# Unit and dimension about De broglie wave

• I
• KT KIM
In summary, the de Broglie wave formula tells us that the wavelength (lambda) is equal to Planck's constant (h) divided by the momentum (p). In practice, this would be expressed as "something [meters]." The units of h/mv can be broken down into joule-second (J*s) for Planck's constant and kilogram-meter per second (kg*m/s) for momentum. Simplifying this, we get units of meters for the wavelength. This is related to the Mass-Energy relation, but it may be difficult to understand how this conversion works.

#### KT KIM

From the de Broglie wave formula we know,
Rhamda=h/p

In actual examples of course the answer would be 'something [meters]'

I am having hardtime to understand how unit of h/mv
[J*s]/[kg]*[m/s] turn into wavelength unit
[m]

I studied the Mass-Energy relation part earlier,
But still can't get the precise way how actually it works and modifes its unit.

KT KIM said:
From the de Broglie wave formula we know,
Rhamda=h/p

In actual examples of course the answer would be 'something [meters]'

I am having hardtime to understand how unit of h/mv
[J*s]/[kg]*[m/s] turn into wavelength unit
[m]

I studied the Mass-Energy relation part earlier,
But still can't get the precise way how actually it works and modifes its unit.
What is 1 Joule, expressed in SI base units?

• KT KIM
KT KIM said:
I am having hardtime how unit of h/mv
[J*s]/[kg]*[m/s] turn into wavelength unit
[m]

I studied the Mass-Energy relation part earlier,
But still can't get the precise way how actually it works and modifes its unit.
Planck's constant is h -its units are joule-sec;
Joule is expressed as energy - its units are kg.m^2.s ^-2 ; so j--s will be kg.m^2.s^-1
so h/ p will be kg.m^2.s^-1 /(kg.m.s^-1 ) = m (in meters) which is the unit of Lambda the wavelength.

drvrm said:
Planck's constant is h -its units are joule-sec;
Joule is expressed as energy - its units are kg.m^2.s ^-2 ; so j--s will be kg.m^2.s^-1
so h/ p will be kg.m^2.s^-1 /(kg.m.s^-1 ) = m (in meters) which is the unit of Lambda the wavelength.
Thanks!

## 1. What is the De Broglie wave?

The De Broglie wave is a concept in quantum mechanics that describes the wave-like nature of matter. It is named after physicist Louis de Broglie, who proposed that all particles, including electrons and atoms, have a wave-like nature.

## 2. What are the units of the De Broglie wavelength?

The units of the De Broglie wavelength are meters (m) or any unit of length, as it represents the physical distance of the wavelength.

## 3. How is the De Broglie wavelength related to the momentum of a particle?

The De Broglie wavelength is inversely proportional to the momentum of a particle. This means that as the momentum increases, the wavelength decreases, and vice versa.

## 4. What is the significance of the De Broglie wavelength?

The De Broglie wavelength is significant because it provides a way to describe the wave-like behavior of matter, which was previously only attributed to light. It also helps to explain certain phenomena in quantum mechanics, such as diffraction and interference patterns.

## 5. How is the De Broglie wavelength calculated?

The De Broglie wavelength can be calculated using the equation: λ = h / p, where λ is the De Broglie wavelength, h is Planck's constant, and p is the momentum of the particle. This equation is based on the De Broglie hypothesis that states that the wavelength of a particle is inversely proportional to its momentum.