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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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1  
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
1  
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
1  
Also, when you do powers in LaTeX you need to do x^{yz}: $x^{yz}$. –  Michael Brown Feb 6 '13 at 16:39
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1 Answer

@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.

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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
    
Also wikipedia disagrees with you: en.wikipedia.org/wiki/Particle_in_a_box –  Michael Brown Feb 9 '13 at 1:33
    
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
    
In fact, if I put the OP's numbers into the Hyperphysics calculator it gives the same answer 51.198... eV. –  Michael Brown Feb 11 '13 at 3:52
    
Quickly googling finds many references that disagree with you. This is such a basic problem that it is unlikely to find so many sources independently mistaken, but you don't have to take my word for it. Solve the Schrodinger equation yourself and apply the boundary conditions. –  Michael Brown Feb 11 '13 at 4:15
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