Recent content by Feynman.12

I have attached a worked solution in which I came to the right answer so I believe it must be right. However, I still don't understand intuitively why ##p = h/\lambda## does not imply ##\Delta p = h/\Delta \lambda##

Here is my working

Homework Statement Show that ## \Delta\lambda\Delta\ x>lamdba^2/4*pi## The Attempt at a Solution When I substitute de Broglie's p=h/lambda I get the equation of ##\frac {\Delta\x}{\Delta\lambda} > 1/(4*pi )##

mindblow moment. Thankyou for your help!

I am currently an undergraduate physics and applied mathematics student, and have wanted to go ahead in my course to learn about particle physics and general relativity. However, these topics, along with Quantum field theory which I want to learn about later, are taught in tensor notation. So...

Okay, I understand that! I was able to try and attempt to solve this with the new knowledge, however I got stuck. I derive an answer that is ##\displaystyle \ \frac{\Delta \lambda}{\lambda_0}=\frac{-\Delta v}{v_f}## - If my mathematics is correct, that would mean that ##\displaystyle \...

I can't find where I have done this. How would you do this question?

Sorry, my attempt is as follows. In the book (Eisberg, Resnick - quantum physics of atoms, molecules, solids, nuclei and particles, pg. 82, question 10) it has the answer as that given above, however, my attachment proves that wrong. Is there anywhere I may have made a mistake?

Homework Statement Show that for a nonrelativistic particle, a small change in speed leads to a change in de Broglie wavelength given from The Attempt at a Solution I have tried to expand the left hand side of the equation, but found that it gave the answer of v0/delta v. My definition of...

Thank you for your help!

Sorry ignore the "as v" in my above reply.

Will it be more correct to say that the addition of gamma in the relativistic momentum will cause the momentum to increase as v, and thus the relativistic de Broglie wavelength to decrease?

That is true. And it makes sense that the classical wavelength will be higher than the relativistic wavelength. This would be because the mass of the electron will increase and thus the momentum, and as the wavelength is inversely proportional to the momentum, then the relativistic wavelength...

Homework Statement Determine at what energy, in electron volts, the Nonrelativistic expression for the de Broglie wavelength will be in error of 1% for an electron. 2. The attempt at a solution For the error to be 1%, that means that the classical wavelength/relativistic wavelength will be...