# Minimal Kinetic energy for particle in a box

This is driving me crazy! The question goes as follows:

A proton is enclosed in a zone of length 2pm along the x-axis. The minimal kinetic energy of the proton lies closest to:

• 5000eV
• 0.5eV
• 50eV
• 500eV
• 5eV (this is supposed to be the correct answer)

So first I calculate the energy level for a particle in a box which is given by the equation $E_n = p_n^2/(2 m) = (n^2 h^2)/(8 m L^2)$

Here $L$ is the length of the zone. So I find $E_1=94007.53 eV$..

So.. How do I relate this to the minimal kinetic energy?

-- EDIT 1 --

After the first hint from someone, I changed my calculations (I used the mass of an electron instead of a proton..) NOW I get $E_1 = 51eV$, which is still wrong..

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It looks like you used the electron mass rather than the proton mass... – Michael Brown Feb 6 '13 at 16:11
Owh god.. You're right -.- I've made like 30+ exercises on different subjects.. All of them, by coincidence, were related to electrons, so I'm so used to using that value as the mass of a sub-atomic particle that I didn't even think twice when using it here.. THANK YOU! – Spyral Feb 6 '13 at 16:15
It's still not right though.. The new answer I become is $E_1 = h^2/(8*1.6726*10^-27*(2*10^(-12))^2)=51 eV$ – Spyral Feb 6 '13 at 16:20
Yes, I get the same. I'm thinking the suggested answer is wrong - you're doing the right thing. – Michael Brown Feb 6 '13 at 16:36
Also, when you do powers in LaTeX you need to do x^{yz}: $x^{yz}$. – Michael Brown Feb 6 '13 at 16:39

@Spyral The error in your calculation is that you have missed out the division by $\pi^2$ ($\pi$ = 3.14...). The equation for the energy in the ground state (n = 1) is given by the equation
$E = h^2/(8\pi^2 m L^2)$
$h$ = Planck's constant; $m$ = mass of the proton in your case, $L$ = the length of the size of the box. If you use this equation (derived from Schrodinger equation) you will find the right answer, ~ 5eV. I hope this helps.
I get that the $\pi^2$ cancels out: $\frac{p^{2}}{2m}=\frac{\hbar^{2}}{2m}\left(\frac{2\pi}{2L}\right)^{2}=\frac{h^2‌​}{2m\left(2\pi\right)^{2}}\frac{\pi^{2}}{L^{2}}=\frac{h^2}{2m\left(2\right)^{2}} \frac{1}{L^{2}}$ Where is my mistake? – Michael Brown Feb 9 '13 at 1:28
The wavefunction must have a node at $x=0$ and $x=L$, so $k L=\pi n$ where $n$ is an integer. So I get $p = \hbar k = \hbar \pi / L = h/2L$. Thus $p^2 / 2m = \frac{h^2}{8 m L^2}$. Your wavefunction is incorrect since it has $k L = 1$, which does not give a node at $x=L$: $\sin( k x ) = \sin(1)\neq 0$. It's not just wiki: hyperphysics also disagrees with you: hyperphysics.phy-astr.gsu.edu/hbase/quantum/schr.html#c3, as does my calculation which uses the usual definition of the momentum and the requirement that the wavefunction is zero at $x=L$. – Michael Brown Feb 11 '13 at 3:47