MA103
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Q: Electron diffraction can be used to investigate both the arrangement of atoms and the dimensions of nuclei. Explain how changing the energy of the electrons can be used to achieve this.
I need help again, but this time, not alpha-scattering experiment but electron diffraction.
What I think is that changing the energy of the electrons is the same as changing the velocity of the electrons. If you know the velocity then you know the wavelength of the electrons, due to de Broglie's equation (lambda=h/mv). And if you know the velocity and the angle of diffraction, you can work out the distance between the atoms by the equation m*lambda=2*d*sin pheta. For the diameter of the nuclei it's almost similar except the end bit: you know the velocity of electrons, so you know the wavelength by de Broglie's equation. From the equation d*sin pheta min = 1.22*lambda you can work out the d, the diameter of the nucleus provided you got the wavelength (which has been calculated) and pheta min, the angle of the first minimum from the centre. I'm not sure about this. Also, am I answering the question?
I need help again, but this time, not alpha-scattering experiment but electron diffraction.
What I think is that changing the energy of the electrons is the same as changing the velocity of the electrons. If you know the velocity then you know the wavelength of the electrons, due to de Broglie's equation (lambda=h/mv). And if you know the velocity and the angle of diffraction, you can work out the distance between the atoms by the equation m*lambda=2*d*sin pheta. For the diameter of the nuclei it's almost similar except the end bit: you know the velocity of electrons, so you know the wavelength by de Broglie's equation. From the equation d*sin pheta min = 1.22*lambda you can work out the d, the diameter of the nucleus provided you got the wavelength (which has been calculated) and pheta min, the angle of the first minimum from the centre. I'm not sure about this. Also, am I answering the question?