- #1
Pengwuino
Gold Member
- 5,124
- 20
The question is:
So I have the equation to find the energy in an infinite well…
[tex] \[
\begin{array}{l}
E_n = n^2 \frac{{\pi ^2 (\hbar c)^2 }}{{2mc^2 L^2 }} \\
L = 2*10^3 nm \\
m = 5.11*10^5 eV \\
\hbar c = 197.33eV*nm \\
\end{array}
\][/tex]
Using this, I find n=1 to equal 9.40*10^-8 eV
The book says I am good there.
The problem is when I have to find the second energy level. I figured you use n=2 which means its just 2^2 or 4 times the energy but the book says its just 2 times the energy of the first energy level. What am I not understanding here?
We can approximate an electron moving in a nanowire as a one-dimensional infinite square-well potential. Let the wire be 2 um long. The nanowire is cooled to a temperature of 13K, and we assume the electron’s average kinetic energy is that of a gas molecule at this temperature ( = 3/2 kT). (a) What are the three lowest possible energy levels of the electrons? (b) What is the approximate quantum number of the electrons moving in the wire?
So I have the equation to find the energy in an infinite well…
[tex] \[
\begin{array}{l}
E_n = n^2 \frac{{\pi ^2 (\hbar c)^2 }}{{2mc^2 L^2 }} \\
L = 2*10^3 nm \\
m = 5.11*10^5 eV \\
\hbar c = 197.33eV*nm \\
\end{array}
\][/tex]
Using this, I find n=1 to equal 9.40*10^-8 eV
The book says I am good there.
The problem is when I have to find the second energy level. I figured you use n=2 which means its just 2^2 or 4 times the energy but the book says its just 2 times the energy of the first energy level. What am I not understanding here?