What is the de Broglie wavelength of this electron?

In summary, to find the de Broglie wavelength of an electron with a kinetic energy of 500 eV, we first convert the energy to joules, then use the relationship between kinetic energy and momentum to solve for momentum. Finally, we use the formula for de Broglie wavelength to calculate the result.
  • #1
gnozahs
3
0

Homework Statement


What is the de Broglie wavelength of this electron?
Given is 500 eV


Homework Equations


KE= 1/2 mv^2
1 eV=1.602x19^-19 J


The Attempt at a Solution


I converted the 500 eV to 8.01x10^-17 via the given conversion. Now I'm stuck.
 
Physics news on Phys.org
  • #2
The deBroglie wavelength is given by the relation

[tex]\lambda = \frac{h}{p}[/tex]

where [itex]h[/itex] is the plank constant and [itex]p[/itex] is momentum. I'm assuming the number in eV you were given is the kinetic energy of the electron--you can look up its mass to get it in terms of momentum through

[tex] T \text{(kinetic energy)}=\frac{p^2}{2m}[/tex]

Can you figure how to get the wavelength of it from this?
 
  • #3


The de Broglie wavelength of an electron can be calculated using the equation: λ = h/mv, where h is the Planck's constant, m is the mass of the electron, and v is the velocity of the electron.

To calculate the velocity of the electron, we can use the kinetic energy equation: KE = 1/2 mv^2. Rearranging for v, we get v = √(2KE/m).

Substituting the given kinetic energy of 500 eV (8.01x10^-17 J) and the mass of an electron (9.11x10^-31 kg), we get v = 1.05x10^7 m/s.

Now, we can plug in the values for h, m, and v in the de Broglie wavelength equation to get:

λ = (6.626x10^-34 J·s) / (9.11x10^-31 kg) * (1.05x10^7 m/s) = 7.28x10^-11 m or 0.0728 nm.

Therefore, the de Broglie wavelength of this electron is 0.0728 nm.
 

1. What is the de Broglie wavelength?

The de Broglie wavelength is the wavelength associated with a moving particle, such as an electron. It is named after French physicist Louis de Broglie who first proposed the concept in his doctoral thesis in 1924.

2. How is the de Broglie wavelength calculated?

The de Broglie wavelength is calculated using the equation λ = h/mv, where λ is the wavelength, h is Planck's constant, m is the mass of the particle, and v is the velocity of the particle. This equation is derived from the de Broglie hypothesis that all matter has wave-like properties.

3. What is the significance of the de Broglie wavelength?

The de Broglie wavelength is significant because it shows that all matter, regardless of size or mass, has both particle and wave-like properties. This helped to unify the theories of quantum mechanics and classical mechanics, leading to a better understanding of the behavior of particles at the subatomic level.

4. How does the de Broglie wavelength relate to the Heisenberg uncertainty principle?

The de Broglie wavelength is related to the Heisenberg uncertainty principle, which states that it is impossible to know both the position and momentum of a particle simultaneously. The de Broglie wavelength is a fundamental property of a particle that contributes to this uncertainty, as the more accurately we know the momentum (through its wavelength), the less accurately we know its position.

5. What factors can affect the de Broglie wavelength of an electron?

The de Broglie wavelength of an electron can be affected by its velocity, mass, and the medium it is traveling through. As the velocity increases, the wavelength decreases, and as the mass increases, the wavelength also increases. Additionally, the wavelength can change if the electron is passing through a medium with varying refractive index, such as in the case of diffraction or interference experiments.

Similar threads

  • Other Physics Topics
Replies
5
Views
854
  • Quantum Interpretations and Foundations
Replies
3
Views
848
  • Quantum Interpretations and Foundations
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
11
Views
2K
  • Introductory Physics Homework Help
Replies
14
Views
2K
  • High Energy, Nuclear, Particle Physics
Replies
6
Views
2K
  • Classical Physics
Replies
5
Views
950
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Biology and Chemistry Homework Help
Replies
1
Views
2K
  • Quantum Interpretations and Foundations
Replies
1
Views
1K
Back
Top