### Magnetic dipole moment of a revolving electron

The magnetic dipole moment of a revolving electron (Or Magnetic Moment due to Orbital Angular Momentum):
An electron revolving in an orbit about the nucleus of an atom behaves like a current carrying loop. It is called a minute current-loop and produces a magnetic field. Every current loop is associated with a magnetic moment.
 Magnetic Dipole Moment of a Revolving Electron

Let us consider, that the magnetic moment associated with a loop carrying current $i$ and having area $A$ is:

$\mu_{L}= i.A \qquad(1)$

The current due to a revolving electron is

$i=\frac{e}{T}$

Where

$T$- The period of revolution of electron motion around the nucleus i.e $T=\frac{2 \pi r}{v}$
$e$- Charge on an electron So from the above equation

$i=\frac{e}{\frac{2 \pi r}{v}}$

$i=\frac{ev}{2 \pi r} \qquad(2)$

The area of the current loop is:

$A=\pi r^{2} \qquad(3)$

Now put the value of $i$ and $A$ in equation $(1)$

$\mu_{L}= \left( \frac{ev}{2 \pi r} \right) \left( \pi r^{2} \right)$

$\mu_{L}= \frac{evr}{2} \qquad(4)$

$\mu_{L}= \left(\frac{evr}{2}\right) \left( \frac{m}{m} \right)$

$\mu_{L}= \left(\frac{e}{2m}\right) \left( mvr \right)$

$\mu_{L}= \left(\frac{e}{2m}\right) L \qquad(5) \qquad (\because L=mvr)$

Where $L$- The orbital angular momentum of the electron and another value of $L$ is

$L=\frac{nh}{2\pi} \qquad(6)$

$\mu_{L}= \left(\frac{e}{2m}\right) \frac{nh}{2\pi}$

$\mu_{L}= n \: \left(\frac{eh}{4m \pi}\right)$

Where $n=1,2,3......$ is the principle quantum number.

This equation gives the magnetic moment associated with the orbital motion of the electron.

Bohr Magneton:
Bohr Magneton is defined as the angular momentum of an eletcron in ground state.
We know that:
• Principle Quantum Number$(n)=1$
• Charge of a electron $(e)=1.6\times10^{-19} C$
• Planck Constant $(h)= 6.623 \times 10^{-34} J-sec$
• Mass of electron $(m)= 9.1 \times 10^{-31} Kg$

• So the magnetic moment of an electron in the ground state:

$\mu_{B}= n \: \left(\frac{eh}{4m \pi}\right)$

Now subtitute the value of $n,e,h,m$ in above equation:

$\mu_{B}= \frac{1 \times 1.6\times10^{-19} \times 6.623 \times 10^{-34} }{4 \times 3.14 \times 9.1 \times 10^{-31} }$

$\mu_{B}= 9.274 \times 10^{-24} A-m^{2}$

The magnetic moments due to orbital motion of electrons in higher orbits are multiples of the Bohr magneton value.

### 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 then 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 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 then 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 light w

### 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

### 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