# Calculate the quantum number of an electron given its velocity

1. Nov 26, 2012

### MatrixMan87

1. The problem statement, all variables and given/known data
An electron moves with speed v=10-4c inside a one dimensional box (V=0) of length 48.5 nm. The potential is infinite elsewhere. The particle may not escape the box. What approximate quantum number does the electron have?

2. Relevant equations
En = n2$\frac{\pi^{2}\hbar^{2}}{2ml^{2}}$

KE = $\frac{mv^{2}}{2}$

9.1e-31 kg / electron
6.2e18 eV / J

3. The attempt at a solution

I know that I need to find the kinetic energy of the electron and compare it with the various energies given by the first equation above in order to determine the value of n. However, I keep getting values for E that are far too small:

KE = $\frac{mv^{2}}{2}$ = $\frac{1.9\times10^{-31} kg \times (3\times10^{4} m/s)^{2}}{2} \times 6.2\times10^{18} eV/J$ = $5.301\times10^{-4} eV$

This is far lower than the energies I get out of the first equation above, so I think that I'm missing something up to this point.

Last edited: Nov 26, 2012
2. Nov 27, 2012

### Staff: Mentor

I get a different value for the kinetic energy (~2.5 meV).
What did you get as ground state energy?

3. Nov 27, 2012

### vela

Staff Emeritus
You can make your life quite a bit easier if you learn to throw in convenient factors of c and get used to working with eV instead of joules. For example,
$$K = \frac{1}{2}mv^2 = \frac{1}{2}(mc^2)\left(\frac{v}{c}\right)^2,$$ where the rest energy of the electron in eV is $mc^2=511000\text{ eV}$. You can now easily see what the kinetic energy should be without a calculator.

The combination $\hbar c = 197\text{ ev nm}$ is a useful one to memorize as well. Again throwing in a few factors of c, we find that the n-th energy level is given by
$$E_n = \frac{n^2\pi^2\hbar^2}{2ml^2} = \frac{n^2\pi^2(\hbar c)^2}{2(mc^2)l^2}.$$ You should be able to see the units work out quite easily if your express $l$ in nm, yielding an answer in eV.

4. Dec 9, 2012

### MatrixMan87

Thanks Vela, that's exactly what I needed.

I was able to equate the kinetic energy equation with the orbital energy level equation and solve for n, like so:

http://imgur.com/7a7uu