How is the Electric Field in Energy Bands Calculated?

jisbon
Messages
475
Reaction score
30
Homework Statement
The energy band in a semiconductor sample is given by
##6.67x^2##
eV, where x is the distance and is valid from 0 to 3m.
(i)Sketch energy band diagram, indicating valence and conduction bands, energy gap and direction of x.
(ii)Determine the electric field in terms of V and m as a function of distance x and indicate the direction of the electric field in the energy band diagram.
(iii)Calculate the electric field at x=3m (in terms of and m)
Relevant Equations
-
This is my attempt at this question, and I'm probably wrong, will need some help/guidance from the experts here :/

i)
1584579037432.png

(ii)

Since energy band given by ##6.67x^2##, can I assume that electric field is simply the energy difference from 0-3m divided by 3m? In this case, would the answer simply be (6.67*3*3)-(6.67*0*0)/3= 20.01V/m?

(iii)At x=3, will the electric field be simply 20.01*3=60.03?

Cheers
 
Physics news on Phys.org
jisbon said:
(iii)At x=3, will the electric field be simply 20.01*3=60.03?

Cheers
No. To get field, you need to take derivative of potential, not simply divide by distance. After taking derivative, the answer is E(x)=2*x*6.67
 
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