What is the approximate quantum number of the electron?

[jk4]

[SOLVED] particle in a box

Homework Statement

An electron moves with a speed of v = $$10^{-4}$$c inside a one-dimensional box of length 48.5nm. The potential is infinite elsewhere. What is the approximate quantum number of the electron?

Homework Equations

$$E_{n} = \frac{n^{2}\pi^{2}\hbar^{2}}{2mL^{2}}$$
n = 1, 2, 3, . . .

$$E = \gamma mc^{2}$$

The Attempt at a Solution

I was trying to solve it by finding the total energy of the electron, then using the first equation I stated using the total energy as "En". Then I would try and solve for n but I get a very different number. The answer is supposed to be 4.

 

[Dick]

n=4 is right. What did you get for the E corresponding v=10^(-4)c? What do you get if you put n=4 into En? BTW v<<c. You could just use (1/2)mv^2 for the energy.

 

[tnho]

n=4 is right. What did you get for the E corresponding v=10^(-4)c? What do you get if you put n=4 into En? BTW v<<c. You could just use (1/2)mv^2 for the energy.

yes. I got 3.969 for using $$E_n=\frac{1}{2}mv^2$$.
Besides, $$E_{n} = \frac{n^{2}\pi^{2}\hbar^{2}}{2mL^{2}}$$ is a result from Schrodinger Equation which is non-relativistic. I am not sure if this form preserve in Dirac Equation or not.

 

[jk4]

EDIT:
ok I got it, thanks for the help.

 

[Dick]

Put units on dimensionful numbers ok. I calculate E = (0.5)(9.1095 E -31*kg)(10E-4 x c)^2 and I get 4.09E-22J. We are using the same numbers, unless you are using a different value of c??

 

[tnho]

EDIT:
ok I got it, thanks for the help.

$$\frac{1}{2}mv^2 = \frac{n^{2}\pi^{2}\hbar^{2}}{2mL^{2}}$$

then

$$n=\frac{mvL}{\pi \hbar}$$

put $$m=9\times 10^{-31}, v=3\times 10^{4}, L=48.5\times 10^{-9} \mbox{ and } \hbar = 1.055\times 10^{-34}$$

then you can get $$n\approx 4$$

get it?

 

[wombat4000]

wow - this really helped - thanks!