Louis De Broglie: Research and Atomic Theories

Louis De Broglie’s History

Louis De Broglie was born on August 15, 1892, and was a French physicist. De Broglie was most famous for his discovery of the wavelength of electrons (specified below). In his early days, De Broglie studied theoretical physics at 18 while also earning his degree in history. He described his mind as “a pure theoretician than that of an experimenter or engineer, loving especially the general and philosophical view.” De Broglie was interested in concepts and the mysteries of atomic physics was what drew him into the field.

In 1924, he would develop his theory of waves accompanying particles, which would later lead to his generalization that all types of matter have wave properties. He won the 1929 Nobel prize for Physics, and his later career would consist of teaching theoretical physics as a professor at the Henri Poincaré Institute.

De Broglie’s Wavelength

A wavelength is represented as λ, equivalent to the velocity of a wavetrain divided by the mass and constant. Examples of how it looks will be provided, but first, where exactly does that formula come from? The De Broglie wavelength came alive in 1924 when Louis De Broglie theorized that particles also have wave properties.

Although in everyday objects we cannot see or feel the wavelengths due to the small size, in subatomic particles they’re extremely visible and active. De Broglie suggested that any particle of matter that contains momentum also has a wavelength linked to it. In simple terms, particles can also be expressed as waves. De Broglie’s generalization about wave-particle duality helped contribute to the universal atomic theory, and changed the study of quantum mechanics.

The Atomic Model

De Broglie’s wave-particle theory also contibuted to the Bohr model for the hydrogen atom. The Bohr model is a structural model of an atom, and electrons are visualized as circular orbits around the nucleus of the atom. Eventually, the Bohr model turned out to be accurate for the hydrogen atom only, but De Broglie’s wave-particle duality theory helped lessen the gaps in Bohr’s model and presented the electron as a particle with wave-like ring around it. De Broglie’s hypothesis came to the conclusion that a certain number of wavelengths have to fit within the circumference of the orbit of the electrons within the hydrogen atom, permanently changing the way the hydrogen model looks.

The Bohr Model visual vs The Broglie-Bohr Model visual.
Notice the difference between the visualization of the electron in a hydrogen atom. The De Broglie-Bohr model highlights the wave-like properties of the particle whereas the Bohr model doesn’t display the wave properties.

De Broglie’s Equation

De Broglie's Equation in simple form.

The equation states that the wavelength of an object is equal to Planck’s constant divided by mass and velocity. Down below is a simple example problem to work this equation out!

Example Problems

PROBLEM 1: Find the wavelength of an electron moving at 4.44 x 104 m/s.

EXPLANATION: De Broglie’s equation is λ = h/mv, and we know it’s asking for the wavelength, which is λ. Three other factors equate to the wavelength, the mass of an electron, the velocity of the electron, and Planck’s constant. This equation gives us the velocity, Planck’s constant, and the mass of an electron is 9.11 x 10-31 kg.

λ = (6.626 x 10-34 J·s) / ((4.44 x 104 m/s)(9.11 x 10-31 kg))

λ = (6.626 x 10-34 J·s) / (4.04484 x 10-26 kgm/s)

λ = 1.638 x 10-8 m

The wavelength is 1.638 x 10-8 m.

PROBLEM 2: An electron has the wavelength of 1.638 x 10-8 m. Using this information, find the speed (velocity) of the electron.

EXPLANATION: Likewise, we have three factors to solve this, except in this situation we have the wavelength, mass, and Planck’s constant. To find the velocity, we would have to reverse the regular procedure we do to find the wavelength.

1.638 x 10-8 m = (6.626 x 10-34 J·s) / ((9.11 x 10-31 kg) * v)

(1.638 x 10-8 m)((9.11 x 10-31 kg) * v)) = (6.626 x 10-34 J·s)

(1.492218 x 10-38 )(v) = (6.626 x 10-34 J·s)

(v) = (6.626 x 10-34 J·s) / (1.492218 x 10-38 )

v = 44403.6997275

In the end, our velocity is 4.44 104 m/s when converted to scientific notation, and you can plug it in to see if the wavelengths match up!