Electrons are incident on a crystal

[aidantof]

Q: Electrons are incident on a crystal which can be thought of as a series of reflecting planes with spacing d=0.074nm. What Kinetic Energy should the electrons be given in order that at least four diffracton maxima are observable?

Attempt:
The equations i think i have to use are nλ=2dsinθ, P=h/λ and then K=P^2/2m (m being mass of the electron) and form the question n=4.

i'm supposed to get an answer of 1100.0eV.

my problem is that i can't work out λ as I'm not given the angle θ, is there another set of equations i should be using? how do i get about this?

thanks aidan.

 

[mgb_phys]

The wavelength of the electron ( or anything else ) depends on it's momentum, wavelength = h/momentum.

If you look at wikipedia for electron diffraction the equations are derived for momentum in terms of accelerating voltage.

 

[m2003]

Dear Aidan,

Thank you for reaching out for assistance with your question on Bragg Diffraction. It appears that you are on the right track with the equations you have mentioned. However, in order to determine the kinetic energy of the electrons, we will need to know the wavelength of the electrons, which can be calculated using the equation nλ=2dsinθ. As you have correctly mentioned, n=4 in this case.

In order to find the angle θ, we can use the fact that the diffraction maxima are observable when the incident angle is equal to the Bragg angle, given by the equation nλ=2dsinθ. Therefore, we can rearrange this equation to solve for θ, which will give us the required angle for the incident electrons.

Once we have the value of θ, we can then use the equation K=P^2/2m, where P=h/λ, to calculate the kinetic energy of the electrons. Substituting the values of h, λ, and m (mass of the electron), we should be able to obtain the required kinetic energy of 1100.0eV.

I hope this helps in solving your problem. If you need any further clarification or assistance, please do not hesitate to ask. Good luck with your calculations!

Sincerely,

Scientist