De Broglie Relationship

Introduction

Louis de Broglie introduced his research work in the year 1924, in which he suggested that electrons (e^-) have both waves as well as particle properties, like light. He reorganised the conditions of the Plank-Einstein relation so that they could be applied to all types of matter. This nature has been explained as dual matter behaviour. He derived a relationship between matter momentum as well as wavelength based on his findings. The De Broglie Relationship is the name given to this type of relationship. Bohr's model did not explain many concepts relevant to the spectrum of numerous atoms as well as the splitting of spectral lines in magnetic plus electric fields (e.f.). To deal with the challenges of Bohr's atomic model, steps were taken to create a comprehensive atomic model. Every moving body has a wavelike appearance. Because of their large mass, the wavelengths affiliated with ordinary objects are short. As a result, their wave properties cannot be identified. On the other hand, the wavelengths associated with electrons or other subatomic particles can be identified with experimental data.

What is the De Broglie Equation

The de Broglie equation (eq.) is an eq. used to explain the wave attributes of matter, particularly the wave nature of the electron (e^-): λ=h/mv, where λ is the wavelength, h is Planck's constant, and m is the mass of a particle moving at velocity v. He proposed that particles can show wave properties. When matter waves were noticed in George Paget Thomson's cathode ray diffraction study as well as the Davisson-Germer study, which applied to electrons (e^-), the de Broglie hypothesis was confirmed. Ever since, it has been demonstrated that the de Broglie equation (eq.) applies to elementary particles, neutral atoms, as well as molecules.

Significance of De Broglie Equation

According to de Broglie, all mass in motion its particle nature. However, a car or ball in motion does not appear to have a particle nature. To demonstrate, de Broglie calculated the wavelengths of a cricket ball.

Let us suppose,

$$\mathrm{Mass\: of\: the\: ball = 150g (150 ✕ 10^{-3}kg),}$$

Velocity = 35 m/s, and

$$\mathrm{h=6.626 ✕ 10^{-34}Js}$$

Now, inserting these no. into the equation

$$\mathrm{λ= hmv}$$

$$\mathrm{λ=6.626 ✕ 10^{-34}/150 ✕ 10^{-3}✕ 35 }$$

We obtain,

$$\mathrm{λ\: ball=1.2621✕ 10^{-34}} $$

$$\mathrm{m=1.2621 ✕ 10^{-24} Å}$$

Because Å is a very small unit, the value is in the power of 10^-24. From this point forward, the moving cricket ball is a particle. The concern now is whether this ball has a wave nature. Because the value of the ball is immeasurable, the result will be a resounding no. As a result, de Broglie's theory of wave-particle duality holds for objects in motion 'up to' but not equal to the size of electrons.

De Broglie Wavelength for an Electron

$$\mathrm{It\: is\: given\: that, m_e =9.1 ✕ 10^{-31} kg, and \: v_e=218 ✕ 10^6 m/s }$$

Placing these values into the formula

$$\mathrm{λ=h/mv=9.1 ✕ 10^{-31}/218 ✕ 10^6}$$

$$\mathrm{λ=3.2 Å }$$

This value is quantifiable. As a result, electrons (e^-) have wave-particle duality. Furthermore, all large bodies have a wave nature, whereas microscopic objects such as electrons (e^-) have a wave-particle nature.

The Conclusion of De Broglie Hypothesis

De Broglie's theory is among the reasons that violate classical mechanics laws & provide the foundation for quantum mechanics. Every moving particle has a wavelength affiliated with it, which describes the matter's dual behaviour. Several uses of this equation (eq.) are well-founded in the fields of chemistry but also physics.

We can deduct from the de Broglie equation (eq.) for a particle, i.e.,

λ=h/p=h/mv,

we conclude that −

If v=0, then λ=∞, and,

If λ=∞, then v=0

This means that waves are only affiliated with material particles in motion. As a result, the charge of these waves is unaffected.

Conclusion

Any component of a material object's behaviour or attributes that differ in space or time in compliance with the mathematical equations (Eqs.) that describe waves is referred to as a De Broglie wave, also identified as a matter-wave. The de Broglie wave equations (Eqs.) can be used to determine the wavelength of any moving object. The wavelength of an electron (e^-) increases as its speed decreases. He established a specific relationship between the wavelength as well as the momentum of a particle (p). λ=h/mv=h/p. The wavelength is given here, as well as the momentum is given by p. Every moving object has a wave nature to it. The wavelengths affiliated with common objects are so short due to their large masses that their wave attributes cannot be defined.

FAQs

1 .Which has the faster speed for the same de Broglie wavelength: electron (e^-) or a proton?

The particle has the same de-Broglie wavelength. As a result, their momentum will be equal. The product of mass & velocity is momentum. As a result, the speed of the particle varies inversely with its mass. A proton with a higher mass will travel faster, while an electron (e^-) with a lower mass will travel faster.

2. Give the relationship between an object's de-Broglie wavelength as well as kinetic energy.

The de-Broglie wavelength of an electron (e^-) moving with kinetic energy K is expressed as, $\mathrm{λ=h/\sqrt{2mk}} $. The relationship between the wavelength of the wave affiliated with the moving electron (e^-) and the momentum of the electron, as per de Broglie, is λ=h/mv. De-Broglie established a relationship between a body's wavelength as well as momentum. The connection is, λ=h/mv=h/p

3. Determine the relationship between the de Broglie wavelength as well as temperature.

We understand that a particle's average KE is −

K=3/2 k_bT

Where k_bdenotes Boltzmann's constant & T denotes temperature in Kelvin.

A particle's kinetic energy is $\mathrm{\frac{1}{2}}$ mv²

$$\mathrm{A\: particle's\: momentum\: p=mv=\sqrt{2mk} = \sqrt{2m(3/2)} k_b T=\sqrt{2mk_b T}} $$

$$\mathrm{de\: Broglie\: wavelength,\: A=h/p=h\sqrt{2mk_b T}} $$

4. Is the de Broglie hypothesis applicable to macroscopic matter?

The de Broglie relation applies to both microscopic but also macroscopic objects. A macro-sized 100Kg car travelling at 100m/s, e.g., will have a

$$\mathrm{ Wavelength of λ=h/mv=6.63 \times 10^{-34}/100 \times 100=6.63 \times 10^{-30} m}$$

The wavelengths of high-energy -radiations are only 10^-12 m.

High frequencies correspond to very small wavelengths. To generate mass, waves under a certain wavelength or above a certain frequency develop particle-antiparticle annihilation. As a result, wave nature or de Broglie wavelength is not visible in the macroscopic matter.

5. For an electron (e^-) In the 10^th Bohr's orbit of an H-atom, the De-Broglie wavelength is [h=6.625✕ 10^-34j/s,m_e=9.110✕ 10^-31 kg, and a_0=0.529 Å].

$$\mathrm{A_0=0.529 Å} $$

The wavelength of e^- de-Broglie present in

$$\mathrm{1^{st}\: Bohr\: orbit\: =2\Pi 0.529 Å} $$

$$\mathrm{10^{th}\: Bohr\: orbit =2\Pi 0.529 Å10} $$

$$\mathrm{2\Pi 5.29 Å} $$

The quantization of angular momentum

$$\mathrm{mva_0=h/2\Pi} $$

$$\mathrm{2\Pi a_0=h/mv=λ}$$

$$\mathrm{2\Pi a_0=2\Pi 5.29 Å} $$

$$\mathrm{=2\Pi 0.529 Å=33.2 Å} $$