Physics Vidyapith ✍️

Davisson and Germer's Experiment on Electron Diffraction: Davisson and Germer's experiment verifies the wave nature of electrons with the help of diffraction of the electron beam as wave nature exhibits the diffraction phenomenon. Principle: The principle of Davisson and Germer's experiment is based on the diffraction phenomenon of the electron beam by crystal and it verifies the de-Broglie relation. Theoretical Formula: If a narrow beam of electrons is accelerated by a potential difference $V$ volts, the kinetic energy $K$ acquired by each electron in the beam is given by $K=eV \qquad(1)$ Where $e$ is the charge of an electron The de-Broglie wavelength is given by $\lambda = \frac{h}{\sqrt {2m_{\circ} K \left( 1+ \frac{E_{K}}{2m_{\circ}c^{2}} \right)}}$ If $E_{K} \lt \lt 2m_{\circ}c^{2}$, then the term $\frac{E_{K}}{2m_{\circ}c^{2}}$ will be negligible. So above equation can be written as $\lambda = \frac{h}{\sqrt {2m_{\circ} K}} \qquad(

A free electron cannot absorb all the energy of a Photon in mutual interaction . Let us consider that a photon of energy $h \nu$ and momentum $\frac{h\nu}{c}$ collides with the free electron of mass $m$ at rest and the photon transfers its total energy and momentum to the electron. If $v$ is the velocity of the electron after a collision, its energy will be $ \frac{1}{2}mv^{2}$ and momentum $mv$. If total incident energy $E$ absorb by an electron then applying the laws of conservation of energy and momentum, we have Total energy before the collision= Total energy after the collision $h \nu = \frac{1}{2}mv^{2} \qquad(1)$ According to de Broglie Hypothesis, the momentum of a particle $P=\frac{h}{\lambda}$ $P=\frac{h \nu}{c} \qquad \left(\because \lambda=\frac{c}{\nu} \right)$ $mv=\frac{h \nu}{c} \qquad \left(\because P = mv \right)$ $h \nu = mvc \qquad(2)$ From equation $(1)$ and equation $(2)$ $\frac{1}{2}mv^{2}= m v c$ $\frac{1}{2}mv^{2}=m v c$

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

The Quantization Condition in Bohr Theory of Hydrogen Atom: $L=\frac{nh}{2 \pi}$ For the angular momentum L, the electron moves arbitrarily only in a stationary circular orbit. According to De Broglie's hypothesis, this condition can be easily obtained. For this purpose, there are following assumptions given below: 1.) The motion of the electron in a stationary circular orbit is represented by a standing matter-wave. If the wavelength of the wave is $\lambda$ then the De Broglie relation $\lambda=\frac{h}{mv} \qquad(1)$ Where $m \rightarrow$ The mass of the electron and $v \rightarrow$ The Velocity in the orbit. 2.) The circular orbit contains an integral number of wavelengths, i.e. $2 \pi r_{n}= n \lambda $ $\frac{2 \pi r_{n}}{\lambda}= n \qquad(2)$ Where $n=1,2,3............$ and $r_{n}$ is the radius of the orbit. Substituting the value of $\lambda$ in equation$(2)$ $\frac{2 \pi r_{n} m v}{h} =n$ $mvr_{n}

Planck's quantum hypothesis with its application and extension to explain the black body radiation like the photo-electric effect, the Compton effect, the variation of specific heat of solid with temperature and the spectrum of hydrogen is now called the Old quantum theory. Through these phenomena are successfully explained by the theory, there are numerous drawbacks of the theory. A few of them are as follows. 1.) Bohr's quantization rules are arbitrary. The theory does not provide a physical explanation for the assumptions. 2.) The old quantum theory cannot be applied to explain the spectra of helium and of more complex atoms. 3.) It can provide only a qualitative and incomplete explanation of the intensities of the spectral lines. 4.) It can not explain the dispersion of light. 5.) The theory of non-harmonic vibrations of systems cannot be applied to explain the vibrations of systems.

Potential Step: 1.) In region $(I)$  i.e.$(-\infty \leq \: x \lt \: 0)$ 2.) In region $(II)$ i.e. $(0 \leq \:x \: \leq +\infty)$ Case $(1)$ When $E \lt V_{\circ}$ A.) Expression for the Amplitudes $B$ and $C$ in terms of the Amplitude $A$ B.) Expression for the wave function in the region $(I)$ and $(II)$ C.) Expression for the Probability Current Densities D.) Reflection and Transmission Coefficients Case $(2)$ : When $E \gt V_{\circ}$ A.) Expression for the Amplitudes $B$ and $C$ in terms of the Amplitude $A$ B.) Expression for the wave function in the region $(I)$ and $(II)$ C.) Expression for the Probability Current Densities

The wave function of a free particle: Suppose, A particle of mass $m$ is in motion along the x-axis. Suppose no force is acting on the particle so that the potential energy of the particle is constant. For convenience, the constant potential energy is taken to b zero. Therefore, the time-independent Schrodinger equation for a free particle is: $-\frac{\hbar^{2}}{2m} \frac{d^{2} \psi}{dx^{2}}=E \psi \qquad(1)$ Since the particle is moving freely with zero potential energy, its total energy $E$ is the kinetic energy given by $E=\frac{p^{2}_{x}}{2m}$ Where $p_{x}$ is the momentum of the particle which is moving along the x-axis. $\frac{d^{2} \psi}{dx^{2}} + \frac{2mE}{\hbar^{2}} \psi =0$ Let $k^{2}=\frac{2mE}{\hbar^{2}} \qquad(2)$, Now substitute this value in the above equation that can be written as $\frac{d^{2} \psi}{dx^{2}} + k^{2} \psi =0 \qquad(3)$ The solution of the above equation $(3)$ $\psi (x) = A e^{ikx} + B e^{-ikx} \qquad(4)$ Here $A$ and $

The Compton Effect 1 Compton Experiment Setup 2 Theory of the Compton Effect 3 Theoretical Derivation of Compton Effect (Equation of Compton Shift) 4 Limitation of Compton Effect 5 Compton Recoil Electron- 5(a)Relation between $\theta$ and $\phi$ 5(b) Kinetic Energy of the Recoil Electron 1.) Compton Experiment Setup: The Compton effect is used to verify the particle nature of matter by applying the photoelectric effect. The setup of the Compton experiment as shown in the figure below which consists of the following parts i.) X-ray source ii.) Collimator iii.) Target iv.) Bragg's Spectrometer i.) X-ray Source: The X-ray source is used to produce the monochromatic X-ray ii.) Collimators: The collimator

Classical mechanics is a branch of physics that deals with the motion of macroscopic bodies or objects $(i.e \: the \: size \: range \: greater \: then \: 10^{-8}m)$ under the influence of forces. While it was groundbreaking when first developed by Sir Isaac Newton in the 17th century, it has certain limitations that were discovered over time. In this answer, we will discuss these limitations in more detail: Classical Mechanics is not applicable to extremely small objects: Classical mechanics assumes that particles have a definite position and momentum, which is not true in the quantum world. This limitation became apparent in the early 20th century with the discovery of quantum mechanics. Quantum mechanics is a branch of physics that deals with the behavior or motion of particles on an atomic and subatomic level $(i.e \: the \: size \: range \: is \: in \: between \: 10^{-8}m \: to \: 10^{-15}m)$ i.e microscopic particles. It has been successful in explaining phenomena s

Classical world vs Quantum world: The classical world and the quantum world are two fundamentally different ways of describing the behavior of matter and energy. In the classical world, the laws of physics are described by classical mechanics, which is based on the concepts of position, velocity, and acceleration of objects. Classical mechanics is deterministic, meaning that if you know the initial conditions of a system, you can predict its future behavior with complete accuracy. This is the world we experience in our everyday lives, and it is characterized by a continuous, smooth flow of events. In contrast, the quantum world is described by quantum mechanics, which is based on the behavior of particles on a subatomic scale. In the quantum world, particles do not have well-defined positions and velocities but rather exist in a superposition of many possible states. Moreover, measurements of quantum particles do not give deterministic results, but rather give probabilities

A non-relativistic free particle of mass $m$ moving in the positive $x$-direction with speed $v_{x}$ has kinetic energy $E=\frac{1}{2} m v^{2}_{x}$ and momentum $p_{x}=mv_{x}$ The energy and momentum are associated with a wave of wavelength $\lambda$ and frequency $\nu$ given by $\lambda = \frac{h}{p_{x}}$ and $\nu=\frac{E}{h}$ The propagation constant $k_{x}$ of the wave is $k_{x}= \frac{2\pi}{\lambda}=\frac{2\pi}{\left(\frac{h}{p_{x}} \right)}=\frac{p_{x}}{\left(\frac{h}{2\pi} \right)}=\frac{p_{x}}{\hbar}$ and the angular frequency $\omega$ is $\omega = 2\pi \nu = \frac{2\pi E}{\hbar}=\frac{E}{\hbar}$ A plane wave traveling along the $x$ axis in the positive direction may be represented by $\psi(x,t)=A e^{-i\left(k_{x} \: x - \omega t\right)}$ Now subtitute the value of $\omega$ and $k_{x}$ in above equation then we get $\psi(x,t)=A e^{i\left( \frac{p_{x}}{\hbar} \: x - \frac{E}{\hbar} \: t\right)}$ $\psi(x,t)=A e^{\frac{i}{\hbar}\l

Derivation: The time dependent Schrodinger quation for the wave function function $psi(x,y,z,t)$ is $-\frac{\hbar^{2}}{2m}\nabla^{2} \psi + V\psi=i\hbar\frac{\partial \psi}{\partial t} \qquad(1)$ Since wave function, $\psi$ is complex quantity i.e. $\psi=\psi_{1}+i \: \psi_{2} \qquad(2)$ Where $\psi_{1}$ and $\psi_{2}$ are real functions of $x,y,z,t$. Substituting this form for $\psi$ in equation $(1)$, we get $-\frac{\hbar^{2}}{2m}\nabla^{2} \left( \psi_{1}+i \: \psi_{2} \right) + V\left( \psi_{1}+i \: \psi_{2} \right) \\ \qquad = i\hbar\frac{\partial }{\partial t} \left( \psi_{1}+i \: \psi_{2} \right)$ Equation real and imaginary parts on either side of this equation, we obtain the following two equations: $-\frac{\hbar^{2}}{2m}\nabla^{2} \psi_{1} + V\psi_{1}=-\hbar\frac{\partial \psi_{2}}{\partial t} \qquad(3)$ $-\frac{\hbar^{2}}{2m}\nabla^{2} \psi_{2} + V\psi_{2}=\hbar\frac{\partial \psi_{1}}{\partial t} \qquad(4)$ Mutiplying equation $(4)$ by $-

1.) Derivation of Probability Current Density for a free particle: Let a particle of mass $m$ is moving in the positive $x$- direction in the region from $x_{1}$ to $x_{2}$. For the one-dimensional motion of the particle, the wave function is $psi(x,t)$ Let $dA$ be the area of the cross-section of the region. The probability of finding a particle in the region is $\int_{x_{1}}^{x_{2}} \psi(x,t) \: \psi^{*}(x,t) \: dx \: dA \qquad(1)$ and the probability density of finding the particle in the region is $P=\psi(x,t) \: \psi^{*}(x,t) \qquad(2)$ If the probability of finding the particle in the region decreases with time, the rate of decrease of the probability that the particle is in the region from $x_{1}$ to $x_{2}$ per unit area is called the probability current density out of the region. Therefore, the probability current density $S_{2} - S_{1}$ out of the region in the positive $x$-direction is given by $S_{2} - S_{1} = - \frac{1}{dA} \left[- \frac{d}{dt

Ehrenfest's Theorem Statement: The theorem states that Quantum mechanics gives the same results as classical mechanics for a particle for which average or expectation values of dynamical quantities are involved. Proof of theorem: The proof of the theorem for one-dimensional motion of a particle by showing that 1) $\frac{d \left < x \right >}{dt} = \frac{\left < p_{x} \right > }{m}$ 2) $\frac{d \left < p_{x} \right >}{dt} = \left < F_{x} \right >$ 1.) To Show that: $\frac{d \left < x \right > }{dt} = \frac{\left < p_{x} \right > }{m}$ Let $x$ is the position coordinate of a particle of mass $m$, at time $t$ The expectation value of position $x$ of a particle is given by $\left < x \right > = \int_{- \infty}^{+ \infty} \psi^{*} (x,t) . x \: \psi (x,t) dx \qquad (1)$ Differentiating the above equation $(1)$ with respect to $t$ $\frac{d \left < x \right > }{dt} = \int_{- \infty}^{+ \infty} x \frac{