De Broglie Relationship

The twentieth century was a period that witnessed the quest to understand the dual nature of radiation and matter as well as new frontiers in quantum mechanics. De Broglie formulated the wave-particle duality, which challenged traditional theories about matter being made up of particles. In other words, like light exhibits particle-like and wave-like properties, particles including electrons can also exhibit wavelike characteristics. In 1924, this radical proposal appeared for the first time in a doctoral thesis by De Broglie. In addition to photons alone, the hypothesis postulated that every particle moving must bear an associated wavelength known today as de Broglie’s wavelength thus providing a remarkable insight into the wave character of matter and inaugurating a new way of thinking about atomic and molecular behavior. Symbolized as λ (lambda), the de Broglie wavelength is inversely proportional to momentum. It gives information about constructive and destructive interference locations during the movement of the particle across space.

In this article, we will cover the concept of De Broglie wavelength and several related parameters. This concept falls under the broader category of Atomic structure, which is a crucial chapter in Class 11 chemistry. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), National Eligibility Entrance Test (NEET), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more.

Chemically speaking; it has significant implications for electron behaviours within atoms/molecules. Let us study in detail about the De Broglie wavelength and related formula for a better understanding of the topic followed by some related problems.

De-Broglie Wavelength- Quantum Leap:

Louis de Broglie made the wave theory of matter in 1924 and it was established as the de Broglie equation, which states the wavelength of the particle (λ) is inversely proportional to the momentum (p) of the particle. It is given in terms of Planck’s constant through the formula λ = h/p λ=h/p In this formula h represents Planck’s constant wherein p stands for the momentum of a particular particle. This equation has a certain aspect of wave-particle duality that describes quantum mechanics; nevertheless, electrons and atom particles exhibit wave properties. The equation presented by de Broglie is called the de Broglie equation proposing a method by which matter possesses wave characteristics thus affecting the fundamental approach toward the particle nature of matter in the atomic and subatomic domain. Their use cuts across all the disciplines in sciences including Physics, Chemistry, Material Sciences, and Others.

The de Broglie’s prediction was confirmed experimentally when it was found that an electron beam undergoes diffraction, a phenomenon characteristic of waves.

It needs to be noted that according to de Broglie, every object in motion has a wave character. This wavelength is quite significant for the subatomic particles which have very small masses. The wavelengths associated with ordinary objects are however so short that their wave properties cannot be detected as they have large masses.

Bohr's model and de Broglie's relation: Number of standing waves made by an electron in n^th Bohr orbit

According to Bohr's model,

$
m v r=\frac{n h}{2 \pi}
$

According to de Broglie's Relation,
$
p=\frac{h}{\lambda}
$

Combining the two

$2 \pi r=n \lambda$

So, the number of waves made by any electron in the nth orbit is equal to the principal quantum number of the orbit, i.e. n.

De-Broglie Equation-

de-Broglie proposed that just like light, matter should exhibit both particle and wave-like properties. This means that just as the photon has momentum as well as wavelength, electrons should also have momentum as well as wavelength, and he proposed the following mathematical relationship:

There is the de Broglie equation, which was given by Louis de Broglie and it defines the relationship between wavelength and momentum of the particle It is expressed as λ = h/mv. It is mathematically symbolized as λ = h/p; here h is Planck’s constant which is a natural constant. This equation implies that all particles, right from electrons to atoms, are regarded as having characteristics of both waves and particles. More precisely, it measures the wave properties of matter; it entails that particles with high momenta possess short wavelengths. This is in line with the de Broglie operation which is a substantial bar in quantum mechanical theory, which is intended to explain matter wave behaviour to give an account of the constitution of atomic and sub-atomic dimensions. This is because it retains several effects on so many science disciplines such as physics, chemistry as well as material science utilizing it to consistencies occurrences like electron interferences, atomic absorption, and movements of molecules among others.

$\lambda=\frac{h}{m v}=\frac{h}{p}=\frac{h}{\sqrt{2 m K E}}=\frac{h}{\sqrt{2 m q V}}$

where m is the mass of the particle, v is its velocity, p is its momentum,

KE is the Kinetic Energy of the particle,

V is the voltage across which the Charged particle having charge q is accelerated.

Recommended topic video on (De Broglie Relationship )

Some Solved Examples

Example 1: The de-Broglie wavelength of a particle of mass 6.63 g moving with a velocity of 100 ms^-1 is:

1) (correct) $10^{-33} \mathrm{~m}$
2) $10^{-35} \mathrm{~m}$
3) $10^{-31} \mathrm{~m}$
4) $10^{-25} \mathrm{~m}$

Solution

As discussed in concept:

De-Broglie wavelength:

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

- wherein

where m is the mass of the particle

v: its velocity

p: its momentum

So,

$\begin{aligned} & \lambda=\frac{6.625 \times 10^{-34}}{6.63 \times 10^{-3} \times 100} \\ & \lambda=10^{-33} \mathrm{~m}\end{aligned}$

Hence, the answer is the option (1).

Example 2: A stream of electrons from a heated filament was passed between two charged plates kept at a potential difference V esu. If e and m are the charge and mass of an electron, respectively, then the value of h/λ (where λ is wavelength associated with electron wave) is given by

1) $\mathrm{meV}$
2) $2 \mathrm{meV}$
3) $\sqrt{m e V}$
4) (correct) $\sqrt{2 m e V}$

Solution

We know that

De-Broglie wavelength:

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

- where

m is the mass of an electron.

v - its velocity.

p - its momentum.

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

The kinetic energy of an electron in $\mathrm{eV}$
$
\begin{aligned}
& K \cdot E=\frac{p^2}{2 m} \\
& \therefore e V=\frac{p^2}{2 m}
\end{aligned}
$

where

V is the potential difference between two charge plates.

e is the charge of an electron.

Or $p=\sqrt{2 m e V}$
from equations (1) and (2)
$
\frac{h}{\lambda}=\sqrt{2 m e V}
$

Hence, the answer is an option (4).

Conclusion:

To sum up, de Broglie’s wavelength has had an overwhelming impact on how we understand the movement of matter at atomic and subatomic levels. By proposing that particles like electrons have a wave nature, Louis de Broglie changed the entire game of quantum mechanics and opened its doors to new revelations about the nature of particles and their interactions with each other. The de Broglie wavelength presents us with a very powerful way of describing moving objects which surpasses classical mechanics or wave theory. In chemistry, it has helped us to study atoms and molecules by providing electron behaviour insights thereby shaping our knowledge of atomic structure, chemical bonding, as well as spectroscopy. Just beyond this fact, however, it has found extensive use in many different areas ranging from quantum computing to materials science where the wavy motions are exploited for various advancements in technology. Still, as we continue delving into the depths of quantum physics, the de Broglie’s wavelength will remain one of its keystones giving us direction into what matters and stimulating our curiosity to further explore where light seems not able to penetrate.