Skip to main content

de-Broglie Concept of Matter wave

Louis de-Broglie thought that similar to the dual nature of light, material particles must also possess the dual character of particle and wave. This means that material particles sometimes behave as particle nature and sometimes behave like a wave nature.

According to de-Broglie –
A moving particle is always associated with a wave, called as de-Broglie matter-wave, whose wavelengths depend upon the mass of the particle and its velocity.

According to Planck’s theory of radiation–

$E=h\nu \qquad(1) $

Where
h – Planck’s constant
$\nu $ - frequency

According to Einstein’s mass-energy relation –

$E=mc^ {2} \qquad (2)$

According to de Broglie's hypothesis equation $ (1)$ and equation $(2)$ can be written as –

$mc^ {2} = h \nu$

$mc^ {2} = \frac{hc}{\lambda }$

$\lambda =\frac{h}{mc}\qquad(3) $

$\lambda =\frac{h}{P}$

Where $P$ –Momentum of Photon

Similarly from equation $(3)$ the expression for matter waves can be written as


$\lambda=\frac{h}{mv}=\frac{h}{P}\qquad(4)$

Here $P$ is the momentum of the moving particle.

1.) de-Broglie Wavelength in terms of Kinetic Energy

$K=\frac{1}{2} mv ^{2}$

$K=\frac{m^{2}v^{2}}{2m}$

$K=\frac{P^{2}}{2m}$

$P=\sqrt{2mK}$

Now substitute the value of $P$ in equation $ (4)$ so

$\lambda =\frac{h}{\sqrt{2mK}} \qquad (5)$

2.) de-Broglie Wavelength for a Charged particle

The kinetic energy of a charged particle is $K = qv$

Now substitute the value of $K$ in equation$(5)$ so

$\lambda =\frac{h}{\sqrt{2mqv}}$

3.) de-Broglie Wavelength for an Electron

The kinetic energy of an electron

$K=ev$

If the relativistic variation of mass with a velocity of the electron is ignored then $m=m_{0}$ wavelength

$\lambda =\frac{h}{\sqrt{2m_{0}ev}}$
So wavelength of de-Broglie wave associated with the electron in non-relativistic cases

4.) de-Broglie wavelength for a particle in Thermal Equilibrium

For a particle of mass $m$ in thermal equilibrium at temperature $T@

$K=\frac{3}{2}kT$

Where $K$ – Boltzmann Constant

$\lambda =\frac{h}{\sqrt{2m.\frac{3}{2}kt}}$

$\lambda =\frac{h}{\sqrt{3mKT}}$

Properties of matter wave →

  1. Matter waves are generated only if the material's particles are in motion.

  2. Matter-wave is produced whether the particles are charged or uncharged.
  3. The velocity of the matter wave is constant; it depends on the velocity of material particles.

  4. For the velocity of a given particle, the wavelength of matter waves will be shorter for a particle of large mass and vice-versa.

  5. The matter waves are not electromagnetic waves.

  6. The speed of matter waves is greater than the speed of light.

    According to Einstein’s mass-energy relation

    $E=mc^{2}$

    $h\nu = mc^{2}$

    $\nu =\frac{mc^{2}}{h}$

    Where $\nu$ is the frequency of matter-wave.

    We know that the velocity of matter-wave

    $ u =\nu \lambda $
    Substitute the value of $\nu$ in the above equation

    $u =\frac{mc^{2}}{h}. \lambda $
    $u =\frac{mc^{2}}{h} . \frac{h}{mv}$

    $u =\frac{c^{2}}{v}$

    Where $v$ → particle velocity which is less than the velocity of light.

  7. The wave and particle nature of moving bodies can never be observed simultaneously.

Comments

Popular Posts

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

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