- #1

- 16

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the wavelenght:

Lambda=h/p = (hc) / (sqrt[2mc^2)xeV)

= 1240 / sqrt(2x3727.38x10^6x5.5x20^6)

???

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- Thread starter diegoarmando
- Start date

- #1

- 16

- 0

the wavelenght:

Lambda=h/p = (hc) / (sqrt[2mc^2)xeV)

= 1240 / sqrt(2x3727.38x10^6x5.5x20^6)

???

- #2

- 11

- 0

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

- 11

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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?

(hands wave around)

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

Is this x larger than the diameter of the nucleus?

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- #4

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