De Broglie Wavelenght of a 5.5Mev

  • #1
Calculate the de Broglie wavelenght of a 5.5 Mev alpha particle emitted from Am (241) nucleus, could this particle exist with inside the Am (241) nucleus (diameter = 1.6x10^-14m)?


the wavelenght:
Lambda=h/p = (hc) / (sqrt[2mc^2)xeV)
= 1240 / sqrt(2x3727.38x10^6x5.5x20^6)

???
 

Answers and Replies

  • #2
Well...

The alpha particle is nonrelativistic since KE/mc^2 = 10^-3, so you can use the classical formula for momentum p = sqrt(2mKE).

Use this momentum to find the deBroglie wavelength.
 
  • #3
Use the uncertainty principle to find out the spatial resolution of a particle with this momentum.

(hands wave around)

x ~= h/(4*pi*p)

Is this x larger than the diameter of the nucleus?
 
Last edited:
  • #4
I find 6x10^-15 m for lambda, so the diameter of Am is 1.6x10^-14, so the wave length is less than the diameter, is that mean the wavelength can be exist inside the Am nucleus?
 

Related Threads on De Broglie Wavelenght of a 5.5Mev

  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
4
Views
7K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
2
Views
930
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
5K
  • Last Post
Replies
4
Views
736
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
4
Views
2K
Top