# Quantum physics - the wave properties of particles - HELP !

1. Mar 17, 2006

### Yael

hey everyone,
i have a question i'm trying to solve here for class
any help would be SO apreciated !

"In the Davisson-Germer experiment, 54.0-eV electrons were diffracted from a nickel lattice. if the first maximum in the diffraction pattern was observed at 50 degrees (as in the figure).
what was the lattice spacing a between the vertical rows of atoms in the figure? (it is not the same as the spacing between the horizontal rows of atoms)"

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2. Mar 17, 2006

### Chi Meson

Use the given kinetic energy of the electrons, along with the mass of electrons, to find the momentum. Then ask deBroglie about the wavelength.

3. Mar 17, 2006

### Yael

hmmm Can you just explain how do i use the kinetic Energy (54.0-ev) to find the momentum ?

and once i do find the momentum - De Broglie is for the wavelength. the question is for finding the "a" spacing in the figure...

Thanks !

4. Mar 17, 2006

### Hootenanny

Staff Emeritus
Use kinetic energy to find the velocity of the electrons. Then use this velocity to find the momentum. You can then find the DeBroglie wavelength. You need the wavelegth to calculate the spacings. I think you can use the formula for young's double slit.

5. Mar 17, 2006

### Yael

when using the kinetic energy to find V... do i use the expression for relativistic kinetic energy where v/c approaches 1? or do i simply use the classical expression of 1/2 mv squared?
i know this is basic but... :-S

thanks !

6. Mar 17, 2006

### Yael

on the same subject - the expression for diffraction is :
2dsinTheta = m times lamda.
where d is the spacing between the horizontal lines and not the vertical ones :-/
how do i get to "a" ?

7. Mar 18, 2006

### Hootenanny

Staff Emeritus
I've just had a quick glance through any examples of electron scattering I have and none of them take into account relativistic effects. As for calculating 'a', I don't know how you could do it, you will probably have to calculate d then work it out from there. I'll ask the other homework helpers to have a look at it.

8. Mar 18, 2006

### gulsen

What I see from the picture is $$a \cos(\theta) = d$$.
As for wavelength, relativistically:
$$\lambda = \frac{h}{p} = \frac{h}{\sqrt{E^2- (mc^2)^2}/c}$$
where $$E = K + mc^2$$

9. Mar 18, 2006

### Staff: Mentor

The first thing to do is understand Bragg's law, which is what that equation describes. That will allow you to figure out the separation (d) between the lattice planes and how that relates to the labeled distance (a).

Look here: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/bragg.html

The Davison-Germer experiment is also discussed here: http://hyperphysics.phy-astr.gsu.edu/hbase/davger.html#c1

10. Mar 18, 2006

### Chi Meson

Please correct me if I am wrong, but is a 54 eV electron going fast enough for relativistic analysis?

11. Mar 18, 2006

### Hootenanny

Staff Emeritus
I wouldn't imagine that 54 eV is enough to take into account relativistic effects either. By my reconing the electron would be travelling at $4.37\times 10^6 m\cdot s^{-1}$, which is only about 1.46% of the speed of light...

12. Mar 18, 2006

### Chi Meson

That's a gamma of 1.0001, so you can ignore reletivistic effects for this one. So momentum = SQRT(2Km)

13. Mar 18, 2006

### Yael

From what i've understood so far :
a = d sin θ (???)

Thanks for all the help so far everyone.

14. Mar 18, 2006

### Hootenanny

Staff Emeritus
Where did you get that from?

15. Mar 18, 2006

### Yael

oh i was looking at the link that Doc Al posted : http://hyperphysics.phy-astr.gsu.edu...davger.html#c1 [Broken]

to be honest i'm all mixed up with how to get to the "a" :-S
i can't seem to get how gulson got: a cos θ = d

Last edited by a moderator: May 2, 2017
16. Mar 18, 2006

### Hootenanny

Staff Emeritus
I can see how he got $a\cos\theta = d$, but I'm not sure it's right. You can make a right angled triangle with a being the base and d being the hyp. I'm not sure if this is the correct method. It's probably best to wait until someone with more knowledge comes online.

17. Mar 18, 2006

### Staff: Mentor

The angle between the normal to the scattering plane (along which "d" is measured) and the horizontal plane (along which "a" is measured) is given by $\theta$. "a" is the hypotenuse of a right triangle, with "d" as the side adjacent to that angle. Thus $d = a \cos\theta$.

18. Mar 18, 2006

### Yael

so...
1. at first stage i use the classic expression of kinetic energy (½ mv²) with the 54 ev electron to find the velocity
2. after i obtain V i turn to de broglie to find the wavelength.
3. with the wavelength - i use bragg's law (2dsinθ = mλ) to get to 'd'.
4. use d to get 'a' through: d = a cosθ

hmmm sounds good?

19. Mar 18, 2006

### Yael

getting into atomic physics i'm trying to answer this question but can't seem to understand exactly what they mean by it :
"Does the light emitted by a neon sign constitute a continuous spectrum or only a few colors? Defend your answer"

20. Mar 18, 2006

### Hootenanny

Staff Emeritus
You've got it

I think another way of looking at it is, "Does a neon sign emitt all wavelengths of visibil light, or does it only emitt a few different wavelengths?" That's my take on it anyway.