Calculating Theta for Lowest Wavelength Electron Beam

[atomqwerty]

Homework Statement

An electron beam (energy 119,9 eV) contacts perpendicularly on a Ni plate, with interatomic spacing d= 0,215 nm. At which angle Theta will be obtained an electron beam with the lower wavelenght possible?

Homework Equations

$$n\lambda=d sin(\theta)$$ (Bragg's for perpendicular incident beam)

The Attempt at a Solution

I've tried to calculate the wavelenght of the electron beam, by using the fact that E= hf = h c/lamba. Then, with the wavelenght, the Bragg law give us the angle Theta (with n=1), that will be refracted the electron beam, buy Bragg's Law says that the lower wavelenght implies the highest order n, that implies in itself, the major abgle, Theta = 90º.

I don't get it...

thanks

 

[mark726]

for your help

Thank you for your question. It is an interesting problem and I would be happy to provide some guidance.

Firstly, you are on the right track by using the Bragg's law for perpendicular incident beams. However, I would like to point out that the equation you have used, n\lambda=d sin(\theta), is for x-ray diffraction, and may not be applicable to electrons. Instead, you can use the de Broglie wavelength equation, \lambda=h/p, where h is Planck's constant and p is the momentum of the electron.

Next, it is important to note that the angle theta in the Bragg's law is the angle of incidence, not the angle of refraction. This means that the electron beam will be refracted at an angle of 90-theta, as shown in the diagram below:

[insert diagram of Bragg's law for electrons]

So, to find the angle theta that will give the lowest possible wavelength, we need to maximize the value of sin(theta). As you correctly pointed out, this will happen at the highest order n, which corresponds to the highest possible value of sin(theta). Therefore, the angle theta that will give the lowest possible wavelength is when sin(theta) = 1, which occurs at theta = 90 degrees.

I hope this helps clarify any confusion you had. Please let me know if you have any further questions. Good luck with your research!
Scientist