Physics

PostPhysics by **Memento Mori** » Sun Oct 19, 2008 9:55 pm

Any idea how to do this? I missed 2 lectures last week.

Calculate the de Broglie wavelength of a beam of electrons, which is accelerated by a potential difference of 40 kV. This beam then passes through a thin polycrystalline metallic foil and a diffraction pattern is observed on a screen 300 mm behind the foil. If the innermost diffraction ring is of radius 8 mm, use Bragg’s law (2dsinθ = nλ, n = 1) to estimate the inter-planar spacing d of the atoms contributing to this ring. [Note that the scattering angle is 2θ]

PostRe: Physics by **Weapon of Choice** » Sun Oct 19, 2008 10:00 pm

A) 3 Apples. We got the answer at the end of the lesson, you skiiving shite!

PostRe: Physics by **Memento Mori** » Sun Oct 19, 2008 10:06 pm

My train was delayed.

PostRe: Physics by **That** » Sun Oct 19, 2008 10:13 pm

Memento Mori wrote:Any idea how to do this? I missed 2 lectures last week.

Calculate the de Broglie wavelength of a beam of electrons, which is accelerated by a potential difference of 40 kV. This beam then passes through a thin polycrystalline metallic foil and a diffraction pattern is observed on a screen 300 mm behind the foil. If the innermost diffraction ring is of radius 8 mm, use Bragg’s law (2dsinθ = nλ, n = 1) to estimate the inter-planar spacing d of the atoms contributing to this ring. [Note that the scattering angle is 2θ]

Go and ask your lecturer, or a friend on the same course, for help.

PostRe: Physics by **Memento Mori** » Sun Oct 19, 2008 10:15 pm

Karlprof wrote:
Memento Mori wrote:Any idea how to do this? I missed 2 lectures last week.

Calculate the de Broglie wavelength of a beam of electrons, which is accelerated by a potential difference of 40 kV. This beam then passes through a thin polycrystalline metallic foil and a diffraction pattern is observed on a screen 300 mm behind the foil. If the innermost diffraction ring is of radius 8 mm, use Bragg’s law (2dsinθ = nλ, n = 1) to estimate the inter-planar spacing d of the atoms contributing to this ring. [Note that the scattering angle is 2θ]

Go and ask your lecturer, or a friend on the same course, for help.

Well it's in tomorrow and it's my first lesson at 9. I was planning on replicating someone else's if I can't make any progress.

PostRe: Physics by **Poncho** » Sun Oct 19, 2008 10:15 pm

Sorry, I don't speak Korean.

PostRe: Physics by **Memento Mori** » Sun Oct 19, 2008 10:16 pm

That was just question 1. The others aren't any easier.

PostRe: Physics by **Pilch** » Sun Oct 19, 2008 10:47 pm

Memento Mori wrote:Any idea how to do this? I missed 2 lectures last week.

Calculate the de Broglie wavelength of a beam of electrons, which is accelerated by a potential difference of 40 kV. This beam then passes through a thin polycrystalline metallic foil and a diffraction pattern is observed on a screen 300 mm behind the foil. If the innermost diffraction ring is of radius 8 mm, use Bragg’s law (2dsinθ = nλ, n = 1) to estimate the inter-planar spacing d of the atoms contributing to this ring. [Note that the scattering angle is 2θ]

(Edited to remove my embarrassing haste-related errors)

The wavelength is calculated as:

wavelength = Planck's constant / momentum

First thing to do is to calculate the energy of the accelerated electrons. This is:

charge * potential difference = 6.408*10^-15 J

energy = ((momentum)^2) / 2*mass, so, momentum = sqrt(2*mass*energy) = 1.080...*10^-22kgm/s

So, the de Broglie wavelength is 6.13*10^-12m.

For the second part, you first have to find the diffraction angle, which is achieved using:

2*theta = arctan(8/300), so theta = 0.0133

Then all you have to do is rearrange Bragg's law to find d using the numbers you already have. This yields an inter-planar spacing of d = 0.23nm. This sounds about right to me, but I've done this very quickly and haven't checked my working at all, so you might want to go through it yourself and make sure that you agree with me.

PostRe: Physics by **Memento Mori** » Sun Oct 19, 2008 10:51 pm

charge * potential difference = 6.408*10^-21 J

How'd you get that line? If electron charge is 1.6 x 10^-19 and pd is 40 x 1000, I get energy as 6.4 x 10^-15.

PostRe: Physics by **Pilch** » Sun Oct 19, 2008 10:54 pm

Yep, my bad. I did 40mV rather than 40kV. Told you I might have made mistakes! You'll have to work that through then, I'm afraid.

PostRe: Physics by **Pilch** » Sun Oct 19, 2008 10:56 pm

So now I get a wavelength of 6.14*10-12m. Do you get the same? And an inter-planar spacing of 0.23nm.

PostRe: Physics by **Memento Mori** » Sun Oct 19, 2008 10:59 pm

Pilch wrote:So now I get a wavelength of 6.14*10-12m. Do you get the same? And an inter-planar spacing of 0.460nm.

Yep. Thanks man.
Different interplanar spacing though.

PostRe: Physics by **Pilch** » Sun Oct 19, 2008 11:00 pm

I get 0.23nm now. How about you?

Whoops! You managed to quote my erroneous post, which I very hastily edited.

PostRe: Physics by **Memento Mori** » Sun Oct 19, 2008 11:02 pm

Nope. I get a different diffraction angle to you.

PostRe: Physics by **Memento Mori** » Sun Oct 19, 2008 11:03 pm

I get 2.33 x10-4 as the angle.

PostRe: Physics by **Pilch** » Sun Oct 19, 2008 11:05 pm

Is your calculator set to radians?

PostRe: Physics by **Memento Mori** » Sun Oct 19, 2008 11:06 pm

Pilch wrote:Is your calculator set to radians?

Nope. If arctan 2 theta = 8/300
then 2 theta= tan 8/300 yeah?

PostRe: Physics by **Pilch** » Sun Oct 19, 2008 11:08 pm

Oh, sorry. I wrote that wrong as well. I meant arctan(8/300) = 2*theta. Such was my haste to get to the answer! Does that make it work now?

PostRe: Physics by **Memento Mori** » Sun Oct 19, 2008 11:09 pm

Pilch wrote:Oh, sorry. I wrote that wrong as well. I meant arctan(8/300) = 2*theta. Such was my haste to get to the answer! Does that make it work now?

No worries. I appreciate this man.

PostRe: Physics by **Memento Mori** » Sun Oct 19, 2008 11:10 pm

I get d= 2.3 x 10^-10. Thanks man.

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