De Broglie Wavelength - Online Tutor, Practice Problems & Exam Prep

The de Broglie wavelength concept connects an object's wave-like behavior to its momentum, utilizing Planck's constant (6.626 x 10^-34 J·s). According to the de Broglie equation, wavelength (λ) is inversely proportional to the product of the object's mass (kg) and velocity (m/s), indicating that as mass or velocity increases, the wavelength decreases. This principle is fundamental in quantum mechanics, providing insight into the wave-particle duality of matter, particularly at the subatomic level. Understanding the de Broglie wavelength is crucial for comprehending phenomena such as electron diffraction and the behavior of particles in potential wells.

λ=hmv

De Broglie Wavelength

Now the de Broglie wavelength associates the wave nature of a moving object with its velocity through Planck's constant. Now when we say object, this is a photon, a subatomic particle or literally anything with a velocity.

Now here we have our de Broglie wavelength formula. Here we're going to say that this formula is used when we have the velocity and wavelength of an object. So here we're going to say that wavelength, which is represented by Lambda in meters, equals Planck's constant, which is equal to 6.626×10-34 joules times seconds divided by mass of the object in kilograms times the velocity in meters per second.

So just realize here that based on the formula, the wavelength of an object is going to be inversely proportional to its mass and velocity.

De Broglie Wavelength Example 1

So here we have to find the wavelength of a proton with a speed of 6.25 * 10 to the five meters per second. Here the mass of a proton is recorded as 1.67 * 10 to the -27 kilograms. All right, so it's going to be

wavelength = Planck's constant mass × velocity

Now remember here that Planck's constant is Joules times seconds, and what we need to realize here is that Joule 1 Joule is equal to kilograms times meter squared over second squared. So if we're doing Joules times seconds, that really means kilograms times meters squared over second squared times seconds. So here one of these seconds will cancel out with this second.

So the units for Planck's constant would be

6.626 × 10 - 34 kg × m 2 / s

We do this to see how the kilograms of my subatomic particle in this case can cancel out. So we have

1.67 × 10 - 27 kg

and it's traveling at a velocity of

6.25 × 10 5 m / s

O here. What cancels out are meters, cancels out with meters. And then we're going to have here seconds cancel out with seconds, kilograms cancel out with kilograms. So we're going to have 1m left. So when we plug that into our calculator, we'll get

6.35 × 10 - 13 m

for our wavelength of our proton within this question.

What is the velocity (in m/s) of an electron that has a wavelength of 3.13 x 10⁵ pm? (Mass of an electron = 9.11 x 10^-31 kg).

• A. 1543 m/s
• B. 3220 m/s
• C. 2320 m/s
• D. 3541 m/s

The faster an electron is moving, the _________ its kinetic energy, and the __________ its wavelength.

• A. higher, shorter
• B. higher, longer
• C. lower, longer
• D. lower, shorter

Consider an atom traveling at 3.00 x 10¹⁵ m/s. The de Broglie wavelength is found to be 7.1316 x 10^-39. Determine the mass (in g) of the atom.

• A. 3.10 × 10^–6 g
• B. 6.20 × 10^–8 g
• C. 3.10 × 10^–8 g
• D. 4.45 × 10^–7 g

Here’s what students ask on this topic:

How do you calculate the de Broglie wavelength?

To calculate the de Broglie wavelength of a particle, you use the de Broglie equation, which relates a particle's wavelength to its momentum. The formula is given by:

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

Where:

• $\lambda$ (lambda) is the de Broglie wavelength
• $h$ is the Planck constant (approximately 6.626 x 10^-34 Js)
• $p$ is the momentum of the particle

The momentum ($p$) of a particle is the product of its mass ($m$) and velocity ($v$), so $p=mv$. Therefore, the de Broglie wavelength can also be calculated by:

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

If you're dealing with an electron or any other charged particle that's been accelerated by a voltage ($V$), you can relate its kinetic energy to the voltage by the equation $\mathrm{KE}=eV$, where $e$ is the elementary charge (approximately 1.602 x 10^-19 C). Since the kinetic energy ($\mathrm{KE}$) is also equal to $\frac{1}{2}m{v}^2$, you can solve for $v$ and then use that to find the momentum, which you can then plug into the de Broglie equation to find the wavelength.

What is the De Broglie wavelength?

The De Broglie wavelength is a fundamental concept in quantum mechanics, introduced by the French physicist Louis de Broglie in 1924. It relates to the wave-particle duality of matter, which proposes that all particles exhibit both wave-like and particle-like properties. The De Broglie wavelength is the wavelength associated with a particle and is given by the De Broglie equation:

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

where $\lambda$ (lambda) is the De Broglie wavelength, $h$ is the Planck constant (approximately 6.626 x 10^-34 Js), and $p$ is the momentum of the particle (the product of its mass $m$ and velocity $v$).

This concept is particularly significant when dealing with microscopic particles, such as electrons, where the wave-like behavior becomes evident. For example, electrons can exhibit diffraction patterns, which is a wave phenomenon, when passed through a double-slit apparatus. The De Broglie wavelength has been instrumental in the development of quantum mechanics and has applications in various fields such as electron microscopy and the study of atomic and subatomic systems.

How does the de Broglie wavelength of an electron change when its momentum increases?

The de Broglie wavelength of an electron is inversely proportional to its momentum. This relationship is given by the de Broglie equation, which states that the wavelength (λ) is equal to the Planck constant (h) divided by the momentum (p) of the particle:

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

Momentum is the product of the mass and velocity of a particle (p = mv). Therefore, when the momentum of an electron increases, which can happen if its velocity increases (assuming its mass remains constant), its de Broglie wavelength decreases. Conversely, if the momentum of the electron decreases, its de Broglie wavelength increases.

In a more intuitive sense, you can think of it like this: as the electron moves faster and its momentum goes up, it behaves more like a particle with a shorter wavelength. When it moves slower and its momentum goes down, it behaves more like a wave with a longer wavelength.