considered to be spread out through the entire volume of the atom and electrons were thought to be distributed

throughout this volume like seeds embedded in a watermelon. Thompson‟s model of atom could not account

for the deflection of the particles passing near the surface of

the atom and particularly by the backward deflection of an

particle.

In 1911, Ernest Rutherford performed an experiment to

investigate the model of the atom. The apparatus consisted of a

source of particles, thin gold foil and a moveable detector.

A fine beam of high energy particles was made to fall on a

thin gold foil.

The scattered particles were detected at different

angle by a moveable detector. It was discovered that most of the

particles passed through the gold foil without any

deflection, or at very small angles of deflection. A very small fraction of particles were scattered through

large angles approaching 180˚.

A graph is plotted between the number of particles scattered and the scattering angle as shown in

the figure.

The results obtained were very surprising to Rutherford. He concluded that there is a very small

region inside an atom which is massive and whole mass of the atom is concentrated at this region which has

positive charge. This small region was give the name „nucleus‟. This is the reason why most of the

particles passed through the gold foil at small scattering angles. Only those particles could be deflected

through large angles which suffered head-on collision with the nuclei.

Therefore, according to Bohr and Rutherford, the model of atom may consist of a nucleus which is

massive and positive charged part, while the electrons are moving around the nucleus in allowed circular

orbits.



54.2 Some Nuclear Properties

(i) Nuclear Systematic

Nuclei are made up of protons and neutrons. But these particles are not true elementary particles;

these are made up of other particles called quarks.

The number of protons in a nucleus is called the atomic number or charge number. It is represented by

Z. The number of neutrons is called the neutron number. It is represented by N. The total number of nucleon

is called the mass number. It is represented by .

The charge on one proton is , while the neutron is a neutral particle. The nuclei,

having same charge number Z but different mass number is called isotopes. The nuclei which are not stable,

used to emit , and rays are called radioactive nuclides.

(ii) The Nuclear Force

A nucleus is packed with protons and neutrons. As proton is a positively charged particle, so, there

must be electrostatic repulsion. But the nucleus is very much rigid and stable. So, there must be some other

force inside the nucleus which is responsible for the stability and rigidity of the nucleus. This force is called

Strong Nuclear Force.

The electrostatic force (Coulomb force) is a long range force, but the Strong Nuclear Force is a short

range force, having the range of . This force binds every nuclear pair: proton-proton, neutron-neutron

and proton-neutron pair with in the tiny nuclear volume. For the short range of the order of , the

coulomb force is much smaller than the Strong Nuclear Force. If the separation between the nucleons is

increased beyond , the strong nuclear force drops to zero rapidly, then the Coulomb‟s force of

repulsion will be able to break the nucleus.

(iii) Nuclear Radii

The Bohr radius is , while the radius of the nucleus is of the order of .

So, the nuclei are smaller than the atoms by a factor of . The size and structure of the nuclei can be studied

by scattering experiments using incident beam of high energy electrons.

The energy of incident electrons must be greater than 200 MeV. These

experiments measure the diffraction pattern diffraction pattern of the

scattered particles and of deduce the shape of the scattering object (the

nucleus). The nucleus does not have sharply defined surface, however

have a characteristic radius „R‟. the density has constant value inside

the nucleus, but it falls to zero through fuzzy surface zone. The mean

radius is given by:

Where A is the mass number and is a constant whose value is 1.2 fm. For example, the has the radius

.



(iv) Nuclear Mass and Binding Energies

Atomic masses are measured in atomic mass units ( ). The atomic mass unit is defined as

times

the atomic mass of . Thus

One atomic mass unit „ ‟ is equivalent to the energy:

This means that we can write as .

(v) Nuclear Spin and Magnetism

Just like atoms, the nuclei have intrinsic angular momentum (nuclear spin) whose maximum value

along z-axis is

, where is total intrinsic angular momentum or nuclear spin quantum number.

In atomic magnetism, the Bohr magneton is given by:

Where is the mass of electron and T stand for tesla. Similarly for nuclear magneton , we have:

The magnetic moment of the heavier nuclei can be analyzed in terms of the magnetic moments of its

constituent protons and neutrons.

54.3 Binding Energy

It is the amount of energy required to tear a nucleus into its constituent nucleons. Or when nucleons

are fused to form a nucleus, then some energy is released which is called binding energy.

Example

The nucleus of Deuteron (a heavy hydrogen atom) consists of a protons and a neutron bounded

together by a strong nuclear force. The energy that we must add to deuteron to tear it apart in to its

constituent nucleons is called the energy. If , and are the masses of deuteron, neutron and proton

respectively, then according to the law of conservation of energy:

(

)

----------------- (1)

Here and

are the masses of the hydrogen and deuteron atom, respectively.



[

] ---------------- (2)

In which,

.

As , and

. So by substituting the

values (2) and replacing by its equivalent , we find the binding energy to be:

.

54.3.1 Binding Energy per Nucleons

This is the binding energy of the deuteron. If the binding energy is divided by mass number A,

then binding energy per nucleon (

) is obtained. Figure show a graph between energy per nucleon (

) and

mass number A.

The (

) is high for middle mass nucleons. It means

that these nucleons are tightly bound and the nucleus

is much stable. The region of greatest stability

corresponds to mass number about 50 to 80.

The biding energy per nucleon curves drops at both

high and low mass numbers has the practical

consequence of the greatest importance. The dropping

of the binding energy curve at high mass numbers tells

us energy can be released in the nuclear fission of a

single massive nucleus into two smaller fragments.

The dropping of the binding energy curve at low mass

numbers tells us that energy will be released if the two nuclei of small mass numbers combined to form a

single middle mass nucleus. This process, the reverse of fission, is called nuclear fusion.

Sample problem 3. Find the total energy required to split into its constituent protons and

neutrons. (b) Find the energy per nucleon. The atomic mass of is

Solution. Number of proton (Atomic Number)

Atom Mass

Number of Neutrons (Neutron #)

Binding Energy

Mass Defect [ ]

Now

(b)Binding energy per Nucleon



Problem. Find the energy per nucleon . The atomic mass of

is

Solution.

Number of proton (Atomic Number)

Atom Mass

Number of Neutrons (Neutron #)

Binding Energy

Mass Defect [ ]

Now

Binding energy per Nucleon

54.4 Radioactive Decay

The elements having charge number greater than 82, are not stable. They disintegrate and emit

particles or radiation spontaneously. Such elements are radioactive and this process of spontaneous emission

of particles or radiations is called radioactive decay or natural radioactivity.

Experiment

The radioactive sample is placed in a cavity in a lead block

above which a photographic plate is held. The whole apparatus is

placed inside a vacuum chamber, in which magnetic field is also

applied, as shown in the figure.

Three images are obtained on the photographic plate. It means

that three types of radiations are emitted by the radioactive sample,

which particles, particles and rays.

(i) particles

particles is a helium nucleus which consist of two protons and two neutrons. Its charge number is

2 and mass number is 4. It is highly ionizing particle, but its range is small. Emissions of particles from an

atom reduces its mass number A by four units and atomic number Z by two units. As a result, the product

element moves two places backward in the periodic table.

Explanation

The radionuclide decays spontaneously according to the scheme:

With the half life of . In this process, an energy of 4.27 MeV is emitted appearing

as the kinetic energy of the alpha particle and the recoiling residual nucleus

.



In order to explain the alpha decay, a model is used in which the particle is assumed to be exist

preformed inside the nucleus before it escapes.

The figure shows the approximate potential energy function U(r) for the particle and the residual

nucleus as the function of their separation. It is a combination of a potential well associated with the

attractive strong nuclear force that acts in the nuclear interior ( ) and a Coulomb‟s potential associated

with the repulsive electrostatic force that acts between the two particles after the decay has occurred ( ).

The line that intersect the potential energy curve at point and , is the measure of emitted during

one alpha decay. The energy of 4.27 MeV is emitted appearing as the kinetic energy of the alpha particle

and the recoiling residual nucleus

.

The decay of the alpha particle is accompanied by the emission of energy. Now the question arises

that why did not decay shortly after they were created?

The answer to this question is given is background of the

potential barrier consideration. We can visualize this barrier as a

spherical shell whose inner radius is and whose outer radius is

, its volume being forbidden to the particle under the law of

classical physics.

But according to the quantum mechanics, there is a chance

of tunneling through barrier. It is this tunneling due to which an

alpha particle gets a chance to come out of nucleus. But its

probability is extremely little, (1 out of ). It is one about in

years. That is why nucleus has such long half life.

If energy of alpha particle is comparatively high, the potential barrier will be appear to it thinner and

lower. Hence in this case, tunneling will occur more readily, and half life is reduced considerably. For

example, alpha particle emitted by has energy of 6.81 MeV.

Sample problem 6. Find the energy released during the alpha decay of . The needed atomic masses

are

Solution.

The reaction equation is :

(ii) particles

A particle is a positive or negative electrons which is emitted from the nucleus of a radioactive

element. When a proton in the nucleus transforms into a neutron, a positive particles is emitted.

When a neutron inside a nucleus changes into protons, then negative particles are emitted:



Emission of negative particles leaves the mass number of the product nucleus unchanged, but the

atomic number (charge number) increases by one unit. So the product element moves on place ahead in the

periodic table.

Similarly, emission of positive particles leaves the mass number of the product nucleus unchanged

but the atomic number (charge number) decreases by one unit. So the product element moves one place

backward in the periodic table.

Explanation: Emission of electron or a positron from a nucleus is known as beta decay. Beta particles don't

exist in nucleus. They are emitted as soon as they are formed by disintegration of a neutron or a proton in a

nucleus. The examples of beta decay are as follows:

Here and

are electron and positron.

The kinetic energy of beta particle does not remain constant, but it changes over a range as shown in the

figure. The maximum value of kinetic energy of the beta particle is 0.653 MeV, which is also the total

disintegration energy for the case of beta decay from Cu. The kinetic energy of the beta particle changes over

the range because the disintegration energy is shared by

beta particle and neutrino.

Beta particle is not emitted alone. But with

every beta particle another particle is emitted, called

neutrino. Total energy of beta particle and neutrino is

quantized. But the disintegration energy is shared by

the beta particle and neutrino in any proportion, i.e.,

beta particle may take any energy between 0 and maximum. The rest of energy is carried by neutrino. Thus

the above mentioned equations of beta decay take the modified form as given below:

̅

Here and ̅ are neutrino and anti-neutrino respectively. Neutrino is emitted with positron ( ) and

anti-neutrino is emitted with electron ( ). The neutrino and anti-neutrino are very light particles having no

charge and mass nearly of the order of

of mass of electron. So they interact with matter weakly and are

very difficult to detect.

(iii) Rays

rays are not material particles but they are electromagnetic rays moving with the velocity of light.

These are the most energetic electromagnetic radiations having shortest wavelength and largest frequency.

ray can produce photoelectric effects, Compton effect and pair production. These radiations have largest

range. rays are emitted from the excited nuclei of radioactive elements.

Emission of rays does not change the mass number and atomic number of the product nucleus.

ray



54.5 Laws of Radioactivity and Half Lift of Radioactive Element

It is the time during which one half of the atoms of the parent element decay in to daughter element.

The half life of a radioactive element may vary from fraction of a second to millions of year.

The radioactive decay obeys the following two statistical laws:

The number of atoms that decay at any instant is proportional to the number of atoms present at that

instant.

No sample of radioactive element can ever completely decay in a finite time.

Let is the number of atoms initially present at time . Let N is the number of atoms at any time

t. If are the number of atoms decay during the interval of time , then the rate of decay

is directly

proportional to the number of atoms present. i.e.,

Where is the decay constant. It is the characteristics of radioactive element.

Initially at , there are radionuclide of parent element and N is the number of atoms at any time t.

Integrating above differential equation over appropriate limits, we have:

∫

∫

| | | |

(

)

---------------- (1)

This is known as radioactive decay law. It is clear from equation (1), that radioactivity follows an exponential

law and it takes an infinite time to decay a radioactive element completely.

When

, then

Putting values in equation (1), we have:

( )



(

)

This is the expression of half life of a radioactive element.

Question: Show that decay rate of radionuclide follows an exponential law.

Ans. The rate of decay (

) i.e., activity R can be find out by differentiating

the equation of law of radioactivity :

Where Decay rate at .

It is clear from equation that radioactive follows an exponential law and it takes an infinite time to

decay a radioactive element completely.

Problem 23. The half life of a radioactive isotope is 140 days. How many days would it take fit the

activity of the sample to fall to one forth of its initial decay rate?

Solution:

Law of radioactivity:

Problem 24. The half life of a particular radioactive isotope is 6.5 hr. if there are initially

of this isotope in a paricualr sample, how many atoms of this isotope remains after 26 hrs.

Solution:



As

Problem 25. A radioactive isotope of gold with a decay constant of . Calculate its: (a)

Half life (b) what fraction of the original will remain after three half lives? (c) After 10 days

Solution.

(a)

(b)

As

(c)

for

As

54.6 Mean Life of Radioactive Element

As the possible life of a radioactive element varies from 0 to , so the total life of all initially

present atoms in a given sample can be find out by the expression ∫

.

The mean or average life for a sample can be described by the expression:

∫

∫

∫

| |

Also,

as

Putting values in equation (1), we have:



∫ (

)

∫

∫

[|

|

∫

]

[

∫

] [

∫

]

∫

|

|

[ ]

[

]

[ ]

This is the expression of mean life of a radioactive element.

Question: Derive the relationship between Half Life and Mean Life of a radioactive element.

Ans. The half life and mean life of radioactive element is described by formulae

and

, respectively. Now,

This expression shows the mean life of a radioactive element is greater than its half life.

54.7 Units for Measuring Ionizing Radiation

Becquerel

SI unit of radioactivity is Becquerel. One Becquerel is one disintegration per second.

The Curie

The unit of activity or rate of decay of a radioactive source is Curie Ci. It is defined as the activity of

one gram of radium in equilibrium.

The Roentgen

It is the unit of exposure which may be defined as the exposure of beam of or to

produce , the air being dry and at standard temperature and pressure.



The Rad

This is acronym for radiation absorbed dose and is a measure of the dose actually delivered to a

specified object. The specified part of a body (the hand, say) is said to have received and absorbed dose of 1

rad when have been delivered to it by ionizing radiation.

The rem

This is acronym for roentgen equivalent in man and is a measure of dose equivalent. The dose

equivalent (in rem) is found by multiplying the absorbed dose (in rad) by quality factor QF. According to the

recommendation of the National Council on Radiation Protection, no individual who is exposed to radiations

should receive a dose equivalent greater than 500 m rem (0.5 rem) in any one year.

54.8 Radioactive Dating

The age of a sample can be determined by radioactive dating. Suppose an initial radio nuclide I decays

to a final product F with a known half-life

At particular time , we start with initial nuclei with product (final) nuclei equal to zero. After

a time t, the initial nuclei at reduce to „ ‟ with the product (final) nuclei „ ‟, where .

To determine the age of radioactive sample we make use of exponential law of radioactivity:

where the decay constant

Taking logarithm on both sides, to the base e, we get:

(

)

(

)

(

)

(

)

Using formula of half life, we have :

(

)

Put

(

)

(

)

It is clear from expression that the age of radioactive sample can be determined by the ratio

. This

method can be used to determine the time since the formation of the solar system. Example include the ration

of to , to and to . The terrestrial rocks, moon rocks are analyzed by these

method, all seem to have common age of around year, which we take to be the age of our solar

system.



Sample Problem 9. In a sample of rock, the ratio of to nuclie is found to be 0.65. What is the

age of the rock? of is years.

Ratio =

(

)

54.9 Energy in Nuclear Reactions

A nuclear reaction can be represented by:

Here is the target nucleus, is the projectile nucleus, is the residual nucleus and is the emerging

nucleus. The projectile particle a may be charged particle which can be accelerated Van de Graph accelerator

or cyclone, or it may be a neutron from the nuclear reactor.

When the projectile particle penetrates a target nucleus, then a nuclear reaction takes place. The

reaction energy Q is defined as (rest mass energies)

------------- (1)

Here ,

,

If „ ‟ represents the kinetic energy, then the reaction energy „ ‟ is given by:

------------- (2)

Equation (1) and (2) are only valid, when Y and b are in their ground state.

Exothermic: If the reaction energy has positive value i.e., , then such nuclear reaction is known as

exothermic .

Endothermic: If the reaction energy Q is negative i.e., , then such a nuclear reaction is known as

endothermic. Such a reaction will not “go” unless a certain minimum kinetic energy (the threshold energy) is

carried into the system b the projectile. Its means that endothermic reaction needs some certain energy for its

performance.

Scattering: A nuclear reaction is called scattering, if the particles „a‟ and „b‟ are identical and so „X‟ and „Y‟

are also identical.

Elastic Scattering: If the kinetic energy of the system before the reaction is equal to the kinetic ener

Documents

B.Sc. Physics · 2020. 12. 28. · CHAPTER # 1: LIGHT AND QUANTUM PHYSICS CHAPTER # 2: WAVE NATURE OF MATTER CHAPTER # 3: NUCLEAR PHYSICS CHAPTER # 4: ENERGY FROM NUCLEUS B.Sc. Physics