- #1
mmh37
- 59
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This is supposed to be an easy question, but I appear to be slightly lost. Can anyone give me a hint on what to do here?
when waves of wavelength lamda are diffracted by a circular disc of diameter D the first minimum in the intensity of the scattered waves occurs at a scattering angle z given by
[tex]sin(z) = 1.22 * lamda / D [/tex]
First Minima occur (when scattered from Carbon and Oxygen nuclei)...
for Oxygen (16 O) with E = 420 MeV : z= 45°
for Oxygen (16 O) with E = 360 MeV: z= 53 °
for Carbon (12 C) with E= 420 MeV: z = 50.5°
USE THE ABOVE DATA TO ESTIMATE THE RADII OF THE CARBON AND OXYGEN NUCLEI!
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NOTE: ... that before I had do derive an expression for the momentum of the particle when it's kinetic energy is very much greater than it's rest mass energy mc^2
using the energy momentum invariant and neglecting the m^2c^4 term I said that:
[tex] E^2 = p^2*c^2 + m^2*c^4 [/tex]
leads to
[tex] p = E/c [/tex]
when waves of wavelength lamda are diffracted by a circular disc of diameter D the first minimum in the intensity of the scattered waves occurs at a scattering angle z given by
[tex]sin(z) = 1.22 * lamda / D [/tex]
First Minima occur (when scattered from Carbon and Oxygen nuclei)...
for Oxygen (16 O) with E = 420 MeV : z= 45°
for Oxygen (16 O) with E = 360 MeV: z= 53 °
for Carbon (12 C) with E= 420 MeV: z = 50.5°
USE THE ABOVE DATA TO ESTIMATE THE RADII OF THE CARBON AND OXYGEN NUCLEI!
______________________________________________________________________________
NOTE: ... that before I had do derive an expression for the momentum of the particle when it's kinetic energy is very much greater than it's rest mass energy mc^2
using the energy momentum invariant and neglecting the m^2c^4 term I said that:
[tex] E^2 = p^2*c^2 + m^2*c^4 [/tex]
leads to
[tex] p = E/c [/tex]
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