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Showing posts from May, 2023

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

Absorption of all the energy of a incident photon by a free electron

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 $(2)$ $\frac{1}{2}mv^{2}= m v c$ $\frac{1}{2}mv^{2}=m v c$

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

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 dividi

Bohr's Model of Atom

Bohr's Atomic Model Postulates: Prof Neil in 1913 Bohr solve the difficulties of Ernest Rutherford's atomic model by applying Planck's quantum theory, For this, he proposed the following three Postulates: 1.) Electrons can revolve only in those orbits in which their angular momentum is an integral multiple of $\frac{h}{2 \pi}$. These orbits have discrete energy and definite radii. So it is called the " stable orbits ". If the mass of the electron is $m$ and it is revolving with velocity $v$ in an orbit of radius $r$, then its angular momentum will be $mvr$. According to Bohr's postulate, $mvr=\frac{nh}{2\pi}$ Where $h$ is Planck's universal constant This Bohr's equation is called the " Bohr's quantization Condition " 2.) When the electrons revolve in stable orbits then they do not radiate the energy in spite of their acceleration toward the center of the orbit. Hence atom remains stable and is said to exis

Bohr's Quantization Condition

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}

Drawbacks of Old Quantum Theory

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.

One dimensional Step Potential Barrier for a Particle

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

Wave function of a particle in free state

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 $