Synthesis of Nanomaterials

In the field of nanomaterials, there are two main approaches to their synthesis and fabrication.

A.) Top-Down Approach

B.) Bottom-up Approach

These two approaches are based on the methods used to create or assemble nanoscale materials and structures.

A.) Top-Down Approach: In this approach, large-scale materials (i.e. Bulk materials) are broken down into smaller and smaller components until they reach the desired nanoscale dimensions i.e. firstly the bulk material is converted into powder form and then the powder's form is converted into nanoparticles. There are various physical methods (like Arc discharge method, Electron beam lithography, Mechanical grinding, etc) used to convert the bulk material into powder form and powder form is converted into nanoparticles by chemical methods (like Sol-Gel Process, Electrochemical method, Microemulsion etc.). One of the challenges with the top-down approach is that it may lead to a lack of control over the final nanomaterial's properties and may also generate waste during the process.

Top-down approaches are based on grinding the material. Thus these processes are subtractive in nature. The parts of mechanical devices used to shape the objects are stiff and hard, so these approaches are not suitable for soft samples. The top-down approach assumes that nanodevices must be produced piece by piece in a series of steps, much like manufactured goods are made. Scanning probe microscopy (SPM) is an important technique used both for the synthesis and characterization of nanomaterial by a top-down approach. Atomic force microscopy (AFM) tips can be used as a nanoscale "write head" to deposit a resist, which is then followed by an etching process to remove material in a top-down approach.

Some common techniques used in the top-down approach include:

a.) Mechanical milling: High-energy mechanical forces are applied to break down large particles into smaller ones, eventually reaching the nanoscale range.

b.) Lithography: Techniques like electron beam lithography or photolithography are used to create patterns on a substrate, defining the nanoscale features.

c.) Etching: Chemical or plasma etching is used to remove material from a larger sample, creating nanoscale structures.

d.) Nanolithography: Specialized lithography techniques are used to directly write or print nanoscale patterns on material.

Advantages of the Top-down Approaches:

1.) This Top-down approach is often suitable for large-scale production of nanomaterials, making it economically viable.

2.) Many top-down approaches have been widely used and optimized for bulk materials, which can be adapted for nanomaterial synthesis.

3.) This approach allows precise control over the size and shape of the resulting nanomaterials.

4.) Suitable for laboratory experimentation.

Disadvantages Top-down Approaches:

1.) It is required large installations and huge capital is required for building their setup.

2.) These approaches are quite expensive.

3.) The growth process is slow and hence these approaches are not suitable for large-scale production.

4.) The biggest problem in the approach is the imperfection of surface structure.

5.) The mechanical and chemical processes involved in the top-down approach can introduce defects in the nanomaterials, affecting their properties.

6.) This approach generates a significant amount of waste during the material reduction process, which can be environmentally problematic.

7.) As the approach starts with larger structures and reduces their size, it may be challenging to achieve precise control over nanoscale features, leading to variations in properties.

B.) Bottom-up Approaches: In this approach, It involves atom-by-atom, molecule-by-molecule, or cluster-by-cluster manipulation for the synthesis of nanostructures. It means that atoms are assembled into molecules, molecules are assembled into clusters, and the clusters are assembled to form the nanoparticle. This approach relies on self-assembly and controlled growth to create the desired nanoscale structure.

The bottom-up approach is based on the principle of molecular recognition (that is self-assembly). Self-assembly means growing more and more things of one's kind from themselves. The principle of self-assembly (shake and bake) involves assembling precursors in random positions and orientations, providing energy (shaking) to enable them to explore configuration space. The hugeness of this space suggests that a convergent pathway is inherent in the process in order to allow it to be completed in a reasonable time. Once the precursors are in position, "baking" may be required to strengthen the bonds connecting them and fix the final object permanently.

Some common techniques used in the bottom-up approach include:

a.) Chemical Synthesis: Chemical reactions are used to build up nanomaterials from atomic or molecular precursors, allowing precise control over the final product's size and shape.

b.) Self-Assembly: Molecules or nanoparticles are designed to interact in a way that leads to their spontaneous arrangement into specific nanoscale structures.

c.) Vapor Deposition: Nanoscale materials are grown layer by layer on a substrate by allowing precursor gases to react on its surface.

d.) Sol-Gel Process: Nanomaterials are formed by hydrolyzing metal alkoxides in a solution, followed by gelation and controlled drying.

The bottom-up approach allows for greater control over the nanomaterial's properties and can potentially lead to unique properties not found in their bulk counterparts. However, it can be more challenging to scale up these processes for mass production compared to the top-down approach.

Advantages of Bottom-down Approaches:

1.) The bottom-up approach allows for precise control at the atomic or molecular level, leading to well-defined nanomaterials with specific properties.

2.) Since the materials are assembled from individual atoms or molecules, there are fewer defects compared to the top-down approach.

3.) Bottom-up synthesis can result in unique nanomaterials with properties not easily achievable through top-down approaches.

Disadvantages of Bottom-down Approaches:

1.) Bottom-up approaches can be more complex and expensive, especially for large-scale production.

2.) Some bottom-up approaches may not be easily scalable for mass production, limiting their industrial applicability.

3.) Achieving high purity and reproducibility in bottom-up synthesis can be challenging due to the intricate processes involved.

Both the top-down and bottom-up approaches have their advantages and disadvantages, and researchers often choose the most suitable method based on the specific properties and applications of the nanomaterial they aim to create. Additionally, a combination of these approaches, known as "hybrid approaches," can be used to achieve even more complex nanoscale structures with tailored properties.

The synthesis of nanomaterial according to technique:

1.) Physical Methods

2.) Chemical Methods

3.) Biological Methods

1.) Physical Methods: Nanoparticles may be synthesized using a number of physical methods which are listed below. These methods are of two types viz mechanical type and vapor deposition type. These methods work at very high temperatures. The highest working temperatures are usually greater than $350^{\circ}C$.

• Arc Discharge Method
• Electron beam Lithography
• Mechanical Grinding
• Inert Gas condensation
• Ion Implantation
• Ball Milling
• Spray pyrolysis
• Vapour phase synthesis

2.) Chemical Methods: These are very simple methods for synthesizing nanoparticles and are very less expensive. The highest working temperatures are usually below $350^{\circ}C$. Large quantities of materials can be produced with a variety of sizes and shapes of particles. A number of chemical methods are shown below.

• Coprecipitation Method
• Microemulsion Method
• Electrochemical Method (Electrolysis)
• Chemical reduction of metal Salts Pyrolysis
• Phytochemical (irradiation Method)
• Solvothermal synthesis
• Sol-Gel process
• Sonochemical methods

3.) Biological Methods: Biological Methods are based on the use of microorganisms (Fungi, yeast, Bacteria, etc.) or plant extracts (and enzymes) or the use of templates such as DNA, viruses and membranes, etc. This type of synthesis is environmentally friendly and least toxic and therefore called green synthesis. Various green methods for the synthesis of nanoparticles are shown below.

• Using Plants and their extracts
• Using microorganisms (Bacteria, Fungi, and actinomycetes)
• Using algae
• Using enzymes and biomolecules
• Using industrial and agricultural wastes

Read More

Basics and types of Nanomaterials

What is a nanoparticle?

Those particles that have size ranges between $1$ to $100 \: nanometres$ are called a nanoparticle. The particles are undetectable by the human eye. There are significant differences in the properties (like magnetic, electrical, Structural, Mechanical, and optical properties) of nanoparticles and bulk materials.

What is nanomaterial?

Those materials that have at least one dimension should be in nanometres i.e. $10^{-9}m$ are called nanomaterials. The prefix 'nano' means a billionth $(10^{-9})$.

Types of nanomaterials

There are two types of nanomaterial that can be classified:

A.) On the basis of dimension

B.) On the basis of material

A.) On the basis of dimensions: According to Siegel, nanostructured materials are classified on the basis of dimension:

1.) Three-dimensional nanomaterials (Bulk Nanomaterial)

2.) Two-dimensional nanomaterials (Quantum Well)

3.) One-dimensional nanomaterials (Quantum Wire)

4.) Zero-dimensional nanomaterials (Quantum Dot)

1.) Three-dimensional nanomaterials (Bulk Nanomaterial): These nanomaterials have not confined to the nanoscale range in any dimension. These materials have three arbitrary dimensions above the nanoscale i.e. $100 nm$. The bulk three-dimensional nanomaterials are composed of a multiple arrangement of nano-size crystals in different orientations. The three-dimensional nanomaterials or bulk nanomaterials can be used as bundles of nanowires, dispersion of nanoparticles, and nanotubes as well as multi-nano layers (polycrystals) in which the $0D$, $1D$, and $2D$ structural elements are in very close contact with each other and form interfaces.

2.) Two-dimensional nanomaterials (Quantum Well): These nanomaterials have one dimension in the nanoscale. It is also called a quantum well. This means that the particles of material are confined only along one dimension. The 2D nanomaterials exhibit plate-like shapes. It includes nanofilms, nanolayers, and nanocoatings with nanometre thickness.

3.) One-dimensional nanomaterials (Quantum Wire): These nanomaterials have two dimensions in the nanoscale. It is also called quantum wire. This means that the particles of material are confined in two dimensions. This leads to needle-shaped nanomaterials. It includes nanofibers, nanotubes, Nanorods, and nanowires.

4.) Zero-dimensional nanomaterials (Quantum Dot): These nanomaterials have all three dimensions in the nanoscale i.e., no dimensions are greater than $100 nm$. It is also called quantum dots. This means that the particles of material are confined in all three dimensions. It includes Nanospheres and nanoclusters.

B.) On the basis of materials: Nanomaterials can be categorized on the basis of material into four types such as:

1.) Inorganic-based nanomaterials (Metal-based materials ):

2.) Carbon-based nanomaterials:

3.) Organic-based nanomaterials (Dendrimers):

4.) Composite-based nanomaterials.

1.) Inorganic-based nanomaterials (Metal-based materials ): Generally, inorganic-based nanomaterials include different metal and metal oxide nanomaterials.

Examples of metal-based inorganic nanomaterials - silver $(Ag)$, gold $(Au)$, aluminum $(Al)$, cadmium $(Cd)$, copper $(Cu)$, iron $(Fe)$, zinc $(Zn)$, and lead $(Pb)$ nanomaterials.

Examples of metal oxide-based inorganic nanomaterials- zinc oxide $(ZnO)$, copper oxide $(CuO)$, magnesium aluminum oxide $(MgAl_{2}O_{4})$, titanium dioxide $(TiO_{2})$, cerium oxide $(CeO_{2})$, iron oxide $(Fe_{2}O_{3})$, silica $(SiO_{2})$, and iron oxide $(Fe_{3}O_{4})$, etc.

(2) Carbon-based nanomaterials: Carbon-based nanomaterials are graphene, fullerene, single-walled carbon nanotube, multi-walled carbon nanotube, carbon fiber, activated carbon, and carbon black.

(3) Organic-based nanomaterials (Dendrimers): The organic-based nanomaterials or dendrimers (i.e. Dendrimers are repetitively branched molecules. Dendrimers name comes from the Greek word ‘dendron’ which means tree.) are formed from organic materials that do not include carbon materials, for instance, dendrimers, cyclodextrin, liposome, and micelle.

(4) Composite-based nanomaterials: The composite nanomaterials can be any combination of all nanomaterials like metal-based, carbon-based, metal oxide-based, and organic-based nanomaterials. These composite nanomaterials have very complicated structures like a metal-organic framework.

Read More

Electric field intensity due to thick hollow non-conducting sphere

Electric field intensity at different points in the field due to uniformly charged thick hollow non-conducting sphere: Let us consider, A hollow non-conducting sphere of inner radius $r_{1}$ and outer radius $r_{2}$ in which $+q$ charge is evenly distributed evenly in the entire volume of the sphere. If $\rho$ is the volume charge density then electric field intensity at different points on the electric field of the thick hollow non-conducting sphere:

1. Electric field intensity outside the thick hollow non-conducting sphere
2. Electric field intensity on the surface of the thick hollow non-conducting sphere
3. Electric field intensity at an internal point of the non-thick hollow conducting sphere

1. Electric field intensity outside the thick hollow non-conducting sphere:

Let us consider, A point $P$ is outside the sphere which is at a $r$ distance from the center point $O$ of the sphere. The direction of electric flux is radially outward in the sphere. So the direction of the electric field vector and the small area vector will be in the same direction i.e. ($\theta =0^{\circ}$). Here $\overrightarrow {dA}$ is a small area element so the small amount of electric flux will pass through this area i.e. →

$ d\phi_{E}= \overrightarrow {E}\cdot \overrightarrow{dA}$

$ d\phi_{E}= E\:dA\: cos\: 0^{\circ} \quad \left \{\because \theta=0^{\circ} \right \}$

$ d\phi_{E}= E\:dA \qquad (1) \quad \left \{\because cos\:0^{\circ}=1 \right \}$

The electric flux passes through the entire Gaussian surface, So integrate the equation $(1)$

$ \oint d\phi_{E}= \oint E\:dA $

$ \phi_{E}=\oint E\:dA\qquad (2)$

According to Gauss's law:

$ \phi_{E}=\frac{q}{\epsilon_{0}}\qquad (3)$

From equation $(2)$ and equation $(3)$, we can write as

$ \frac{q}{\epsilon_{0}}=\oint E\:dA$

$ \frac{q}{\epsilon_{0}}= E\oint dA$

For entire Gaussian spherical surface is

$\oint {dA}=4\pi r^{2}$.

So from the above equations

$ \frac{q}{\epsilon_{0}}= E(4\pi r^{2})$

$ E=\frac{1}{4\pi \epsilon_{0}}\frac{q}{r^{2}} \qquad(4)$

From the above equation, we can conclude that the behavior of the electric field at the external point due to the uniformly charged thick hollow non-conducting sphere is the same as the point charge i.e. like the entire charge is placed at the center.

Since the sphere is a non-conductor so the charge is distributed in the entire volume of the sphere. So charge on the thick hollow sphere-

$q=\frac{4}{3} \pi \left(r_{2}^{3}-r_{1}^{3} \right) \rho $

Substitute this value of charge $q$ in the above equation $(4)$, Therefore,

$ E=\frac{1}{4\pi \epsilon_{0}}\frac{4\pi \left(r_{2}^{3}-r_{1}^{3} \right)\: \rho}{3r^{2}}$

$ E=\frac{\rho}{3 \epsilon_{0}}\frac{\left(r_{2}^{3}-r_{1}^{3} \right)}{r^{2}}$

This equation describes the electric field intensity at the external point of the thick hollow non-conducting sphere.

2. Electric field intensity on the surface of the thick hollow non-conducting sphere:

If point $P$ is placed on the surface of the thick hollow non-conducting sphere i.e. ($r=r_{2}$). so electric field intensity on the surface of the thick hollow non-conducting sphere:

$ E=\frac{\rho}{3\epsilon_{0}}\frac{\left(r^{3}-r_{1}^{3} \right)}{r^{2}}$

3. Electric field intensity at an internal point of the thick hollow non-conducting sphere:

If point $P$ is placed inside the sphere at the distance $r$ from the origin $O$, the electric flux which is passing through the Gaussian surface

$ \phi_{E}= E.4\pi r^{2}$

Where $\phi_{E}=\frac{q'}{\epsilon_{0}}$

$ \frac{q'}{\epsilon_{0}}=E.4\pi r^{2}$

Where $q'$ is part of charge $q$ which is enclosed with Gaussian Surface

$ E=\frac{1}{4 \pi \epsilon_{0}} \frac{q'}{r^{2}} \qquad \qquad (5)$

The charge is distributed uniformly in the entire volume of the sphere so volume charge density $\rho$ will be the same as the entire sphere i.e.

$ \rho=\frac{q}{\frac{4}{3}\pi \left(r_{2}^{3}-r_{1}^{3} \right)}=\frac{q'}{\frac{4}{3}\pi \left(r^{3}-r_{1}^{3} \right)}$

$ \frac{q}{\left(r_{2}^{3}-r_{1}^{3} \right)}=\frac{q'}{\left(r^{3}-r_{1}^{3} \right)}$

$ q'=q\frac{\left(r^{3}-r_{1}^{3} \right) }{\left(r_{2}^{3}-r_{1}^{3} \right)}$

Put the value of $q'$ in equation $(5)$, so

$ E=\frac{1}{4\pi \epsilon_{0}}\frac{q} {r^{2}} \frac{\left(r^{3}-r_{1}^{3} \right) }{\left(r_{2}^{3}-r_{1}^{3} \right)}$

Where $q=\frac{4}{3} \pi \left(r_{2}^{3}-r_{1}^{3} \right) \rho $. So above equation can be written as:

$ E=\frac{1}{4\pi \epsilon_{0}}\frac{\frac{4}{3} \pi \left(r_{2}^{3}-r_{1}^{3} \right) \rho } {r^{2}} \frac{\left(r^{3}-r_{1}^{3} \right) }{\left(r_{2}^{3}-r_{1}^{3} \right)}$

$ E=\frac{1}{4\pi \epsilon_{0}}\frac{\frac{4}{3} \pi \left(r_{2}^{3}-r_{1}^{3} \right) \rho } {r^{2}} \frac{\left(r^{3}-r_{1}^{3} \right) }{\left(r_{2}^{3}-r_{1}^{3} \right)}$

$E=\frac{ \rho }{3 \epsilon_{0}} \frac{\left(r^{3}-r_{1}^{3} \right)}{r^{2}}$

Read More

Characteristics of Electromagnetic Wave

Electromagnetic Wave:

An electromagnetic wave is the combined effect of an electric field and magnetic field which carry energy from one place to another.

When an electric field and the magnetic field are applied perpendicular to each other then a wave propagates perpendicular to both the electric field and the magnetic field. This wave is called the electromagnetic wave.

Characteristics of Electromagnetic Wave:

1.) Electric and Magnetic Fields: Electromagnetic waves are produced through the mutually perpendicular interaction of electric and magnetic fields. The propagation of the wave is also perpendicular to both the electric field and the magnetic field.

2.) Wave Nature of electromagnetic waves: Electromagnetic waves are characterized by their wave-like behavior, so they exhibit the properties such as wavelength, frequency, amplitude, and velocity. This wave-like behavior of electromagnetic waves can undergo phenomena like interference, diffraction, and polarization.

3.) The spectrum of an electromagnetic wave: In the spectrum of electromagnetic waves, All the wavelengths and frequencies of the waves are included such as radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays etc.

4.) Speed of electromagnetic wave: The speed of electromagnetic waves in free space or vacuum is equal to the speed of light in free space or vacuum, which is approximately 299,792 $Km/s$.

5.) Transverse waves of an electromagnetic wave: Electromagnetic waves are transverse waves which mean that the oscillations of the electric and magnetic fields occur perpendicular to the direction of wave propagation.

6.) Dual nature of electromagnetic wave: Electromagnetic waves have both wave-like and particle-like behavior. They can be described as a stream of particles called photons, each photon carrying a specific amount of energy (quantum). This duality is described by the wave-particle duality principle in quantum mechanics.

7.) Energy transfer in electromagnetic waves: Electromagnetic waves transport energy through space. The amount of energy carried by each wave depends on its frequency. Higher frequency waves, such as gamma rays and X-rays, carry more energy than lower frequency waves like radio waves.

8.) Absorption, Reflection, and Transmission of Electromagnetic waves: Electromagnetic waves can be absorbed by certain materials, reflected off surfaces, or transmitted through transparent substances. The behavior of waves at boundaries depends on factors such as the angle of incidence, the nature of the material, and the frequency of the wave.

9.) Electromagnetic Induction of electromagnetic wave: When electromagnetic waves interact with conductive materials or circuits, they can induce electric currents or voltages. This principle is the basis for technologies like antennas, wireless communication, and electromagnetic sensors.

10.) Electromagnetic Interactions of electromagnetic waves: Electromagnetic waves can interact with matter in various ways, including absorption, scattering, and emission. These interactions are utilized in fields such as optics, spectroscopy, medical imaging, and telecommunications.

Read More

Electric and magnetic field vector are mutually perpendicular to each other in electromagnetic wave

In electromagnetic waves, the electric field vector and magnetic field vector are mutually perpendicular to each other (Proof)

The general solution of the wave equation for the electric field vector and magnetic field vector are respectively given below

$\overrightarrow{E}= E_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)} \qquad(1)$

$\overrightarrow{B}= B_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)} \qquad(2)$

Here $E_{\circ}$ and $B_{\circ}$ are the complex amplitude of electric field vector $\overrightarrow{E}$ and magnetic field vector $\overrightarrow{B}$ respectively and $\overrightarrow{k}$ is the propagation constant.

Now

$\overrightarrow{\nabla} \times \overrightarrow{E}= \left( \hat{i} \frac{\partial}{\partial x} + \hat{i} \frac{\partial}{\partial x} +\hat{i} \frac{\partial}{\partial x} \right). \left( \hat{i}E_{x} + \hat{j}E_{y} + \hat{k}E_{z} \right) $

$\overrightarrow{\nabla} \times \overrightarrow{E} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ E_{x} & E_{y} & E_{z} \\ \end{vmatrix}$

$\overrightarrow{\nabla} \times \overrightarrow{E} = \hat{i} \left[ \frac{\partial E_{z}}{\partial y} -\frac{\partial E_{y}}{\partial z} \right] - \hat{j} \left[ \frac{\partial E_{z}}{\partial x} -\frac{\partial E_{x}}{\partial z} \right] + \hat{k} \left[ \frac{\partial E_{y}}{\partial x} -\frac{\partial E_{x}}{\partial y } \right] \qquad(3)$

Here $E_{x}$, $E_{y}$ and $E_{z}$ are the component of electric field vector $\overrightarrow{E}$ in $x$, $y$, and $z$ direction. So the component form of the equation $(1)$

$\overrightarrow{E_{x}}= E_{\circ x} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)}$

$\overrightarrow{E_{y}}= E_{\circ y} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)}$

$\overrightarrow{E_{z}}= E_{\circ z} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)}$

We know that:

$\overrightarrow{k}.\overrightarrow{r}=\left( \hat{i}k_{x} + \hat{j}k_{y} + \hat{k}k_{z} \right). \left( \hat{i}x + \hat{j}y +\hat{k}z \right)$

$\overrightarrow{k}.\overrightarrow{r}=x k_{x} + yk_{y} + zk_{z} $

So above equation can be written as:

$\overrightarrow{E_{x}}= E_{\circ x} e^{[i(x k_{x} + yk_{y} + zk_{z}) - \omega t)]} \qquad(4)$

$\overrightarrow{E_{y}}= E_{\circ y} e^{[i(x k_{x} + yk_{y} + zk_{z}) - \omega t)]} \qquad(5)$

$\overrightarrow{E_{z}}= E_{\circ z} e^{[i(x k_{x} + yk_{y} + zk_{z}) - \omega t)]} \qquad(6)$

Now find that derivative from the equation $(4)$, equation $(5)$, and equation $(6)$ then substitute these values in equation $(3)$, So we get

$\left[ \frac{\partial E_{z}}{\partial y} -\frac{\partial E_{y}}{\partial z} \right] = \frac{\partial E_{z}}{\partial y} - \frac{\partial E_{y}}{\partial z} $

$\left[ \frac{\partial E_{z}}{\partial y} -\frac{\partial E_{y}}{\partial z} \right] = \frac{\partial}{\partial y} \left( E_{\circ z} e^{[i(x k_{x} + yk_{y} + zk_{z}) - \omega t)]} \right) -\frac{\partial}{\partial z} \left( E_{\circ y} e^{[i(x k_{x} + yk_{y} + zk_{z}) - \omega t)]} \right) $

$\left[ \frac{\partial E_{z}}{\partial y} -\frac{\partial E_{y}}{\partial z} \right] = \left(i k_{y} E_{z} - i k_{z} E_{y} \right) $

$\left[ \frac{\partial E_{z}}{\partial y} -\frac{\partial E_{y}}{\partial z} \right] = i \left( k_{y} E_{z} - k_{z} E_{y} \right) \qquad(7)$

Similarly

$\left[ \frac{\partial E_{z}}{\partial x} -\frac{\partial E_{x}}{\partial z} \right] = i \left( k_{x} E_{z} - k_{z} E_{x} \right) \qquad(8)$

$\left[ \frac{\partial E_{y}}{\partial x} -\frac{\partial E_{x}}{\partial y} \right] = i \left( k_{x} E_{y} - k_{y} E_{x} \right) \qquad(9)$

Now substitute the value of equation $(7)$, equation $(8)$, and equation $(9)$ in equation $(3)$

$\overrightarrow{\nabla} \times \overrightarrow{E} = i \left[\hat{i} \left( k_{y} E_{z} - k_{z} E_{y} \right) - \hat{j} \left( k_{x} E_{z} - k_{z} E_{x} \right) _ \hat{k} \left( k_{x} E_{y} - k_{y} E_{x} \right) \right]$

$\overrightarrow{\nabla} \times \overrightarrow{E} = i \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ k_{x} & k_{y} & k_{z} \\ E_{x} & E_{y} & E_{z} \\ \end{vmatrix}$

$\overrightarrow{\nabla} \times \overrightarrow{E} = i \left( \overrightarrow{k} \times \overrightarrow{E} \right) \qquad(10)$

According to Maxwell's third equation

$\overrightarrow{\nabla} \times \overrightarrow{E}= - \frac{\partial \overrightarrow{B}}{\partial t}$

$\overrightarrow{\nabla} \times \overrightarrow{E}= - \frac{\partial}{\partial t} \left( B_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)} \right) \quad \left\{From \: equation \: (2)\right\}$

$\overrightarrow{\nabla} \times \overrightarrow{E}= i \omega \overrightarrow{B} \qquad(11)$

From equation $(10)$ and equation $(11)$

$i \left( \overrightarrow{k} \times \overrightarrow{E} \right) = i \omega \overrightarrow{B}$

$ \left( \overrightarrow{k} \times \overrightarrow{E} \right) = \omega \overrightarrow{B} \qquad(12)$

$\therefore$ Magnetic field vector $(\overrightarrow{B})$ is perpendicular to both electric field vector $(\overrightarrow{E})$ and propagation of wave vector $(\overrightarrow{k})$.

Similarly, from $\overrightarrow{\nabla} \times \overrightarrow{B}$, we get

$ \left( \overrightarrow{k} \times \overrightarrow{B} \right) = -\frac{\omega}{c} \overrightarrow{E} \qquad(13)$

Thus, In an electromagnetic wave, the electric field and magnetic field vector are perpendicular to each other and also to the direction of propagation of the wave.

Read More

Transverse Nature of Electromagnetic Wave

Electromagnetic waves are transverse in nature: (Proof)

The general solution of the wave equation for the electric field and magnetic field are respectively given below

$\overrightarrow{E}= E_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)} \qquad(1)$

$\overrightarrow{B}= B_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)} \qquad(2)$

Here $E_{\circ}$ and $B_{\circ}$ are the complex amplitude of electric field vector $\overrightarrow{E}$ and magnetic field vector $\overrightarrow{B}$ respectively and $\overrightarrow{k}$ is the propagation constant.

Now

$\overrightarrow{\nabla}. \overrightarrow{E}= \left( \hat{i} \frac{\partial}{\partial x} + \hat{i} \frac{\partial}{\partial x} +\hat{i} \frac{\partial}{\partial x} \right). \left( \hat{i}E_{x} + \hat{j}E_{y} + \hat{k}E_{z} \right) $

$\overrightarrow{\nabla}. \overrightarrow{E}= \frac{\partial}{\partial x} \left(E_{x} \right)+ \frac{\partial}{\partial y} \left(E_{y} \right) + \frac{\partial}{\partial z} \left(E_{z} \right) \qquad(3)$

Here $E_{x}$, $E_{y}$ and $E_{z}$ are the component of electric field $\overrightarrow{E}$ in $x$, $y$, and $z$ direction. So the component form of the equation $(1)$

$\overrightarrow{E_{x}}= E_{\circ x} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)}$

$\overrightarrow{E_{y}}= E_{\circ y} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)}$

$\overrightarrow{E_{z}}= E_{\circ z} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)}$

Here

$\overrightarrow{k}.\overrightarrow{r}=\left( \hat{i}k_{x} + \hat{j}k_{y} + \hat{k}k_{z} \right). \left( \hat{i}x + \hat{j}y +\hat{k}z \right)$

$\overrightarrow{k}.\overrightarrow{r}=x k_{x} + yk_{y} + zk_{z} $

So above equation can be written as:

$\overrightarrow{E_{x}}= E_{\circ x} e^{[i(x k_{x} + yk_{y} + zk_{z}) - \omega t)]} \qquad(4)$

$\overrightarrow{E_{y}}= E_{\circ y} e^{[i(x k_{x} + yk_{y} + zk_{z}) - \omega t)]} \qquad(5)$

$\overrightarrow{E_{z}}= E_{\circ z} e^{[i(x k_{x} + yk_{y} + zk_{z}) - \omega t)]} \qquad(6)$

Now find that derivative of the equation $(4)$ along the direction of $x$ then

$\frac{\partial E_{x}}{\partial x} = i k_{x} E_{\circ x} e^{[i(x k_{x} + yk_{y} + zk_{z}) - \omega t)]} $

$\frac{\partial E_{x}}{\partial x} = i k_{x} E_{x} $

Similarly, the derivative of the equation $(5)$, and equation $(6)$ along the direction of $y$ and $z$ then

$\frac{\partial E_{y}}{\partial y} = i k_{y} E_{y} $

$\frac{\partial E_{z}}{\partial z} = i k_{z} E_{z} $

Now substitute the value of $\frac{\partial E_{x}}{\partial x}$, $\frac{\partial E_{y}}{\partial y}$, and $\frac{\partial E_{z}}{\partial z}$ in equation $(3)$

$\overrightarrow{\nabla}. \overrightarrow{E}= i k_{x} E_{x} + i k_{y} E_{y} + i k_{z} E_{z}$

$\overrightarrow{\nabla}. \overrightarrow{E}= i \left( k_{x} E_{x} + k_{y} E_{y} + k_{z} E_{z} \right)$

$\overrightarrow{\nabla}. \overrightarrow{E}= i \left( \hat{i}k_{x} + \hat{j}k_{y} + \hat{k}k_{z} \right). \left( \hat{i}E_{x} + \hat{j}E_{y} + \hat{k} E_{z} \right)$

$\overrightarrow{\nabla}. \overrightarrow{E}= i \left( \overrightarrow {k} . \overrightarrow {E} \right) \qquad(7)$

From Maxwell's first equation in free space:

$\overrightarrow{\nabla}. \overrightarrow{E}= 0 \qquad(8)$

From equation $(7)$ and equation $(8)$

$i \left( \overrightarrow {k} . \overrightarrow {E} \right)=0$

$ \overrightarrow {k} . \overrightarrow {E} = 0$

From the above equation, we can conclude that the electric field is perpendicular to the direction of propagation of the wave i.e. $\overrightarrow {E}\perp \overrightarrow {k}$

Similarly, The same result is obtained from $\overrightarrow{\nabla}. \overrightarrow {B}$ i.e. $ \overrightarrow {k} . \overrightarrow {B} = 0$, So we can conclude that the magnetic field is also perpendicular to the direction of propagation i.e. $\overrightarrow {B}\perp \overrightarrow {k}$

Thus, "The electromagnetic waves are transverse in nature"

Read More

Diamagnetic Substances and Its properties

Diamagnetic Substances :

Those substances, which are placed in the external magnetic field then they weakly magnetize in the opposite direction of the external magnetic field, are called diamagnetic substances. The susceptibility $\chi_{m} $ of diamagnetic substances is small and negative. Further, When diamagnetic substance placed in magnetic field then the flux density of the diamagnetic substance is slightly less than that in the free space. Thus, the relative permeability of diamagnetic substance $\mu_{r}$, is slightly less than 1.

Properties of Diamagnetic substances:

1. When a rod of a diamagnetic material is suspended freely between external magnetic poles (i.e. Between North and South Poles) then its axis becomes perpendicular to the external magnetic field $B$ (Figure). The poles produced on the two sides of the rod are similar to the poles of the external magnetic field.

2. In a non-uniform magnetic field, a diamagnetic substance tends to move from the stronger magnetic field to the weaker magnetic field. If a diamagnetic liquid is taken in a watch glass placed on two magnetic poles very near to each other, then the liquid is depressed in the middle as shown in figure below(Figure) where the field is strongest. Now, if the distance between the poles is increased, the liquid rises in the middle, because now the field is strongest near the poles.

3. If the solution of diamagnetic substance is poured into a U-tube and apply the strong magnetic field into one arm of this U-tube then the level of the solution in that arm is depressed. As shown in the figure below:

4. When diamagnetic gas molecules are passed between the poles of a magnet then diamagnetic gas molecules are spread across the field.

5. The susceptibility of a diamagnetic substance is independent of temperature.

Explanation of Diamagnetism on the Basis of Atomic Model:

The property of diamagnetism is generally found in those substances whose atoms (or ions or molecules) have an 'even' number of electrons. These even numbers of electron form pairs. In each pair of electrons, the spin of one electron is opposite to the other. So, the magnetic moment of one electron is opposite to the others because of that, the effect of magnetic dipole moments are neutralized by each other. As such, the net magnetic dipole moment of an atom (or ion or molecule) of a diamagnetic substance is zero.

When a diamagnetic substance is placed in an external magnetic field $B$ then this external magnetic field modifies the motion of the electrons in the atoms (or ions or molecules). Due to this, In each pair of electrons, the spin of one electron is become fast (Lenz's Law) and the other is slow due to that , the net magnetic dipole moment of the paired electron does not zero. Thus, a small magnetic dipole moment is induced in each atom of the substance (or ion or molecule) which is directly proportional to the magnetic field $B$ and opposite to its direction. Hence, the diamagnetic substance is magnetized opposite to the external magnetic field $B$, and the field lines become less dense inside the diamagnetic substance compared to those outside.

If the temperature of the diamagnetic substance is changed, there is no effect on its diamagnetic property. Thus, diamagnetism is temperature-independent.

Read More

Distinction between Spontaneous and Stimulated Emission of Radiation

Some of the differences between spontaneous and stimulated emission of radiation are given as follows:

1. In spontaneous emission, an atom in excited state falls to the ground state on its own without any incident photon while in stimulated emission transition takes place by stimulating photons or by an external agency.

2. In stimulated emission for each incident photon there are two outgoing photons in the same direction while in spontaneous emission the emitted photons move randomly in any direction.

3. The photons emitted in spontaneous emiss ion have a random phase and hence are incoherent while in stimulated emission the emitted photons are in phase and hence are coherent.

4. The rate of spontaneous emission is proportional to only the number of atoms in the excited state while the rate of stimulated emission is proportional to the number of atoms left in the excited state as well as on the energy density of the incident radiation.

5. In stimulated emission of radiation an amplified beam is achieved while in spontaneous emission there is no such amplification.

6. The light emitted through the spontaneous emission is not monochromatic while in stimulated transition monochromatic radiation is obtained.

7. Spontaneous emission is not controllable from outside while stimulated emission is controllable from outside.

8. In spontaneous emission, the net intensity is proportional to the number of radiating atoms while in stimulated emission it is proportional to the square of the number of radiation atoms.

Read More

Minimum Energy Or Zero Point Energy of a Particle in an one dimensional potential box or Infinite Well

Zero Point Energy of a Particle in an Infinite Well Potential Well:

The normalized wave function or eigenwave function:

$\psi_{n}(x) = \sqrt{\frac{2}{L}} sin \left( \frac{n\pi x}{L} \right)$

The probability density

$| \psi_{n}(x)|^{2} = \frac{2}{L} sin^{2} \left( \frac{n\pi x}{L} \right)$

The energy of a particle in a one-dimensional box or infinite potential well:

$E_{n}=\frac{n^{2}h^{2}}{8 mL^{2}}$

Where $n$ is called the quantum number and $n=1,2,3,4..........$ For $n=0, \psi_{n}(x)=0$ and $| \psi_{n}(x)|^{2}=0$. This shows that for $n=0$ $| \psi_{n}(x)|^{2}=0$ will be zero everywhere in the box which means that the probability of finding the particle inside the box is zero. i.e. particle is not present at all inside the box. Thus $n=0$ is not possible.

If $n\neq 0$ then $E \neq 0$. This means that the minimum energy of the particle in the box will not be zero. The minimum energy value will be obtained for the next lowest value of $n$ i.e. for $n=1$, which is

$E_{1}=\frac{h^{2}}{8 mL^{2}}$

This minimum energy of the particle is often called zero point energy which is finite inside the box. According to classical mechanics, the minimum value $E=0$ is also permissible.

Read More

Bohr's Theory of Hydrogen-Like Atoms

A hydrogen-like atom consists of a very small positively-charged nucleus and an electron revolving in a stable circular orbit around the nucleus.

The radius of electrons in stationary orbits:

Let the charge, mass, velocity of the electron and the radius of the orbit is respectively  $e$, $m$, and $v$  and $r$. The $+ze$ is the positive charge on the nucleus where $Z$ is the atomic number of the atom. As We know that when an electron revolves around the nucleus then the centripetal force on an electron is provided by the electrostatic force of attraction between the nucleus and an electron, we have

$\frac{mv^{2}}{r}=\frac{1}{4 \pi \epsilon_{\circ}} \frac{(Ze)(e)}{r^{2}}$

$mv^{2}=\frac{Ze^{2}}{4 \pi \epsilon_{\circ} r} \qquad(1)$

According to the first postulate of Bohr's model of the atom, the angular momentum of the electron is

$mvr=n \frac{h}{2 \pi} \qquad(2)$

Where $n \: (=1,2,3,.....)$ is quantum number.

Now squaring equation $(2)$ and dividing by equation $(1)$, we get

$r=n^{2} \frac{h^{2} \epsilon_{\circ}}{\pi m Z e^{2}} \qquad(3)$

The above equation is for the radii of the permitted orbits. From the above equation, this concluded that

$r \propto n^{2}$

Since, $n =1,2,3,.....$ it follows that the radii of the permitted orbits increased in the ratio $1:4:9:16:,.....$ from the first orbit.

Bohr's Radius:

For Hydrogen Atom $(z=1)$, The radius of the atom of the first orbit $(n=1)$ will be

$r_{1}= \frac{h^{2} \epsilon_{\circ}}{\pi m e^{2}}$

This is called Bohr's radius and its value is $0.53 A^{\circ}$. Since $r \propto n^{2}$, the radius of the second orbit of the hydrogen atom will be $( 4 \times 0.53 A^{\circ}) $ and that of the third orbit $9 \times 0.53 A^{\circ}$

The velocity of electrons in stationary orbits:

The velocity of the electron in permitted orbits can be obtained by the formula of equation $(2)$

$v=n\frac{h}{2 \pi m r}$

Now put the value of $r$ in above eqaution from equation $(3)$, we get

$v=\frac{Ze^{2}}{2 h \epsilon_{\circ}} \left( \frac{1}{n} \right) \quad(4)$

Thus $v \propto \frac{1}{n}$

This shows that the velocity of the electron is maximum in the lowest orbit $n=1$ and as goes on higher orbits velocity decreases.

For Hydrogen Atom $(z=1)$, The velocity of electron to move in the first orbit $(n=1)$ is

$v_{1}=\frac{e^{2}}{2h\epsilon_{\circ}}$

Its value is $2.19 \times 10^{6} m/sec$

Note:

$\frac{v_{1}}{c}= \frac{2.19 \times 10^{6}}{3 \times 10^{8}} =\frac{1}{137}$

Thus, $\frac{v_{1}}{c}$ or $\frac{e^{2}}{2h\epsilon_{\circ}}$ is a pure number. It is called the "Fine Structure Constant" and is denoted by $\alpha$

The energy of electrons in stationary orbits:

The total energy $E$ of a moving electron in an orbit is the sum of kinetic energies and potential energies. The kinetic energy of moving the electron in a stationary orbit is:

$K=\frac{1}{2} m v^{2}$

Now susbtitute the value of $v$ from equation $(1)$, we get

$K=\frac{ze^{2}}{8 \pi \epsilon_{\circ} r}$

The potential energy of a moving electron in an orbit of radius $r$ due to the electrostatic attraction between nucleus and electron is given by

$U=\frac{1}{4 \pi \epsilon_{\circ}} \frac{(Ze)(-e)}{r}$

$U=-\frac{Ze^{2}}{4 \pi \epsilon_{\circ} r} $

The total energy of the electron is

$E=K+U$

$E=\frac{ze^{2}}{8 \pi \epsilon_{\circ} r} -\frac{Ze^{2}}{4 \pi \epsilon_{\circ} r} $

$E=-\frac{ze^{2}}{8 \pi \epsilon_{\circ} r}$

Subtituting the value of $r$ in above equation from equation $(3)$, we get

$E=-\frac{mz^{2}e^{4}}{8 \epsilon^{2}_{\circ} h^{2}} \left( \frac{1}{n^{2}} \right) \qquad(5)$

This is the equation for the energy of the electron in the $n^{th}$ orbit.

Suppose, Excited state energy is $E_{2}$ and lower state energy is $E_{1}$. So the energy difference between these two states is:

$E_{2}- E_{1}= \frac{mz^{2}e^{4}}{8 \epsilon^{2}_{\circ} h^{2}} \left( \frac{1}{n_{1}^{2}} -\frac{1}{n_{2}^{2}} \right) \qquad(6)$

According to the third postulate of Bohr's Atomic model, the frequency $\nu$ of the emitted electromagnetic wave:

$\nu=\frac{E_{2}- E_{1}}{h}$

$\nu=\frac{mz^{2}e^{4}}{8 \epsilon^{2}_{\circ} h^{3}} \left( \frac{1}{n_{1}^{2}} -\frac{1}{n_{2}^{2}}\right)$

The corresponding wavelength $\lambda$ of the emitted electromagnetic radiation is given by

$\frac{c}{\lambda}=\frac{mz^{2}e^{4}}{8 \epsilon^{2}_{\circ} h^{3}} \left( \frac{1}{n_{1}^{2}} -\frac{1}{n_{2}^{2}}\right)$

$\frac{1}{\lambda}=\frac{mz^{2}e^{4}}{8 \epsilon^{2}_{\circ} c h^{3}} \left( \frac{1}{n_{1}^{2}} -\frac{1}{n_{2}^{2}}\right) \qquad(7)$

Where $\frac{1}{\lambda}$ is called "wave number" (i.e. number of waves per unit length). In the last equation$(7)$, the quantity $\frac{m e^{4}}{8 \epsilon^{2}_{\circ} c h^{3}}$ is a constant an it is known as "Rydberg Constant (R)". That is

$R = \frac{me^{4}}{8 \epsilon^{2}_{\circ} c h^{3}} \qquad(8)$

So equation $(7)$ can be written as

$\frac{1}{\lambda}=z^{2} R \left( \frac{1}{n_{1}^{2}} -\frac{1}{n_{2}^{2}}\right) \qquad(9)$

This is Bohr's formula for hydrogen and hydrogen-like atoms $(He^{+}, Li^{++},.......)$.

For hydrogen $Z=1$

$\frac{1}{\lambda}= R \left( \frac{1}{n_{1}^{2}} -\frac{1}{n_{2}^{2}}\right) \qquad(10)$

The value of the Rydberg Constant is

$R = \frac{me^{4}}{8 \epsilon^{2}_{\circ} c h^{3}} = 1.090 \times 10^{7} m^{-1}$

This value fairly agrees with empirical value $(1.097 \times 10^{7} m^{-1})$ obtained experimentally by Balmer.

The total energy in terms of Rydberg's Constant:

The $E$ expression can be written in terms of Rydberg's constant $R$ in a simplified form. So from equation $(5)$ and $(8)$ we get

$E=-Z^{2}\frac{Rhc}{n^{2}} \qquad(11)$

Putting the known values of $R$, $h$ and $c$ taking $1 eV =1.6 \times 10^{-19} \: J$ then we get

$E=-Z^{2}\frac{13.6}{n^{2}} \: eV \qquad(12)$

For a Hydrogen atom, $Z=1$

$E=-\frac{13.6}{n^{2}} \: eV \qquad(12)$

Read More