Construction and Working of Nuclear Reactor or Atomic Pile

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 good absorbers of slow neutrons. Therefore when the rods are pushed into the reactor, the fission rate decreases, and when they are pulled out, the fission grows.

Coolant: The coolant is used to remove the heat from the reactor which is produced inside the reactor. For this purpose air-water or CO₂ is passed through the reactor.

Shield: various type of intense rays are emitted from the reactor, which may be injurious to the people working near the reactor. To protect them from this radiation, thick concrete walls are erected around the reactor.

Safety device: In case of an accident or other emergency, a special set of control rods, called " shut-off rods" enter the reactor automatically. They immediately absorb the neutrons so that the chain reaction stops entirely.

Construction:

It is made up of a large number of uranium rods which are placed in calculated geometrical lattice between layers of pure graphite ( moderator) blocks. To prevent oxidation of Uranium, the rods are covered by close-fitting aluminum cylinders. The control rods are so inserted within the lattice that they will be raised or lowered between the Uranium rods whenever necessary. The whole reactor is encircled by a concrete shield.

Working:

The actual operation of the reactor is started by pulling out the control rods so that they do not absorb many neutrons. Then, the stray neutrons, which are always present in the tractor, start fissioning the U-235 nuclei. In each fission, two or three fast neutrons are produced. These neutrons repeatedly strike the moderator and slowed down. Then, These neutrons start fissioning the U-235 nuclei. So, a chain-reaction of fission starts. The number of neutrons, which is produced in fissioning, is controlled by pushing the cadmium rods into the reactor. These rods absorb a number of the neutrons. Thus, the energy produced in the reactor is kept under control to avoid any explosion. The coolant is pond pumped through the reactor to carry away The Heat generated by the fission of uranium nuclei. the hot CO₂ passes through the heat exchanger and convert cold water into steam. This steam is used to operate turbines for generating electricity.

Numerical Aperture and Acceptance Angle of the Optical Fibre

Angle of Acceptance → If incident angle of light on the core for which the incident angle on the core-cladding interface equals the critical angle then incident angle of light on the core is called the "Angle of Acceptance. Transmission of light when the incident angle is equal to the acceptance angle If the incident angle is greater than the acceptance angle i.e. $\theta_{i}>\theta_{0}$ then the angle of incidence on the core-cladding interface will be less than the critical angle due to which part of the incident light is transmitted into cladding as shown in the figure below Transmission of light when the incident angle is greater than the acceptance angle If the incident angle is less than the acceptance angle i.e. $\theta_{i}<\theta_{0}$ then the angle of incidence on the core-cladding interface will be greater than the critical angle for which total internal reflection takes place inside the core. As shown in the figure below Transmission of lig

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Fraunhofer diffraction due to a single slit

Let $S$ be a point monochromatic source of light of wavelength $\lambda$ placed at the focus of collimating lens $L_{1}$. The light beam is incident normally from $S$ on a narrow slit $AB$ of width $e$ and is diffracted from it. The diffracted beam is focused at the screen $XY$ by another converging lens $L_{2}$. The diffraction pattern having a central bright band followed by an alternative dark and bright band of decreasing intensity on both sides is obtained. Analytical Explanation: The light from the source $S$ is incident as a plane wavefront on the slit $AB$. According to Huygens's wave theory, every point in $AB$ sends out secondary waves in all directions. The undeviated ray from $AB$ is focused at $C$ on the screen by the lens $L_{2}$ while the rays diffracted through an angle $\theta$ are focussed at point $p$ on the screen. The rays from the ends $A$ and $B$ reach $C$ in the same phase and hence the intensity is maximum. Fraunhofer diffraction due to

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Particle in one dimensional box (Infinite Potential Well)

Let us consider a particle of mass $m$ that is confined to one-dimensional region $0 \leq x \leq L$ or the particle is restricted to move along the $x$-axis between $x=0$ and $x=L$. Let the particle can move freely in either direction, between $x=0$ and $x=L$. The endpoints of the region behave as ideally reflecting barriers so that the particle can not leave the region. A potential energy function $V(x)$ for this situation is shown in the figure below. Particle in One-Dimensional Box(Infinite Potential Well) The potential energy inside the one -dimensional box can be represented as $\begin{Bmatrix} V(x)=0 &for \: 0\leq x \leq L \\ V(x)=\infty & for \: 0> x > L \\ \end{Bmatrix}$ $\frac{d^{2} \psi(x)}{d x^{2}}+\frac{2m}{\hbar^{2}}(E-V)\psi(x)=0 \qquad(1)$ If the particle is free in a one-dimensional box, Schrodinger's wave equation can be written as: $\frac{d^{2} \psi(x)}{d x^{2}}+\frac{2mE}{\hbar^{2}}\psi(x)=0$ $\frac{d^{2} \psi(x)}{d x

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Physics Vidyapith ✍️

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