Work energy theorem Statement and Derivation

Work-energy theorem statement:

The work between the two positions is always equal to the change in kinetic energy between these positions. This is known as the work energy Theorem.

Motion of particle between two position
$W=K_{f}-K_{i}$

$W=\Delta K$

Derivation of the Work-energy theorem:

According to the equation of motion:

$v^{2}_{B}=v^{2}_{A}+2as $

$2as=v^{2}_{B}-v^{2}_{A}$

$2mas=m(v^{2}_{B}-v^{2}_{A})$

$mas=\frac{m}{2} (v^{2}_{B}-v^{2}_{A})$

$Fs=\frac{1}{2}mv^{2}_{B}-\frac{1}{2}mv^{2}_{A} \qquad (\because F=ma)$

$W=\frac{1}{2}mv^{2}_{B}-\frac{1}{2}mv^{2}_{A} \qquad (\because W=Fs)$

$W=K_{f}-K_{i}$

Where
$K_{f}$= Final Kinetic Energy at position $B$
$K_{i}$= Initial Kinetic Energy at position $A$

$W=\Delta K$

Alternative Method (Integration Method):

We know that the work done by force on a particle from position $A$ to position $B$ is-

$W=\int F ds$

$W=\int (ma)ds \qquad (\because F=ma)$

$W=m \int \frac{dv}{dt}ds$

$W=m \int dv \frac{ds}{dt}$

$W=m \int v dv \qquad (\because v=\frac{ds}{dt})$

If position $A$ is the initial point where velocity is $v_{A}$ and position $B$ is the final point where velocity is $v_{B}$ then work is done by force under the limit-

$W=m \int_{v_{A}}^{v_{B}} v dv$

$W=m[\frac{v^{2}}{2}]_{v_{A}}^{v_{B}} $

$W=\frac{1}{2}mv^{2}_{B}-\frac{1}{2}mv^{2}_{A}$

$W=K_{f}-K_{i}$

$W=\Delta {K}$

Davisson and Germer's Experiment and Verification of the de-Broglie Relation

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

$\lambda = \frac{h}{\sqrt {2m_{\circ} K}} \qquad(2)$

Now subtitute the value of $K$ from equation $(1)$ to equation $(2)$ then equation $(2)$ can be written as

$\lambda = \frac{h}{\sqrt {2m_{\circ} eV}}$

Now substitute the numerical value of $h$, $m_{\circ}$ and $e$, we get

$\lambda = \frac{6.63 \times 10^{-34}}{2 \times 9.11 \times 10^{-31} \times 1.6 \times 10^{-19} \times V}$

$\lambda = \frac{6.63 \times 10^{-9}}{29.15\times V}$

$\lambda = \frac{12.28}{\sqrt{V}}A^{\circ} \qquad(3)$

This is the theoretical value of $\lambda$ for the known potential difference in volts.

The calculation shows that the wavelength of the waves associated with the beam of electrons is of the same order as that of X-rays. Therefore, if such a beam of electrons is reflected from a crystal, the reflected beam will show the same diffraction and interference phenomena as for X-rays of the same wavelength. This consideration was the basis of Davisson and Germer's experiments. In one set of experiments, the [111] face of the nickel crystal was arranged perpendicular to the incident beam of electrons.

Apparatus:

There are the following apparatus is used in the Davisson and Germer experiment.

1. Electron gun
2. Target
3. Electron detector or Faraday's cylinder
4. Galvanometer
Davisson and Germer’s Experiment Setup
1. Electron gun: The electron gun is a device that is used to produce a highly accelerated and collimated electron beam by applying high potential.

2. Target: It is a single large metal crystal i.e. nickel used as a target. In crystal, the atoms are arranged in regular lattice i.e. [111] so that the surface lattice of the crystal acts as a diffraction grating and the electrons get diffracted by the crystal in different directions. The electron beam is incident normal to the nickel crystal. The crystal can be rotated about an axis perpendicular to the incident beam so that various azimuthal angles could be used.

3. Electron detector or Faraday's cylinder: It is used to detect or measure the intensity of diffracted beams of electrons. It can be moved along a circular scale $S$. This electron detector is connected to a Galvanometer.

4. Galvanometer: The galvanometer is a device that is used to measure the very small amount of current following in the circuit.

The whole apparatus is completely enclosed and highly evacuated.

Working: When a low potential is applied to the electron gun so that a beam of slow electrons emerges from the gun and falls normally on the surface of the crystal. To collect the diffracted electrons, the Faraday cylinder is moved to various positions on the scale $S$ and the corresponding galvanometer measures the current at each position through an electron detector. The observation is repeated for electrons accelerated through different potentials. The current which is a measure of the intensity of the diffracted electron beam, is plotted against the diffracting angle $\phi$ for each accelerating potential as shown in the figure below
Intensity of the Diffracted Electron Beam
It is observed in the curves that at the voltage of $40$ volts, a smooth curve is obtained and a bump begins to appear in the curve for $44$ volts. As the potential difference is further increased, the bump starts shifting upward and becomes most prominent in the curve for $54$ volts at $\phi = 50 ^{\circ}$. Beyond $54$ volts the bump gradually diminishes and becomes insignificant at $68$ volts. The pronounced current peak at $54$ volts and at $50^{\circ}$ provided an evidence that electrons were diffracted by the target and verifies the existence of electron wave nature.

Calculation of Wavelength:

Theoretical Calculation:

The wavelength of electron at $54$ volts can be find by de- Broglie formula as shown in the equation $(3)$

$\lambda= \frac{12.28}{\sqrt{54}}\: A^{\circ}$

$\lambda= 1.671 \: A^{\circ} $

Experimental Calculation:

Experimental calculation is done by Bragg's diffraction equation

$n \lambda = 2d \: sin\theta$

Where
$d \rightarrow$ interplanar Spacing
$n \rightarrow$ Order of plane

For nickel crystal :
$n=1$
$d=0.91 A^{\circ}$

In the experiment, the diffracted electron beam appearing at $\phi=50 ^{\circ}$ aries from wave-like diffraction from the family of Bragg's planes.
The corresponding angle of incidence relative to the family of Bragg's planes is

$\theta = \frac{180 - \phi}{2}$
$\theta = \frac{180 - 50}{2}$
$\theta= 65^{\circ}$
Diffraction of Electron Beam
Now apply these values to Bragg's equation as written above

$\lambda = 2 \times 0.91 \times sin 65^{\circ} $

$\lambda = 2 \times 0.91 \times 0.906 $

$\lambda = 1.65 A^{\circ} $

The theoretical and experimental value at $54$ volts verifies the wave nature of electrons by diffraction of the beam.

Magnetic field at the center of circular loop

Mathematical Analysis of magnetic field at the center of circular loop:

Let us consider, a current-carrying circular loop of radius $a$ in which $i$ current is flowing. Now take a small length of a current element $dl$ so magnetic field at the center of a circular loop due to the length of current element $dl$. According to Biot-Savart Law:
Magnetic field at the Centre of Current-Carrying Conducting Circular Loop
$dB=\frac{\mu_{\circ}}{4 \pi} \frac{i .dl .sin\: \theta}{a^{2}} $

Here $\theta$ is the angle between length of current element $\left( \overrightarrow{dl} \right)$ and radius $\left( \overrightarrow{a} \right)$. These are perpendicular to each other i.e. $\theta = 90^{\circ}$

$dB=\frac{\mu_{\circ}}{4 \pi} \frac{i .dl .sin\: 90^{\circ}}{a^{2}} $

$dB=\frac{\mu_{\circ}}{4 \pi} \frac{i .dl }{a^{2}} \qquad \left(1 \right)$

The magnetic field at the center due to a complete circular loop

$B=\int dB \qquad \left(2 \right)$

From equation $(1)$ and equation $(2)$

$B=\int \frac{\mu_{\circ}}{4 \pi} \frac{i .dl}{a^{2}}$

$B=\frac{\mu_{\circ}}{4 \pi} \frac{i}{a^{2}} \int dl $

For complete loop $\int dl = 2 \pi a$, So from above equation that can be written as

$B=\frac{\mu_{\circ}}{4 \pi} \frac{i}{a^{2}} \left( 2 \pi a \right) $

$B=\frac{\mu_{\circ} i}{2 a}$

This is an equation of the magnetic field at the center of the circular loop.

Ampere's Circuital Law and its Modification

Ampere's Circuital Law Statement: When the current flows in any infinite long straight conductor then the line integration of the magnetic field around the current-carrying conductor is always equal to the $\mu_{0}$ times of the current.

$\int \overrightarrow{B}. \overrightarrow{dl} = \mu_{\circ} i$

Derivation of Ampere's Circuital Law: Let us consider, An infinite long straight conductor in which $i$ current is flowing, then the magnetic field at distance $a$ around the straight current carrying conductor
Ampere’s Circuital Law for Infinite Long Straight Current-Carrying Conductor
$B=\frac{\mu_{\circ}}{2 \pi} \frac{i}{a}$

Now the line integral of the magnetic field $B$ in a closed loop is

$\oint \overrightarrow{B}. \overrightarrow{dl} = \oint \frac{\mu_{\circ}}{2 \pi} \frac{i}{a} dl$

$\oint \overrightarrow{B}. \overrightarrow{dl} = \frac{\mu_{\circ}}{2 \pi} \frac{i}{a} \oint dl$

$\oint \overrightarrow{B}. \overrightarrow{dl} = \frac{\mu_{\circ}}{2 \pi} \frac{i}{a} \left(2 \pi a \right) \qquad \left( \because \oint dl= 2 \pi a \right)$

$\oint \overrightarrow{B}. \overrightarrow{dl} = \mu_{\circ}i$

Modified Ampere's circuital law: When the capacitor is placed in between the conductors then a current flows in the capacitor which is known as displacement current $\left(i_{d} \right)$. So modified Ampere's circuital law:

$\oint \overrightarrow{B}. \overrightarrow{dl} = \mu_{\circ} \left( i + i_{d} \right)$

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