View Full Version : de Broglie Wavelenght of a 5.5Mev
diegoarmando
Aug24-07, 11:09 AM
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)
???
ChaoticOrder
Aug24-07, 11:39 AM
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.
ChaoticOrder
Aug24-07, 11:44 AM
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?
diegoarmando
Aug24-07, 04:37 PM
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?
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