De Broglie Wavelenght of a 5.5Mev

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The de Broglie wavelength of a 5.5 MeV alpha particle is calculated using the formula λ = h/p, with p derived from the classical momentum equation. The calculation shows that the wavelength is approximately 6 x 10^-15 m. This value is smaller than the diameter of the Americium-241 nucleus, which is 1.6 x 10^-14 m. The findings suggest that the alpha particle's wavelength could exist within the nucleus. Therefore, it is feasible for the alpha particle to be contained inside the Am-241 nucleus.
diegoarmando
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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)

?
 
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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.
 
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:
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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