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Showing posts from February, 2023

Eigen value of the momentum of a particle in one dimension box or infinite potential well

Equation of eigen value of the momentum of a particle in one dimension box: The eigen value of the momentum $P_{n}$ of a particle in one dimension box moving along the x-axis is given by $P^{2}_{n} = 2 m E_{n}$ $P^{2}_{n} = 2 m \frac{n^{2} \pi^{2} \hbar^{2}}{2 m L^{2}} \qquad \left( \because E_{n}= \frac{n^{2} \pi^{2} \hbar^{2}}{2 m L^{2}} \right)$ $P^{2}_{n} = \frac{n^{2} \pi^{2} \hbar^{2}}{L^{2}}$ $P_{n} = \pm \frac{n \pi \hbar}{L}$ $P_{n} = \pm \frac{n h}{2L} \qquad \left( \hbar = \frac{h}{2 \pi} \right)$ The $\pm$ sign indicates that the particle is moving back and forth in the infinite potential box. The above equation shows that eigen value of the momentum of the particle is discrete and the difference between the momentum corresponding to two consecutive energy levels is always constant and equal to $\frac{h}{2L}$

Electric field intensity due to uniformly charged solid sphere (Conducting and Non-conducting)

A.) Electric field intensity at different points in the field due to the uniformly charged solid conducting sphere: Let us consider, A solid conducting sphere that has a radius $R$ and charge $+q$ is distributed on the surface of the sphere in a uniform manner. Now find the electric field intensity at different points due to the solid-charged conducting sphere. These different points are: Electric field intensity outside the solid conducting sphere Electric field intensity on the surface of the solid conducting sphere Electric field intensity inside the solid conducting sphere 1.) Electric field intensity outside the solid conducting sphere: If $O$ is the center of solid conducting spherical then the electric field intensity outside of the sphere can be determined by the following steps → First, take the point $P$ outside the sphere Draw a spherical surface of radius r which passes through point $P$. This hypothetical surface is known

Derivation of Planck's Radiation Law

Derivation: Let $ N$ be the total number of Planck’s oscillators and $E$ be their total energy, then the average energy per Planck’s oscillator is $ \overline{E}=\frac{E_{N}}{N} \qquad (1)$ Let there be $ N_{0}, N_{1} ,N_{2} ,N_{3},---N_{n}$ oscillator having energy $ E_{0}, E_{1}, E_{2}, -- E_{n}$ respectively. According to Maxwell’s distribution, the number of oscillators in the $ n^{th}$ energy state is related to the number of oscillators in the ground state by $ N_{n}=N _{0} e^{\tfrac{-nh\nu }{kt} }\qquad (2)$ Where $ n$ is a positive integer. So put $ n= 1,2,3,…….$. The above equation can be written for different energy states. i.e. $ N_{1}= N _{0} e^{\tfrac{-h\nu }{kt} }$ $ N_{2}= N _{0} e^{\tfrac{-2h\nu }{kt} }$ $ N_{3}= N _{0} e^{\tfrac{-3h\nu }{kt} }$ $.............$ $.............$ So, the total number of Planck’s Oscillators – $ N= N _{0} +N_{1}+N_{2}+N_{3}+.... N_{n}$ $ N= N _{0} + N_{0} e^{\tfrac{-h\nu }{kt}}+ N_{0} e^{\

Orthogonality of the wave functions of a particle in one dimension box or infinite potential well

Description of Orthogonality of the wave functions of a particle in one dimension box or infinite potential well: Let $\psi_{n}(x)$ and $\psi_{m}(x)$ be the normalized wave functions of a particle in the interval $(0, L)$ corresponding to the different energy level $E_{n}$ and $E_{m}$ respectively. These wave functions are: $\psi_{n}(x)= \sqrt{\frac{2}{L}} sin \frac{n \pi x}{L}$ $\psi_{m}(x)= \sqrt{\frac{2}{L}} sin \frac{m \pi x}{L}$ Where $m$ and $n$ are integers. In this function are real. Therefore $\psi_{n}^{*}(x) = \psi_{n}(x)$ $\psi_{m}^{*}(x) = \psi_{m}(x)$ Where $m=n$, $\int_{0}^{L} \psi_{n}^{*}(x) \psi_{m}^{*}(x) dx = \frac{2}{L} \int_{0}^{L} sin \frac{m \pi x}{L} . sin \frac{n \pi x}{L} dx$ $\int_{0}^{L} \psi_{n}^{*}(x) \psi_{m}^{*}(x) dx =\frac{1}{L} \int_{0}^{L} \left[ cos \left\{ \frac{(m-n) \pi x}{L} \right\} - cos \left\{ \frac{(m+n) \pi x}{L} \right\} \right] dx $ $\int_{0}^{L} \psi_{n}^{*}(x) \psi_{m}^{*}(x) dx =\frac{1}{L

The electric potential at different points (like on the axis, equatorial, and at any other point) of the electric dipole

Electric Potential due to an Electric Dipole: The electric potential due to an electric dipole can be measured at different points: The electric potential on the axis of the electric dipole The electric potential on the equatorial line of the electric dipole The electric potential at any point of the electric dipole 1. The electric potential on the axis of the electric dipole: Let us consider, An electric dipole AB made up of two charges of -q and +q coulomb is placed in a vacuum or air at a very small distance of $2l$. Let a point $P$ is on the axis of an electric dipole and place at a distance $r$ from the center point $O$ of the electric dipole. Now put the test charged particle $q_{0}$ at point $P$ for the measurement of electric potential due to dipole's charges. So Electric potential at point $P$ due $+q$ charge of electric dipole→ $ V_{+q}=\frac{1}{4\pi \epsilon_{0}} \frac{q}{r-l}$ The electric potential at point $P$ due $-q$ charge of elect

Electric field intensity due to uniformly charged plane sheet and parallel sheet

Electric field intensity due to a uniformly charged infinite plane thin sheet: Let us consider, A plane charged sheet (It is a thin sheet so it will have surface charge distribution whether it is a conducting or nonconducting sheet) whose surface charge density is $\sigma$. From symmetry, Electric field intensity is perpendicular to the plane everywhere and the field intensity must have the same magnitude on both sides of the sheet. Let point $P_{1}$ and $P_{2}$ be the two-point on the opposite side of the sheet. To use Gaussian law, we construct a cylindrical Gaussian surface of cross-section area $\overrightarrow{dA}$, which cuts the sheet, with points $P_{1}$ and $P_{2}$. The electric field $\overrightarrow{E}$ is normal to end faces and is away from the plane. Electric field $\overrightarrow{E}$ is parallel to cross-section area $\overrightarrow{dA}$. Therefore the curved cylindrical surface does not contribute to the flux i.e. $\o

Electric field intensity due to uniformly charged wire of infinite length

Derivation of electric field intensity due to the uniformly charged wire of infinite length: Let us consider a uniformly-charged (positively charged) wire of infinite length having a constant linear charge density (that is, a charge per unit length) $\lambda$ coulomb/meter. Let P is a point at a distance $r$ from the wire at which electric field $\overrightarrow{E}$ has to find. Let us draw a coaxial Gaussian cylindrical surface of length $l$ through point $P$. By symmetry, the magnitude $E$ of the electric field will be the same at all points on this surface and directed radially outward. Thus, Now take small area elements $dA_{1}$, $dA_{2}$ and $dA_{3}$ on the Gaussian surface as shown in figure below. Therefore, the total electric flux passing through area elements is: $ \phi_{E}= \oint \overrightarrow{E} \cdot \overrightarrow{dA_{1}} + \oint \overrightarrow{E} \cdot \overrightarrow{dA_{2}} + \oint\overrightarrow{E} \cdot \overrightarrow{dA_{3}}$ $

Electric field intensity due to point charge by Gauss's Law

Derivation of electric field intensity due to a point charge by Gauss's Law: Let us consider, a source point charge particle of $+q$ coulomb is placed at point $O$ in space. Let's take a point $P$ on the electric field of the source point charge particle. To find the electric field intensity $\overrightarrow{E}$ at point $P$, first put the test charge particle $+q_{0}$ on the point $P$ and draw a gaussian surface which passes through the point $P$. After that take a very small area $\overrightarrow{dA}$ around the point $P$. If the distance between the source charge particle and small area $\overrightarrow {dA}$ is $r$ then electric flux passing through the small area $\overrightarrow{dA}$ → $ d\phi_{E}= \overrightarrow{E} \cdot \overrightarrow{dA}$ $ d\phi_{E}= E\:dA\: cos\theta$ Electric field due to point charge from the figure, the direction between $\overrightarrow{E}$ and $\overrightarrow{dA}$ is parallel to each other i.e. the angle will be $0^{\ci

Normalization of the wave function of a particle in one dimension box or infinite potential well

Description of Normalization of the wave function of a particle in one dimension box or infinite potential well: We know that the wave function for the motion of the particle along the x-axis is $\psi_{n}(x)= A \: sin \left( \frac{n \pi x}{L} \right) \quad \left\{ Region \quad 0 \lt x \lt a \right\}$ $\psi_{n}(x)= 0 \quad \left\{ Region \quad 0 \gt x \gt a \right\}$ The total probability that the particle is somewhere in the box must be unity. Therefore, $\int_{0}^{L} \left| \psi_{n}(x)\right|^{2}dx =1$ Now substitute the value of the wave function in the above equation. Then $\int_{0}^{L} \left| A \: sin \left( \frac{n \pi x}{L} \right) \right|^{2}dx =1$ $\int_{0}^{L} A^{2} \: sin^{2} \left( \frac{n \pi x}{L} \right) dx =1$ $ \frac{A^{2}}{2}\int_{0}^{L} \left[ 1- cos \left( \frac{2n \pi x}{L} \right) \right] dx =1$ $ \frac{A^{2}}{2} \left[ x - \left( \frac{L}{2n\pi} \right) sin \left( \frac{2n \pi x}{L} \right) \right]_{0}^{L} =1$ $ \frac{A^{2}

The electric potential energy of an electric dipole in the uniform electric field

Derivation of the electric potential energy of an electric dipole in the uniform electric field: Let us consider an electric dipole $AB$, which is made up of two charges $q_1$ and $q_2$, which are placed at a distance of $2l$ in the electric field $E$. So force acting on each charge due to the electric field will be $qE$. If the dipole gets rotated a small-angle $d\theta$ against the torque acting on it in the uniform electric field $E$ then the small work done is $dW=\tau. d\theta \qquad (1)$ Force of moment on an electric Dipole The torque (i.e moment of force) on an electric dipole in a uniform electric field $ \tau=p.E\:sin\theta$ Now substitute the value of $\tau$ in equation $(1)$. So work done $ dW=p.E\:sin\theta.d\theta$ If the dipole rotate the angle from $\theta_{1}$ to angle $\theta_{2}$ then workdone $\int_{W_{1}}^{W_{2}}dW=p.E\int_{\theta_{1}}^{\theta_{2}}sin\theta \: d\theta$ $ W_{2}-W_{1}=p.E\left[-cos\theta \right]

de-Broglie Concept of Matter wave

Louis de-Broglie thought that similar to the dual nature of light, material particles must also possess the dual character of particle and wave. This means that material particles sometimes behave as particle nature and sometimes behave like a wave nature. According to de-Broglie – A moving particle is always associated with a wave, called as de-Broglie matter-wave, whose wavelengths depend upon the mass of the particle and its velocity. According to Planck’s theory of radiation– $E=h\nu \qquad(1) $ Where h – Planck’s constant $\nu $ - frequency According to Einstein’s mass-energy relation – $E=mc^ {2} \qquad (2)$ According to de Broglie's hypothesis equation $ (1)$ and equation $(2)$ can be written as – $mc^ {2} = h \nu$ $mc^ {2} = \frac{hc}{\lambda }$ $\lambda =\frac{h}{mc}\qquad(3) $ $\lambda =\frac{h}{P}$ Where $P$ –Momentum of Photon Similarly from equation $(3)$ the expression for matter waves can be written as $\la

Group velocity is equal to particle velocity

Prove that: Group velocity is equal to Particle Velocity Solution: We know that group velocity $V_{g}=\frac{d\omega}{dk}$ $V_{g}=\frac{d(2\pi\nu )}{d(\frac{2\pi }{\lambda })} \qquad \left(k=\frac{2\pi}{\lambda} \right)$ $V_{g}=\frac{d\nu}{d(\frac{1}{\lambda })}$ $\frac{1}{V_{g}}=\frac{d( \frac{1}{\lambda })}{d\nu}\qquad(1)$ We know that the total energy of the particle is equal to the sum of kinetic energy and potential energy. i.e $E=K+V$ Where $K$ – kinetic energy $V$ – Potential energy $E=\frac{1}{2} mv^{2}+V$ $E-V=\frac{1}{2}\frac{(mv)^2}{m}$ $E-V=\frac{1}{2m }(mv)^2$ $2m(E-V)=(mv)^2$ $mv=\sqrt{2m(E-V)}$ According to de-Broglie wavelength- $\lambda =\frac{h}{mv}$ $\lambda =\frac{h}{\sqrt{2m(E-V)}}$ $\frac{1}{\lambda} =\frac{\sqrt{2m(E-V)}}{h}\qquad(3)$ Now put the value of $\frac{1}{\lambda }$ in equation$(1)$ $\frac{1}{V_{g}} =\frac{d}{dv}[\frac{{2m(E-V)}^\tfrac{1}{2}}{h}]$ $\frac{1}{V_{g}} =\frac{d}{dv}[\frac{

Principle, Construction and Working of Venturimeter

Principle of Venturimeter: It is a device that is used for measuring the rate of flow of liquid through pipes. Its principle and working are based on Bernoulli's theorem and equation. Construction: It consists of two identical coaxial tubes $X$ and $Z$ connected by a narrow co-axial tube $Y$. Two vertical tubes $P$ and $Q$ are mounted on tubes $X$ and $Y$ to measure the pressure of the liquid that flows through pipes.. As shown in the figure below. Working and Theory: Connect this venturimeter horizontally to the pipe through which the liquid is flowing and note down the difference of liquid columns in tubes $P$ and $E$. Let the difference is $h$. Let us consider that an incompressible and non-viscous liquid flows in streamlined motion through a tube $X$,$Y$, and $Z$ of the non-uniform cross-section. Now Consider: The Area of cross-section of tube $X$ = $A_{1}$ The Area of cross-section of tube $Y$ = $A_{2}$ The velocity per second (i.e. equal to

Solution of electromagnetic wave equations in conducting media

The electromagnetic wave equations in conducting media: For electric field vector: $ \nabla^{2}.\overrightarrow{E}-\mu \epsilon\frac{\partial^{2} \overrightarrow{E}}{\partial t^{2}} - \sigma \mu \frac{\partial \overrightarrow{E}}{\partial t}=0 \qquad(1)$ For magnetic field vector: $\nabla^{2}.\overrightarrow{B} - \mu \epsilon \frac{\partial^{2} B}{\partial t^{2}}-\sigma \mu \frac{\partial \overrightarrow{B}}{\partial t}=0 \qquad(2)$ The wave equation of electric field vector: $\overrightarrow{E}(\overrightarrow{r},t)=E_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)} \qquad(3)$ The wave equation of magnetic field vector: $\overrightarrow{B}(\overrightarrow{r},t)=B_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)} \qquad(4)$ Now the solution of electromagnetic wave for electric field vector. Differentiate with respect to $t$ of equation $(3)$ $\frac{\partial \overrightarrow{E}}{\parti

Electromagnetic Wave Equation in Conducting Media (i.e. Lossy dielectric or Partially Conducting)

Maxwell's Equations: Maxwell's equation of the electromagnetic wave is a collection of four equations i.e. Gauss's law of electrostatic, Gauss's law of magnetism, Faraday's law of electromotive force, and Ampere's Circuital law. Maxwell converted the integral form of these equations into the differential form of the equations. The differential form of these equations is known as Maxwell's equations. $\overrightarrow{\nabla}. \overrightarrow{E}= \frac{\rho}{\epsilon_{0}}$ $\overrightarrow{\nabla}. \overrightarrow{B}=0$ $\overrightarrow{\nabla} \times \overrightarrow{E}=-\frac{\partial \overrightarrow{B}}{\partial t}$ $\overrightarrow{\nabla} \times \overrightarrow{B}= \mu \overrightarrow{J}$ Modified form: $\overrightarrow{\nabla} \times \overrightarrow{B}= \mu \overrightarrow{J}+\mu \epsilon \frac{\partial \overrightarrow{E}}{\partial t}$ For Conducting Media: Cur