Cut-off energy of Be having HCP structure

  • Thread starter Thread starter Armani
  • Start date Start date
  • Tags Tags
    Energy Structure
Armani
Messages
8
Reaction score
0
Hello,

I have to calculate the cut-off energy of Beryllium that have hcp structure with the paramters: a=2.29Å and c=3.58Å.

WHAT I HAVE DONE SO FAR:
Using plane (100) or (010)

Since the formula is: $$E=\frac{1}{2} \times G^2$$

$$G=\frac{2 \Pi}{2d} $$

$$\frac{1}{d^2} = \frac{4}{3} \times \frac{h^2+h \times k+k^2}{a^2}+\frac{l^2}{c^2}$$ ---> $$d=1.98 \times 10^{-10}$$
So
$$G= 3.168208497 \times 10^{10}$$
and finally
$$E=5.018772540 \times 10^{20}$$

BUT the SOLUTION says that E=5.22 meV. Can someone correct my calculation?
 
Physics news on Phys.org
I moved your thread to our homework section.

Your calculation is missing units. Add them and you'll find your mistake. There is no reasonable energy unit where 1020 would not be way too large.
 
It is not working..maybe I am to confused.
 
What is not working? Did you add the units as suggested?
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top