Solve Step Problem: Potential Homework

  • Thread starter Thread starter tarkin
  • Start date Start date
  • Tags Tags
    Potential
AI Thread Summary
The discussion revolves around solving parts (d) and (e) of a homework problem involving the calculation of T using the equation T = 4k1k2/(k1+k2)². The main challenge is that the variable k, which depends on the mass of the particle, is not provided in the problem. Participants suggest that the mass should cancel out when substituting k into the equation for T, leading to a simplified calculation. The user expresses confusion over the cancellation of mass in the equations and is encouraged to review their calculations. The conversation highlights the importance of correctly substituting variables and understanding their relationships in physics problems.
tarkin
Messages
13
Reaction score
0

Homework Statement


Please see attached image. I'm trying to do part (d), then (e) afterwards.

For (d) I can sketch the graph okay, just trying to calculate T now.

Homework Equations


T= 4k1k2/(k1+k2)2

k = ((2m/ћ)2(E-V))½

The Attempt at a Solution



I tried substituting in k into the equation for T. But k depends on the mass of the particle, which isn't given in the question. So what can I do here? I guess one option is just finding the point on the graph, but that's not really a calculation.
 

Attachments

  • potentialstep.jpg
    potentialstep.jpg
    48.6 KB · Views: 458
Physics news on Phys.org
tarkin said:
I tried substituting in k into the equation for T. But k depends on the mass of the particle, which isn't given in the question. So what can I do here?
Look carefully at the equation for T after having substituted the equation for k.
 
DrClaude said:
Look carefully at the equation for T after having substituted the equation for k.

When I subbed in k I hoped the m's would cancel but I couldn't get them to. Is it just some mistake I'm making - should they indeed cancel?

Thanks
 
tarkin said:
When I subbed in k I hoped the m's would cancel but I couldn't get them to. Is it just some mistake I'm making - should they indeed cancel?
It should. You have k's squared both at the numerator and the denominator.
 
DrClaude said:
It should. You have k's squared both at the numerator and the denominator.

Okay, I'll take another look at my working out. Thanks a million!
 
Thread 'Help with Time-Independent Perturbation Theory "Good" States Proof'
(Disclaimer: this is not a HW question. I am self-studying, and this felt like the type of question I've seen in this forum. If there is somewhere better for me to share this doubt, please let me know and I'll transfer it right away.) I am currently reviewing Chapter 7 of Introduction to QM by Griffiths. I have been stuck for an hour or so trying to understand the last paragraph of this proof (pls check the attached file). It claims that we can express Ψ_{γ}(0) as a linear combination of...
Back
Top