Solving the De Broglie Problem: Calculating Length of a 1-D Box

  • Thread starter Thread starter psingh
  • Start date Start date
  • Tags Tags
    De broglie
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
4 replies · 9K views
psingh
Messages
18
Reaction score
0

Homework Statement


What is the length of a one-dimensional box in which an electron in the n=1 state has the same energy as a photon with a wavelength of 500 nm


Homework Equations




E=h^2/8mL^2 and E=hc/lambda

making it
L=sqrt( (h*lambda)/(8cm) )


The Attempt at a Solution



I plugged in for those numbers and did not come out with the correct number. any suggestions?
 
Physics news on Phys.org
Are you sure that 8 should not be a 4?
 
The eightfold way is right here. 1/8=(L/2)^/2.
I think your first h should be hbar.
 
Hey
The energy if the first state of an indefinite one-dimensional box is:
[tex]E=\frac{\pi^{2}\hbar^{2}}{2mL^{2}}[/tex]
Where m is the mass of the particle and L is the length of the box.
The photon has the energy given by
[tex]E=\hbar\omega=\frac{2\pi\hbar{c}}{\lambda}[/tex]
Where [tex]\lambda[/tex] is the wave length.
And therefore the length L is [tex]L=\sqrt{\frac{\pi\hbar\lambda{c}}{4m}}[/tex]
 
Last edited:
eys_physics said:
And therefore the length L is [tex]L=\sqrt{\frac{\pi\hbar\lambda{c}}{4m}}[/tex]

Which is wrong. It should be [tex]\sqrt{\frac{\pi\hbar\lambda}{4mc}[/tex], which is identical to what the OP psingh had written correctly. Replacing [itex]h[/itex] by [itex]2\pi\hbar[/itex] won't do any good.

Perhaps the OP made some arithmetical mistake...