Q1) Explain dual nature of matter?

A1)

As we know in the Photoelectric Effect, the Compton Effect, and the pair production effect—radiation exhibits particle-like characteristics in addition to its wave nature. In 1923 de Broglie took things even further by suggesting that this wave–particle duality is not restricted to radiation, but must be universal.

In 1923, the French physicist Louis Victor de Broglie (1892-1987) put forward the bold hypothesis that moving particles of matter should display wave-like properties under suitable conditions.

All material particles should also display dual wave–particle behaviour. That is, the wave–particle duality present in light must also occur in matter.

So, starting from the momentum of a photon p = hν/c = h/λ.

We can generalize this relation to any material particle with nonzero rest mass.  Each material particle of momentum  behaves as a group of waves (matter waves) whose wavelength λ and wave vector  are governed by the speed and mass of the particle. De Broglie proposed that the wave length λ associated with a particle of momentum p is given as where m is the mass of the particle and v its speed.

λ =   =    …….(1)    =     …….(2) 

Where ℏ = h/2π. The expression known as the de Broglie relation connects the momentum of a particle with the wavelength and wave vector of the wave corresponding to this particle. The wavelength λ of the matter wave is called de Broglie wavelength. The dual aspect of matter is evident in the de Broglie relation.

λ is the attribute of a wave while on the right hand side the momentum p is a typical attribute of a particle. Planck’s constant h relates the two attributes. Equation (1) for a material particle is basically a hypothesis whose validity can be tested only by experiment.

However, it is interesting to see that it is satisfied also by a photon. For a photon, as we have seen, p = hν/c.

Therefore

= = λ

Q2) What is the de Broglie wavelength associated with (a) an electron moving with a speed of 5.4×106 m/s, and (b) a ball of mass 150 g travelling at 30.0 m/s?

A2)

(a)For the electron:

 Mass m = 9.11×10–31 kg, speed v = 5.4×106 m/s.

Then, momentum

p = m v = 9.11×10–31 kg × 5.4 × 106 (m/s)

p = 4.92 × 10–24 kg m/s

de Broglie wavelength, λ = h/p = 6.63 x 10-34Js/ 4.92 × 10–24 kg m/s

λ= 0.135 nm

(b)For the ball:

Mass m’ = 0.150 kg,

Speed v ’= 30.0 m/s.

Then momentum p’ = m’ v’= 0.150 (kg) × 30.0 (m/s)

p’= 4.50 kg m/s

de Broglie wavelength λ’ = h/p’ =6.63 x 10-34Js/ 4.50 kg m/s =1.47 ×10–34 m

The de Broglie wavelength of electron is comparable with X-ray wavelengths. However, for the ball it is about 10–19 times the size of the proton, quite beyond experimental measurement.

Q3) What is the de Broglie wavelength associated with an electron, accelerated through a potential difference of 100 volts?

A3) Accelerating potential V = 100 V. The de Broglie wavelength λ is

λ= h /p = 1 227/ nm

λ.1 227/  nm = 0.123 nm

The de Broglie wavelength associated with an electron in this case is of the order of x ray wavelengths.

Q4) What is wave packet or wave group?

A4)

We know that in classical physics a particle is well localized in space. We can calculate its position and velocity simultaneously.

But for quantum mechanics, a particle is not well localized in space. We cannot calculate its position and velocity simultaneously.

In Quantum mechanics it is describe that the matter wave associated with the particle.  Wave functions depend on the whole space. Hence they cannot be localized.

If the wave function is made to vanish everywhere except in the neighbourhood of the particle or the neighbourhood of the classical trajectory, it can then be used to describe the dynamics of the particle. That is a particle which is localized within a certain region of space can be described by a matter wave whose amplitude is large in that region and zero outside it. This matter wave must then be localized around the region of space within which the particle is confined.

A localized wave function is called a Wave Packet or Wave Group. A wave packet or wave group therefore consists of a group of waves of slightly different wavelengths with phases and amplitudes so chosen that they interfere constructively over a small region of space and destructively elsewhere.

Wave packets are Not only useful in the explanation of isolated particles that are confined to a certain spatial region but they also play a key role in understanding the connection between quantum mechanics and classical mechanics. The wave packet concept also represents a uniting mathematical tool that describes particle-like behaviour and also its wave-like behaviour.

Q5) Define phase and group velocity?

Or

Define term vp and vg?

A5)

Phase and group velocity are two important and related concepts in wave mechanics. They arise in quantum mechanics in the time development of the state function for the continuous case, i.e. wave packets.

According to the theory of relativity particle velocity (v)  is always less than the speed of light c. but according to the De-Broglie wave velocity must be greater than c. This is an unexpected result. According to this the de-Broglie wave associated with the particle would travel faster than the particle itself thus leaving the particle far behind.

The difficulty was recovered by Schrodinger. He proposed that a material particle in motion is equivalent to a wave packet rather than a single wave. Wave packet comprises of a group of waves, each with slightly different velocity and wavelength.
Such a wave packet moves with its own velocity vg­ called the group velocity.

The individual wave forming the wave packet possesses an average velocity vp called the phase velocity.

It can be shown that the velocity of the material particle v is the same as group velocity

Let us assume two wave trains have same amplitude but different frequency and phase velocities

u (x, t) =A sin(ωt - kx)    …….(1)

u’(x, t) =A sin[(ω+∆ω)t-(k+∆k)x]   …….(2)

Where ω  and (ω+Δω)  are angular frequencies and k and (k +Δk) are propagation constants

The superposition of two waves is of the form

ψ(x,t) = u +u’ = A sin(ωt - kx)sin[(ω+∆ω)t - (k+∆k)x]      …….(3)

As Δ ω and Δk are small therefore (ω+Δω) and (k + Δk) k, equation (3) reduces to

ψ(x,t) = 2Acos[ t -x] sin(ωt - kx)    .…….(4)

This equation represents a vibration of amplitude

2Acos[ t -x]       …….(5)

The phase of the resultant wave moves with the velocity known as phase velocity

vp=          ……..(6)

And the amplitude moves with the velocity known as group velocity

vg = = 

vg =          ….….(7)

Q6) What is Heisenberg’s Uncertainty Principle?

 A6)

According to classical physics, given the initial conditions and the forces acting on a system, the future behaviour (unique path) of this physical system can be determined exactly. That is, if the initial coordinates, velocity , and all the forces acting on the particle are known, the position ,  and velocity are uniquely determined by means of Newton’s second law. So by Classical physics it can be easily derived.

Does this hold for the microphysical world?

Since a particle is represented within the context of quantum mechanics by means of a wave function corresponding to the particle’s wave, and since wave functions cannot be localized, then a microscopic particle is somewhat spread over space and, unlike classical particles, cannot be localized in space. In addition, we have seen in the double-slit experiment that it is impossible to determine the slit that the electron went through without disturbing it. The classical concepts of exact position, exact momentum, and unique path of a particle therefore make no sense at the microscopic scale. This is the essence of Heisenberg’s uncertainty principle.

In its original form, Heisenberg’s uncertainty principle states that: If the x-component of the momentum of a particle is measured with an uncertainty ∆px, then its x-position cannot, at the same time, be measured more accurately than ∆x = ℏ/(2∆px). The three-dimensional form of the uncertainty relations for position and momentum can be written as follows:

This principle indicates that, although it is possible to measure the momentum or position of a particle accurately, it is not possible to measure these two observables simultaneously to an arbitrary accuracy. That is, we cannot localize a microscopic particle without giving to it a rather large momentum.

 We cannot measure the position without disturbing it; there is no way to carry out such a measurement passively as it is bound to change the momentum.

To understand this, consider measuring the position of a macroscopic object (you can consider a car) and the position of a microscopic system (you can consider an electron in an atom). On the one hand, to locate the position of a macroscopic object, you need simply to observe it; the light that strikes it and gets reflected to the detector (your eyes or a measuring device) can in no measurable way affect the motion of the object.

On the other hand, to measure the position of an electron in an atom, you must use radiation of very short wavelength (the size of the atom). The energy of this radiation is high enough to change tremendously the momentum of the electron; the mere observation of the electron affects its motion so much that it can knock it entirely out of its orbit.

 It is therefore impossible to determine the position and the momentum simultaneously to arbitrary accuracy. If a particle were localized, its wave function would become zero everywhere else and its wave would then have a very short wavelength.

Q7) The uncertainty in the momentum of a ball travelling at 20m/s is 1×10−6 of its momentum. Calculate the uncertainty in position? Mass of the ball is given as 0.5kg.

A7)

Given
v = 20m/s,

m = 0.5kg,

h = 6.626 × 10-34 m2 kg / s

Δp =p×1×10−6

As we know that,
P = m×v = 0.5×20 = 10kgm/s
Δp = 10×1×10−6

Δp = 10-5
 

Heisenberg Uncertainty principle formula is given as,

∆x∆p

∆x

∆x

∆x =0.527 x 10-29 m

Q8) Prove non-existence of electrons in the nucleus?

A8)

One of the applications is to prove that electron cannot exist inside the nucleus.

But to prove it, let us assume that electrons exist in the nucleus.

As the radius of the nucleus in approximately 10-14m. If electron is to exist inside the nucleus, then uncertainty in the position of the electron is given by

According to uncertainty principle

∆x ∆p =h/2π

Thus ∆p=h/2π∆x

Or ∆p=6.62 x10-34/2 x 3.14 x 10-14

Or ∆p=1.05 x 10-20 kg m/ sec

If this is p the uncertainty in the momentum of electron then the momentum of electron should be at least of this order that is p=1.05*10-20 kg m/sec.

An electron having this much high momentum must have a velocity comparable to the velocity of light. Thus, its energy should be calculated by the following relativistic formula

E =

Therefore, if the electron exists in the nucleus, it should have an energy of the order of 19.6 MeV.

However, it is observed that beta-particles (electrons) ejected from the nucleus during b – decay have energies of approximately 3 Me V, which is quite different from the calculated value of 19.6 MeV.

Another reason that electron cannot exist inside the nucleus is that experimental results show that no electron or particle in the atom possess energy greater than 4 MeV.

Therefore, it is confirmed that electrons do not exist inside the nucleus.

Q9) Derive the Schrodinger wave equation?

A9)

Schrodinger wave equation, is the fundamental equation of quantum mechanics, same as the second law of motion is the fundamental equation of classical mechanics. This equation has been derived by Schrodinger in 1925 using the concept of wave function on the basis of de-Broglie wave and plank’s quantum theory.

Let us consider a particle of mass m and classically the energy of a particle is the sum of the kinetic and potential energies. We will assume that the potential is a function of only x.

 So We have

E= K+V =mv2+V(x) = +V(x)   ……….. (1)

By de Broglie’s relation we know that all particles can be represented as waves with frequency ω and wave number k, and that E= ℏω and p= ℏk.

Using this equation (1) for the energy will become

ℏω = + V(x)    ……….. (2)

A wave with frequency ω and wave number k can be written as usual as

ψ(x, t) =Aei(kx−ωt)     ……….. (3)

the above equation is for one dimensional and for three dimensional we can write it as

ψ(r, t) =Aei(k·r−ωt)    ……….. (4)

But here we will stick to one dimension only.

=−iωψ   ⇒   ωψ=   ……….. (5)   

=−k2ψ  ⇒   k2ψ = -    ……….. (6)

If we multiply the energy equation in Eq. (2) by ψ, and using(5) and (6) , we obtain

ℏ(ωψ)  = ψ+ V(x) ψ  ⇒   = - + V(x) ψ  ……….. (7)

This is the time-dependent Schrodinger equation.

If we put the x and t in above equation then equation (7)  takes the form as given below

= - + V(x) ψ(x,t)    ……….. (8)

In 3-D, the x dependence turns into dependence on all three coordinates (x, y, z) and the  term becomes ∇2ψ.

The term |ψ(x)|2 gives the probability of finding the particle at position x.

Let us again take it as simply a mathematical equation, then it’s just another wave equation. However We already know the solution as we used this function ψ(x, t) =Aei(kx−ωt) to produce Equations (5), (6) and (7)

But let’s pretend that we don’t know this, and let’s solve the Schrodinger equation as if we were given to us. As always, we will guess an exponential solution by looking at exponential behaviour in the time coordinate, our guess is ψ(x, t) =e−iωtf(x) putting this into Equation (7) and cancelling the e−iωt yields

  = - + V(x) f(x)   ……….. (9)

We already know that E=. However ψ(x, t) is general convention to also use the letter ψ to denote the spatial part. So we will now replace f(x) with ψ(x)

   Eψ = - + V(x) ψ    ……….. (10)

This is called the time-independent Schrodinger equation.

The Schrodinger equation also known as Schrodinger’s wave equation is a partial differential equation that describes the dynamics of quantum mechanical systems by the wave function. The trajectory, the positioning, and the energy of these systems can be retrieved by solving the Schrodinger equation.

All of the information for a subatomic particle is encoded within a wave function. The wave function will satisfy and can be solved by using the Schrodinger equation. The Schrodinger equation is one of the fundamental axioms that are introduced in undergraduate physics.

Q10) The mass of an electron is 9.1×10–31 kg. Its uncertainty in velocity is 5.7×105 m/sec. Calculate uncertainty in its position?

A10)

Given     m= 9×10–31

ΔV= 5.7×105 m/sec                 

Δx=?

h= 6.6×10–34 Joule-Sec.

Δx.Δv ≥h/4πm

Δx≥h/4πmΔv

≥6.6×10–34/9×3.14×9.1×10–31×5.7×105

≥ 0.010×10–8

≥ 1×10–10m

Q11) The mass of a ball is 0.15 kg & its uncertainty in position to 10–10m. What is the value of uncertainty in its velocity?

A11)

Given

m=0.15 kg.     

h=6.6×10-34 Joule-Sec.

Δx = 10 –10 m

Δv=?

Δx.Δv ≥h/4πm

Δv≥h/4πmΔx

≥6.6×10-34/4×3.14×0.15×10–10

≥ 3.50×10–24m

Q12) The mass of a bullet is 10gm & uncertainty in its velocity is 5.25×10–26 cm/sec. Calculate the uncertainty in its position?

A12)

m=10 gm.                    h=6.6×10–27 erg-sec

Δv= 5.25×10–26 cm

Δx=?

Δx.Δv ≥h/4πm

Δx≥h/4πmΔv

≥6.6×10-34/4×3.14×10×5.25×10–26

≥ 0.10×10–2m

≥ 1×10–3 cm

Q13) List the conditions wave function should satisfy?

A13)

The wave function, at a particular time, contains all the information that anybody at that time can have about the particle.  But the wave function itself has no physical interpretation.  It is not measurable.  However, the square of the absolute value of the wave function has a physical interpretation.  We interpret |ψ(x,t)|2 as a probability density, a probability per unit length of finding the particle at a time t at position x.

The wave function ψ associated with a moving particle is not an observable quantity and does not have any direct physical meaning. It is a complex quantity. The complex wave function can be represented as

 ψ(x, y, z, t) = a + ib

and its complex conjugate as

ψ*(x, y, z, t) = a – ib.

The product of wave function and its complex conjugate is

ψ(x, y, z, t)ψ*(x, y, z, t) = (a + ib) (a – ib) = a2 + b2

a2 + b2 is a real quantity.

However, this can represent the probability density of locating the particle at a place in a given instant of time.

The positive square root of ψ(x, y, z, t) ψ*(x, y, z, t) is represented as |ψ(x, y, z, t)|, called the modulus of ψ. The quantity |ψ(x, y, z, t)|2 is called the probability. This interpretation is possible because the product of a complex number with its complex conjugate is a real, non-negative number. 

We should be able to find the particle somewhere, we should only find it at one place at a particular instant, and the total probability of finding it anywhere should be one. 

For the probability interpretation to make sense, the wave function must satisfy certain conditions. 

• The wave function must be single valued at each point.
• The probability of finding the particle at time t in an interval ∆x must be some number between 0 and 1.
• ψ must be finite everywhere.
• ψ must be continuous everywhere and must also be continuous everywhere except where V(x) is infinite.
• ψ (x) must vanish ψ0 as x.
• The wave function should satisfy the normalization condition. Normalization condition of a wave function ψ is mathematical statement of existence of the particle somewhere.  so that if we sum up all possible values ∑|ψ(xi,t)|2∆xi  we must obtain 1.  The total probability of finding the particle anywhere must be one. Normalization condition is given as

dx =1

Only wave function with all these properties can yield physically meaningful   result.

Q14) What is Physical significance of wave function?

A14)

• The wave function ‘Ѱ’ has no physical meaning. it is a complex quantity representing the variation of a matter wave.
• The wave function Ѱ(r,t) describes the position of particle with respect to time .
• It can be considered as ‘probability amplitude’ since it is used to find the location of the particle.
• The square of the wave function gives the probability density of the particle which is represented by the wave function itself.
• More the value of probability density, more likely to find the particle in that region.

Q15) Discuss Particle in a One Dimensional Deep Potential Well

A15)

Let us consider a particle of mass ‘m’ in a deep well restricted to move in a one dimension (say x). Let us assume that the particle is free inside the well except during collision with walls from which it rebounds elastically.

The potential function is expressed as

V= 0 for 0      ………. (1)

V= for x <0, x>L     ………. (2)

Figure : Particle in deep potential well

The probability of finding the particle outside the well is zero (i.e. Ѱ =0) 

 Inside the well, the Schrödinger wave equation is written as

ψ + E ψ =0    …………….(2)    

Substituting   E = k2                                       …………….(3)

writing the SWE for 1-D we get

+ k2 ψ =0       …………….(4)

The general equation of above equation may be expressed as

ψ = Asin (kx + ϕ)      …………….(5)

Where A and ϕ  are constants to be determined by boundary conditions

Condition I: We have ψ = 0 at x = 0, therefore from equation

0 = A sinϕ   

As A then sinϕ =0  or  ϕ=0    …………….(6)

Condition II: Further ψ = 0 at x = L, and ϕ=0 , therefore  from equation (5)

0 = Asin kL

As A then sinkL =0  or  kL=nπ

k =       …………….(7)

where n= 1,2,3,4………

Substituting the value of k from (7) to (3)

)2 = E

 This gives energy of level  

En =    n=1,2,3,4…so on    …………….(8)

 
From equation En is the energy value (Eigen Value) of the particle in a well.

It is clear that the energy values of the particle in well are discrete not continuous.

Using (6) and (7) equation (5) becomes, the corresponding wave functions will be

ψ = ψn = Asin   …………….(9)

The probability density

|ψ(x,t)|2 = ψ ψ*

|ψ(x,t)|2 = A2sin2   …………….(10)

The probability density is zero at x = 0 and x = L. since the particle is always within the well

    …………….(11)

=1

A =

Substituting A in equation (9) we get

ψ = ψn = sin    n=1,2,3,4…..    …………….(12)

The above equation (12) is normalized wave function  or Eigen function belonging to energy value En

Figure: Wave function for Particle

Therefore, according to uncertainty principle it is difficult to assign a position to the electron.

Q16) Discuss Bottom up approach?

A16)

In bottom-up approaches nanomaterials are assembled from basic building blocks, such as molecules or nanoclusters. The basic building blocks, in general, are nanoscale objects with suitable properties that can be grown from elemental precursors. The concept of the bottom-up approach is that the complexity of nanoscale components should reside in their self-assembled internal structure, requiring as little intervention as possible in their fabrication from the macroscopic world.

The bottom-up approach uses atomic or molecular feed-stocks as the source of the material to be chemically transformed into larger nanoparticles. This has the advantage of being potentially much more convenient than the top down approach. By controlling the chemical reactions and the environment of the growing nanoparticle, then the size, shape and composition of the nanoparticles may all be affected. For this reason nanoparticles produced by bottom up, chemically based and designed, reactions are normally seen as being of higher quality and having greater potential applications. This has led to the growth of a host of common bottom up strategies for the synthesis of nanoparticles. Many of these techniques can be tailored to be performed in gas, liquid, solid states, hence the applicability of bottom-up strategies to a wide range of end products. Most of the bottom up strategies requires suitable organometallic complexes or metal salts to be used as chemical precursors, which are decomposed in a controlled manner resulting in particle nucleation and growth. One of the key differences that can be used to subdivide these strategies into different categories is the method by which the precursor is decomposed.

A typical example of bottom-up is processing for nanocomposite magnets from individual high-magnetization and high-coercivity nanoparticles. The assembling critically depends on availability of anisotropic (single crystal) hard magnetic nanoparticles. Anisotropic nanoparticles produced via surfactant-assisted high energy ball milling satisfy the major requirements for this application.

Q17) Discuss Top down approach?

A17)

In top-down approaches, a bulk material is restructured (i.e. partially dismantled, machined, processed or deposited) to form nanomaterials. The aggressive scaling of electronic integrated circuits in recent years can be considered the greatest success of this paradigm. For top-down methods, the challenges increase as devices size is reduced and as the desired component designs become larger and more complex. Also the top-down assembly of nanocomponents over large areas is difficult and expensive.

The top-down method involves the systematic breakdown of a bulk material into smaller units using some form of grinding mechanism. This is beneficial and simple to execute and avoids the use of volatile and poisonous compounds frequently found in the bottom-up techniques. However, the quality of the nanoparticles formed by grinding is accepted to be poor in comparison with the material produced by modern bottom up methods. The main drawbacks include defect problems from grinding equipment, low particle surface areas, asymmetrical shape and size distributions and high energy needed to produce relatively small particles. Apart from these disadvantages, it must be distinguished that the nano-material produced from grinding still finds use, due to the simplicity of its manufacture, in applications including magnetic, catalytic and structural properties.

Q18) Write applications of nanomaterials?

A18)

Nano materials possess unique and beneficial, physical, chemical and mechanical properties; they can be used for a wide verity of applications.

Material technology

• Nanocrystalline aerogel are light weight and porous, so they are used for insulation in offices homes, etc.
• Cutting tools made of Nano crystalline materials are much harder, much more wear- resistance, and last stranger.
• Nano crystalline material sensors are used for smoke detectors, ice detectors on air craft wings, etc.
• Nano crystalline materials are used for high energy density storage batteries.
• Nano sized titanium dioxide and zinc dioxide is used in sunscreens to absorb and reflect ultraviolet rays.
• Nano coating of highly activated titanium dioxide acts as water repellent and antibacterial.
• The hardness of metals can be predominately enhanced by using nanoparticles.
• Nanoparticles in paints change colour in response to change in temperature or chemical environment, and reduce the infrared absorption and heat loss.
• Nano crystalline ceramics are used in automotive industry as high strength springs, ball bearings and valve lifters.

Information technology

• Nanoscale fabricated magnetic materials are used in data storage
• Nanocomputer chips reduce the size of the computer.
• Nano crystalline starting light emitting phosphors are used for flat panel displays.
• Nanoparticles are used for information storage.
• Nanophotonic crystals are used in chemical optical computers.

Biomedical

• Biosensitive nanomaterials are used for ragging of DNA and DNA chips.
• In the medical field, nanomaterials are used for disease diagnosis, drug delivery and molecular imaging.
• Nano crystalline silicon carbide is used for artificial heart valves due to its low weight and high strength.

Energy storage

• Nanoparticles are used hydrogen storage.
• Nano particles are used in magnetic refrigeration.
• Metal nanoparticles are useful in fabrication of ionic batteries.