Physics Vidyapith ✍️

1.) Wein’s laws of Energy distributions→ A.) Wein's Fifth Power law→ The total amount of the energy emitted by a black body per unit volume at an absolute temperature in the wavelength range $\lambda$ and $\lambda + d\lambda$ is given as $E\lambda \cdot d\lambda= \frac{A}{\lambda^{5}}f\left ( \lambda T \right ) \cdot d\lambda \qquad (1)$ Where $A$ is a constant and $f(\lambda T)$ is a function of the product $\lambda T$ and is given as $ f\left ( \lambda T\right )=e^-\frac{hc}{\lambda kT}\qquad (2)$ From equation $(1)$ and $(2)$ $E_\lambda \cdot d\lambda = \frac{A}{\lambda ^{5}}e^\frac{-hc}{\lambda kT} \cdot d\lambda$ $E_\lambda \cdot d \lambda = A \lambda ^{-5} e^\frac{-hc}{\lambda kT} \cdot d \lambda$ Wien’s law energy distribution explains the energy distribution at the short wavelength at higher temperatures and fails for long wavelengths. B.) Wein's Displacement law→ As the temperature of the body is raised the maximum energy shift

We know that phase velocity $V_{p}=\frac{\omega }{k}$ $\omega =V_{p}.k \qquad(1)$ And group velocity $V_{g}=\frac{d\omega}{dk} \qquad(2)$ Substitute the value of $\omega$ from equation$(1)$ in equation $(2)$ $V_{g}=\frac{d}{dk}(V_{p}.k)$ $V_{g}=V_{p}+k.\frac{dV_{p}}{dk}$ $V_{p}=V_{p}+k.\frac{dV_{p}}{d\lambda}.\frac{d\lambda }{dk} \qquad (3)$ But $\lambda=\frac{2\pi }{k}$ The above equation can be obtain from following formula i.e. $k=\frac{2\pi}{\lambda }$ Now put the value of $\lambda$ in equation $(3)$ $V_{g}=V_{p}+k\frac{dV_{p}}{d\lambda}\frac{d}{dk}(\frac{2\pi }{k}$ $V_{g}=V_{p}+k\frac{dV_{p}}{d\lambda}(\frac{-2\pi }{k^{2}}$ $V_{g}=V_{p}-\frac{2\pi}{k}\frac{dV_{p}}{d\lambda }$ $V_{g}=V_{p}-\lambda\frac{dV_{p}}{d\lambda }$ Thus, the above equation represents the relation between group velocity and phase velocity.

Electric Field: That region around any charged particle (source-charge) in which other test charged particle experience an electric force (attraction or repulsion). This region is called electric field of that charged particle. The size and magnitude of the test charged particle is very small i.e tends to zero which does not modify the electric field of source charged particle. The electric field can only experience, not be seen by eyes. The electric field is defined by imaginary lines which are called electric field lines . Electric Field Lines: Electric field lines are imaginary lines that describe the behavior of electric fields. The behavior of the electric field tells about the force acting on free test-charged particles. This force gives the motion to the free test charge particle which is placed in the path of the electric field line. So the test charged particle moves in the direction of the electric field line. Hence, the direction of el

What is an electric charge? Electric charge is an intrinsic property of elementary particles (i.e. electron, proton and neutron etc.) of any substance which gives rise to electric force between them elementary particles. Types of charge: There are two types of electric charge Positive Electric Charge like Proton Negative Electric Charge like Electron The same nature of charged particles repels each other and the opposite nature of charged particles attracts each other. For example, electrons and electrons repel each other, and electrons and protons attract each other. Similarly, proton and proton repel each other. Generally, any substance is electrically neutral because the number of electrons and protons are equal in it but according to the  Free electron Model theory, these substances are classified into three categories. These are: Conductor Semico

Electric Potential: When a test-charged particle is brought from infinity to a point in the electric field then the work done per unit test charge particle is called electric potential . It is represented by $V$. It is a scalar quantity. Let's consider a test-charged particle $q_{0}$ bring from infinity to at a point $P$ in the electric field. If the work done by test charged particle is $W$ then electric potential → $V=\frac{W}{q_{0}}$ Unit of Electric Potential: $Joul/Coulomb$  OR  $N-m/Ampere-sec$ In MKS: $Kg-m^{2}-Ampere^{-1}-sec^{-3}$ Dimension of Electric potential: $[ML^{2}A^{-1}T^{-3}]$ The electric potential at a point due to point charged particle: Let us consider, a source point charge $+q$ is placed in air and vacuum at point $O$.Let's take a point $P$ at distance $r$ from the source point charged particle. Here the test-charged particle $+q_{0}$ is brought from infinity to point $P$.If the test-charged pa

A nuclear reactor is a device within which a self-sustaining controlled chain reaction is produced by fissionable material. it is thus a source of control energy that is utilized for several useful purposes. The reactor has some important part which is given below: Fuel: The fassionable material such as Uranium-235 and Plutonium-239 known as fuel. These materials play an important role in operating the nuclear reactor. Moderator: It slows down the neutrons to thermal energies through the elastic collision between its nuclei and fission neutrons. Thermal neutrons have a very high probability of fissioning Uranium-235 nuclei. Examples: heavy water graphite beryllium oxide. Heavy water is the best moderator. Control Rods: These rods are used to control the fission rate in the reactor. these Rods are fixed in the reactor walls. These rods are made up of the material of cadmium and Boron. These materials are g

Einstein’s Mass-Energy Relation: Einstein's mass energy relation gives the relation between mass and energy. It is also knows as mass-energy equivalence principle. According to Newtonian mechanics, Newton’s second law $f=\frac{dP}{dt}$ Where $P$ is the momentum of the particle. So put $P=mv$ in above equation: $f=\frac{d}{dt}\left ( mv \right )\quad\quad (1)$ According to theory of relativity, mass of the particle varies with velocity so above equation $(1)$ can be written as: $f=m \frac{dv}{dt}+v\frac{dm}{dt}\quad\quad (2)$ When the particle is displaced through a distance $dx$ by the applied force $F$. Then the increase in kinetic energy $dk$ of the particle is given by $dk= Fdx\quad\quad (3)$ Now substituting the value of force $F$ in equation $(3)$ $dk =m\frac{dv}{dt}\cdot dx+v\frac{dm}{dt}\cdot dx \quad (4) $ $dk=mv\cdot dv +v^{2}\cdot dm \:\: (5) \: \left \{ \because \frac{dx}{dt}=v \right \}$ The variation of mass with velocity equation

Concept of Simultaneity (Relative character of Time ): The interval aren't the same for two observes in relative motion. This cause an important incontrovertible fact that two events that appear to happen simultaneously to at least one observer are not simultaneous to another observer in relative motion. Suppose two events occur (or two-time bombs explode) at different places $x_{1}$ and $x_{2}$ but at the same time $t_{0}$ with respect to an observer in a stationary frame (or on the ground). The situation of the different to an observer in moving frame $S'$ or to a pilot of a spaceship moving with velocity $v$ relative to stationary frame $S$ (or ground). To him, according to Lorentz transformation for time. The explosion at $x_{1}$ occurs at $t'_{1} = \frac{t_{0} -x_{1}\frac{v}{c^{2}}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$ $ x_{2}$ occurs at $t'_{2} = \frac{t_{0}- x_{2}\frac{v}{c^{2}}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$ Hence the two events (explosions)

Derivation of Mean Or Average Value of Alternating Current: Let us consider alternating current $i$ propagating in a circuit then the average value of current. $ i_{mean}=\frac{1}{\left ( \frac{T}{2} \right )}\int_{0}^{\frac{T}{2}}i \:dt \qquad (1)$ $ where \quad i = i_{0}sin \omega t\quad(2)$ Now substitute the value of current $i$ in above equation $(1)$  $ i_{mean}= \frac{2}{T}\int_{0}^{\frac{T}{2}}i_0. sin \omega t.dt$ $ i_{mean}= \frac{2.i_{0}}{T}\int_{0}^{\frac{T}{2}}\sin \omega t.dt$ $ i_{mean}= \frac{2 i_{0}}{T}[\frac{-cos\:\omega t}{\omega} ]_{0}^{\frac{T}{2}}$ The value of $\omega$ is $\frac{2 \pi}{T}$ i.e $\omega=\frac{2\pi}{T}$ $ i_{mean}= \frac{2 i_{0}}{T \left (\frac{2\pi}{T} \right )}\left [ -cos \left (\frac{2 \pi}{T} \right ) \left ( \frac{T}{2} \right ) \\ \qquad \qquad \qquad +cos0^\circ \right ] $ $ i_{mean}= \frac{i_{0}}{\pi}\left [ - cos\pi+cos0^{\circ} \right ]$ $ i_{mean}=\frac{i_{0}}{\pi}\left [1+1 \right ]$ $ i_{m

Planck in 1900 suggested the correct explanation of the black body radiation curve. They gave the following assumption → A chamber contains black body energy radiation and simple harmonic oscillators (atoms of Wall, i.e. Black lamp & Platinum coating inside wall, behave as oscillators or resonators) of molecular dimensions which can vibrate with all possible frequencies. The frequency of energy radiation emitted by an oscillator is the same as the frequency of its vibration. An oscillator cannot emit or absorb the energy in a continuous manner it can emit or absorb energy in a small unit (packet) called  Quanta . If an oscillator is vibrating with a frequency $ \nu $ it can only radiate in quanta of magnitude $h\nu $ i.e. “The oscillator can have only discrete energy value $E_{n}$ ” given by– $E_{n}=nh\nu$ Where $n$ – an integer $h$– Planck ’s constant and the value is $6.626\times10^{-34} J-s$ The average energy of Planck’s oscillator of frequenc

If the x-coordinate of the position of a particle is known to an accuracy of $\delta x$, then the x-component of momentum cannot be determined to an accuracy better than $\Delta P_{x}\approx \frac{\hbar }{\Delta x}$. $\Delta P_{x}. \Delta x\approx \hbar$ The above inequality must be satisfied $\Delta P_{x}. \Delta x\geqslant \hbar$ Where $\hbar $ - Planck’s Constant This is the Uncertainty principle with macroscopic objects. Exact statement of the Uncertainty principle → The product of the uncertainties in determining the position and momentum of the particle can never be smaller than the number of the order $\frac{\hbar }{2}$. $\Delta P_{x}. \Delta x\geqslant \frac{\hbar}{2}$ Where $\delta x$ and $\delta P $ are defined as the root mean square deviation from their mean values. The Uncertainty principle can also describe by the following formula → $\Delta x.\Delta p_{x}\approx \frac{\hbar}{2}$ $\Delta x.\Delta p_{x}\geqslant \frac{

Simple harmonic motion: If an object repeats the process or path at a fixed interval of time is known as periodic motion or Uniform circular motion . It is also called the Simple harmonic motion . We know that the wave is the study of infinite S.H.M. Let us consider a particle is moving with uniform velocity in a circular path with radius $A$ i.e. particle is doing simple harmonic motion. Let at any instant $t$ particle move from position $P_{1}$ to $P_{2}$. So vector resolution of $P_{2}$ position is: From below figure- Horizontal Component i.e. $x$ component of position vector $P_{2}$: $x=A sin(\omega t - \phi) \qquad (1)$ Vertical Component i.e. $y$ component of position vector $P_{2}$: $y=A cos (\omega t - \phi) \qquad (2)$ Simple harmonic motion of a particle According to Max Born hypothesis- “Wave is a complex quantity which is represented by a wave function $\varphi$ ”  i.e. wave function mathematically can be represented as

The Potential Energy of a system of two-point like charges→ When the system of two charged particles is configured, in which one charge is at rest of position and another is brought from infinity to near the first charge then the work done acquire by this charged particle is stored in the form of electric potential energy between these charges. Derivation→ Let us consider, If two charge $q_{1}$ and $q_{2}$ in which one charge $q_{1}$ is at the rest of the position at point $P_{1}$ and another charge $q_{2}$ is brought from infinity to a point $P_{2}$ to configure the system then the electric potential at point $P_{2}$ due to charge particle $q_{1}$ → $V=\frac{1}{4\pi\epsilon_{0}}\frac{q_{1}}{r}$ Where $r$ is the distance between the point $P_{1}$ and Point $P_{2}$ Here, Charge $q_{2}$ is moved in from infinity to point $P_{2}$ then the work required is → $W=V q_{2}$ $W= \frac{1}{4\pi\epsilon_{0}}\frac{q_{1}q_{2}}{r}$ Since the electric potential at in

Gauss's Law: Gauss's law for electric flux is given by Carl Friedrich Gauss in 1813. He extended the work of Joseph-Louis Lagrange . This formula was first formulated in 1713 by Lagrange. Gauss's law stated that: The electric flux passing normal through any closed hypothetical surface is always equal to the $\frac{1}{\epsilon_{0}}$ times of the total charge enclosed within that closed surface. This closed hypothetical surface is known as Gaussian surface. Let us consider that a $+q$ coulomb charge is enclosed within the Gaussian's surface. Then according to Gauss's Law, the electric flux will be: $\phi _{E}= \frac{q}{\epsilon_{0}}$ The electric flux of the electric field → $\phi_{E}=\oint \overrightarrow{E}\cdot\overrightarrow{dA}$ Substitute this value of electric flux $\phi_{E}$ in the above formula so we get → $\oint \overrightarrow{E}\cdot\overrightarrow{dA}=\frac{q}{\epsilon_{0}}$ Where $\

Derivation of vector form of Coulomb's law: Let us consider, Two-point charges $+q_{1}$ and $+q_{2}$ are separated at a distance $r$ (magnitude only) in a vacuum as shown in the figure given below. Let $\overrightarrow{F_{12}}$ is the force on charge $+q_{1}$ due to charge $+q_{2}$ and $\overrightarrow{F_{21}}$ is the force on charge $+q_{2}$ due to charge $+q_{1}$. Then $\overrightarrow{F_{12}}=\frac{1}{4\pi \varepsilon _{0}}\frac{q_{1}q_{2}}{r^2}\:\:\hat{r_{21}}\qquad(1)$ Where $\widehat{r}_{21}$ ➝ Unit Vector Pointing from charge $+q_{2}$ to charge $+q_{1}$ $\overrightarrow{F_{21}}=\frac{1}{4\pi \varepsilon _{0}}\frac{q_{1}q_{2}}{r^2}\:\:\hat{r_{12}}\qquad(2)$ Where$\widehat{r}_{12}$ ➝ Unit Vector Pointing from charge $+q_{1}$ to charge $+q_{2}$ From the above figure, we can conclude that the direction of unit vector $\widehat{r}_{12}$ and $\widehat{r}_{21}$ is opposite. i.e. $\hat{r_{12}}=-\hat{r_{21}}\qquad(3)$ So from equ