Wave Nature Of Matter And De Broglie's Equation

Wave Nature of Matter and De Broglie's Equation

As we know light behaves both as a wave and particle. If you observe phenomena like interference, diffraction, or reflection, you will find that light is a wave. However, if you are looking at phenomena like the photoelectric effect, you will find that light has a particle character. De Broglie’s hypothesis stated that there is symmetry in nature and that if the light behaves as both particles and waves, matter too will have both the particle and wave nature. i.e. if a lightwave can behave as a particle then the particle can also behave as waves.

De Broglie’s Equation

De Broglie’s equation is a fundamental concept in quantum mechanics that reveals the wave-particle duality of matter. Proposed by Louis de Broglie in 1924, the equation states that any moving particle or object has an associated wavelength, given by $\lambda =\frac{h}{p}$ where λ is the wavelength, h is Planck's constant (6.626 x 10^-34 Js), and p is the momentum of the particle. This groundbreaking idea suggested that particles such as electrons can exhibit wave-like behaviour, a hypothesis that was later confirmed by experiments demonstrating electron diffraction and interference.

According to De Broglie, A moving material particle can be associated with the wave.

De Broglie proposed that the wavelength $\lambda$ associated with the moving material particle of momentum p is given as

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

where $h=$ plank's constant and $h=6.626\times {10}^{-34}\mathrm{Js}$

Furthermore, we can write De-Broglie wavelength as

$\lambda =\frac{h}{p}=\frac{h}{mv}=\frac{h}{\sqrt{2mK}}$

where

$\begin{array}{rl}& h=\text{plank's constant}\\ & m=\text{mass of particle}\\ & v=\text{speed of the particle}\\ & K=\text{Kinetic energy of particle}\end{array}$

So from De Broglie’s Equation, we can conclude that

$\lambda \alpha \frac{1}{m}$

i.e. Wavelength associated with a heavier particle is smaller than that with a lighter particle.

$\lambda \alpha \frac{1}{\nu }$

i.e when the Particle moves faster, then the wavelength will be smaller and vice versa

if the particle is at rest then De - Broglie wavelength will be infinite $(\lambda =\mathrm{\infty })$

$\lambda \alpha \frac{1}{p}\alpha \frac{1}{v}\alpha \frac{1}{\sqrt{K}}$

De - Broglie wavelength $(\lambda )$ is independent of charge.

De - Broglie Wavelength of Electron

The de Broglie wavelength of an electron is a concept that highlights the wave-particle duality of matter, particularly for subatomic particles like electrons

De Broglie’s Equation is given as $\lambda =\frac{h}{p}=\frac{h}{mv}=\frac{h}{\sqrt{2mK}}$

So for an electron having velocity v attained by it when it is accelerated through a potential difference of V.

then (Kinetic energy gain by the electron)=(work is done on an electron by the electric field)

i.e $K=W_E\Rightarrow \frac{1}{2}{m}_e{v}^2=eV$

So De - Broglie wavelength of Electron is given as ${\lambda }_e=\frac{h}{m_{e}v}=\frac{h}{\sqrt{2m_{e}K}}=\frac{h}{\sqrt{2m_e(eV)}}$

using $h=6.626\times {10}^{-34}\mathrm{Js}$ and $m_e=9.1\times 10\mathrm{kg}$ and $e=1.6\times {10}^{-19}\mathrm{C}$
${\lambda }_e=\frac{12.27}{\sqrt{V}}{A}^{\circ }$
(i.e answer will be in $A^0=$ Angstrom)

Similarly, we can find De - Broglie wavelength associated with charged particle

De - Broglie Wavelength With Charged Particle

The de Broglie wavelength is a concept in quantum mechanics that describes the wave nature of particles.

$\begin{array}{rl}& \lambda =\frac{h}{\sqrt{2mK}}=\frac{h}{\sqrt{2mqV}}\\ & \text{Where}K\to \text{kinetic energy of particle}\\ & q\to \text{charged particle}\\ & V\to \text{potential difference}\\ & \end{array}$

De - Broglie wavelength of the proton

using $m_p=1.67\times {10}^{-27}\mathrm{kg}$ and $q_p=e=1.6\times {10}^{-19}\mathrm{C}$

we get ${\lambda }_{\text{proton}}=\frac{0.286}{\sqrt{V}}{A}^{\circ }$

De - Broglie wavelength of Deuteron

using $m_D=2\times 1.67\times {10}^{-27}\mathrm{kg}$ and $q_D=e=1.6\times {10}^{-19}\mathrm{C}$

we get ${\lambda }_{\text{deutron}}=\frac{0.202}{\sqrt{V}}{A}^{\circ }$

De - Broglie wavelength of an Alpha particle (He²⁺⁾

using $m_{{\alpha }^{2+}}=4\times 1.67\times {10}^{-27}\mathrm{kg}$ and $q_{{\alpha }^{2+}}=2e=2\times 1.6\times {10}^{-19}\mathrm{C}$

we get ${\lambda }_{\alpha -\text{partical}}=\frac{0.101}{\sqrt{V}}{A}^{\circ }$

Electron microscope

An electron microscope is an important application of de-Broglie waves designed to study very minute objects like viruses, microbes and the crystal structure of solids. In the electron microscope, by selecting a suitable value of potential difference V, we can have an electron beam of as small a wavelength as desired. And this de-Broglie wavelength is calculated by using the formula ${\lambda }_e=\frac{12.27}{\sqrt{V}}{A}^{\circ }$.

Solved Examples Based on Wave Nature of Matter and De Broglie's Equation

Example 1: A charged oil drop is suspended in a uniform field of 3 x 10⁴ V/m so that it neither falls nor rises. The charge on the drop will be (take the mass of the charge = 9.9 × 10^-15 kg and g = 10 m/s²)

1) $3.3\times {10}^{-18}\mathrm{C}$
2) $3.2\times {10}^{-18}\mathrm{C}$
3) $1.6\times {10}^{-18}\mathrm{C}$
4) $4.8\times {10}^{-18}\mathrm{C}$

Solution:

Wave-particle duality postulates that all particles exhibit both wave and particle properties

At equilibrium:

Net force = 0

$\begin{array}{rl}& \mathrm{mg}-\mathrm{qE}=0\Rightarrow q=\frac{mg}{E}\\ & q=\frac{9.9\times {10}^{-15}\times 10}{3\times {10}^4}=3.3\times {10}^{-18}\mathrm{C}\end{array}$

Hence, the answer is the option (1).

Example 2: What is the wavelength of a photon with energy 1 ev?

1) $12.4\times {10}^3{A}^{\circ }$
2) $2.4\times {10}^3{A}^{\circ }$
3) $0.4\times {10}^2{A}^{\circ }$
4) $1000A^{\circ }$

Solution:

$\begin{array}{rl}& \lambda =\frac{hc}{E}=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{1\times 1.6\times {10}^{-19}}\times {10}^{10}{A}^{\circ }\\ & =12.375\times {10}^3{A}^{\circ }\end{array}$

Hence, the answer is the option (1).

Example 3: An electron beam is accelerated by a potential difference V to hit a metallic target to produce X-rays. It produces continuous as well as characteristic X-rays. If λ_min is the smallest possible wavelength of X-ray in the spectrum, the variation of log λ_min with log V is correctly represented in :

1)

2)

3)

4)

Solution:

De - Broglie wavelength

$\lambda =\frac{h}{p}=\frac{h}{mv}=\frac{h}{\sqrt{2mE}}$
wherein
$m=$ mass of particle
$v=$ speed of the particle
$E=$ Kinetic energy of the particle
$eV=\frac{hc}{{\lambda }_{\text{min}}}$
$V=\frac{hc}{e{\lambda }_{min}}$
Take logs on both sides
$\log V=\log \frac{hc}{e}-\log {\lambda }_{min}$
$\log {\lambda }_{min}=\log \frac{hc}{e}-\log V$

Hence, the answer is the option (1).

Example 4: Particle A of mass m and initial velocity $v$ collides with a particle B of mass $\frac{m}{2}$ which is at rest. The collision is head-on and elastic. The ratio of the de-Broglie wavelengths λ_A to λ_B after the collision is :

1) $\frac{{\lambda }_A}{{\lambda }_B}=\frac{1}{3}$
2) $\frac{{\lambda }_A}{{\lambda }_B}=2$
3) $\frac{{\lambda }_A}{{\lambda }_B}=\frac{2}{3}$
4) $\frac{{\lambda }_A}{{\lambda }_B}=\frac{1}{2}$

Solution:

m_A = Mass of particle A

m_B= Mass of particle B

v_A= velocity of Particle A

v_b= Velocity of particle B

The velocity of the particle After the collision,

$v_A=\frac{m_A-m_B}{m_A+m_B}\cdot v=\frac{m-\frac{m}{2}}{\frac{3m}{2}}\cdot v=\frac{v}{3}$

The velocity of particle B after the collision,

$\begin{array}{rl}& v_B=\frac{2m_A}{m_A+m_B}\cdot v=\frac{2m}{\frac{3m}{2}}\cdot v=\frac{4v}{3}\\ & \frac{{\lambda }_A}{{\lambda }_B}=\frac{P_B}{P_A}=\frac{\left(\frac{m}{2}\right)\cdot \frac{4v}{3}}{m\cdot \frac{v}{3}}=2:1\end{array}$

Hence, the answer is the option (2).

Example 5: De-Broglie wavelength associated with the electron in the n=4 level is :

1) two times the de-Broglie wavelength of the electron in the ground state

2) four times the de-Broglie wavelength of the electron in the ground state

3) half of the de-Broglie wavelength of the electron in the ground state

4) M1/4^th of the de-Broglie wavelength of the electron in the ground state

Solution:

De - Broglie wavelength

$\lambda =\frac{h}{p}=\frac{h}{mv}=\frac{h}{\sqrt{2mE}}$

wherein

$\begin{array}{rl}& h=\text{plank's constant}\\ & m=\text{mass of particle}\\ & v=\text{speed of the particle}\\ & E=\text{Kinetic energy of particle}\\ & \end{array}$

From de broglie equation: $p=\frac{h}{\lambda }$ and $mvr=\frac{nh}{2\pi }$

$\begin{array}{rl}\Rightarrow & 2\pi r=n\lambda \\ & r=r_0\frac{n^2}{z}\\ \Rightarrow & 2\pi r_0\frac{n^2}{z}=n\lambda \\ \Rightarrow & \lambda =n\cdot \frac{2\pi r_0}{z}\end{array}$

In ground state n = 1 $\therefore \lambda =\frac{2\pi r_0}{z}$

and in n = 4, ${\lambda }^{\prime }=\frac{8\pi r_0}{z}$

Hence, the answer is the option (2).

Summary

The wave nature of matter, proposed by Louis de Broglie in 1924, suggests that particles such as electrons exhibit wave-like properties, encapsulated in the equation λ=h/p. This wave-particle duality is foundational in quantum mechanics and is utilized in technologies like electron microscopes for high-resolution imaging. The de Broglie wavelength of a particle depends on its momentum and is significant in understanding phenomena at the atomic and subatomic levels.