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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