/sci/ - Science & Math

>>6434188

Hint: Consider the total energy of the hydrogen atom, kinetic plus potential

>>6434190

Sorry, for the electron in a hydrogen atom

E= (p^2/2m)-(e^2/r)

p = planke/r

so E = (planke^2/2mr^2) - e^2/r

where go from here?

>>6434208

The energy tends to a minimum

>>6434188
This smells like a homework thread

>>6434230

Im not asking you to solve this for me, solve it for yourself

>>6434208
ok so differentiate wrt r and get

-plank^2/mr^3 + e^2/r^2 = 0

r = h^2/me^2
did i do it?

>>6434230

Will also be posting more questions

>>6434234

Find the value, verify yourself

>>6434239
Yeah I got it.

>>6434208
>p = planke/r

How did you get this value for the momentum of the electron?

>>6435442
de Broglie, I guess.

>>6435459
Too many approximations in this derivation for my liking...

>>6434188
>Not specifying the probability cutoff.
>Implying the radius can still be defined
>Uncertainty principle
>Not a useless heuristic

I like homework questions like this. Not just mindless calculation, but forcing you to link up related concepts.

I'm not going to do it for you, though.

>>6435464
>>6435493

>Can't figure out the question
>Samefags because he doesn't understand the concept
>Doesn't understand approximations are 99% of physics

>>6434188

First question was brought to you in part by Richard Feynman, pulled from Volume 3 of his first year lectures. Next question incoming.

Two circular coaxial loops of radius R each with N tight turns are separated by a distance s=R.

Find the magnetic field at a point midway between the loops (point P in the diagram)

Two questions to keep shit interesting

A uniform solid cylinder of radius r rolls inside a cylindrical surface of radius R (r<R). The cylinder is released from rest at an angle \theta to the vertical and rolls without slipping. Determine the angular speed of the cylinder when it reaches the bottom of the cylindrical surface

>>6435585
0 because there is no moving charge specified.

>>6435622
There is a current i in the diagram.

>>6435599
Can you do this with a conservation of energy calculation? Like take the difference in potential energy of the solid cylinder between before it is released and when it is at the bottom, and then find the angular speed that corresponds to that angular kinetic energy for a rotating body with a moment of inertia like a uniform solid cylinder.

Or does the fact that rolling involves a motion of the CoM in addition to motion about the CoM destroy that argument? Is there some way to account for that?

>>6435622
not op but from the diagram there's a current i in each

use the equation relating energy and momentum and then: delta x * delta p > h bar / 2 pi

>>6434188
Find speed using Newton's second, considering electron as a particle.
Find speed as a wave, using de-broglie, noting that an integer number of wavelengths must fit inside the electron's orbit. (hint: may be simpler to actually find v^2)
equate
???
Profit

>>6435631
Me again, a little bit of googling tells me that the rotational motion and the translational motion each contribute perfectly equal parts to the total kinetic energy, such that the original potential energy mgh = (1/2)(I*omega^2) + (1/2)(m*V^2), i.e. the sum of the rotational and translational kinetic energy.

I see no reason to assume that this is true, but I might be able to justify it to myself with a geometric argument that I haven't thought through yet. Something to do with the fact that the cylinder must travel 1 circumference worth of translational distance for every 1 rotation about its center if it is to roll without slipping.

As far as I can tell this problem should be path independent and the fact that the surface on which it is rolling is curved with a radius R shouldn't make any difference other than to define the initial potential energy mgh (since h is not given, only an angle theta).

>>6435646

>h-bar/2pi

lol

>>6435658

You're going in the right direction

Hint: V is related to /omega at the bottom of the motion

>>6435585

Hint: Biot-Sarvart law

>>6435442

From the uncertainty principle

>>6435684
Ah, I see, so I don't really need to prove anything, because I can just write V in terms of omega because after all, it's certainly equal to 2*pi*r*oemga.

And then we just have an expression relating the initial condition with omega, the desired quantity. So it's solved. I don't feel like actually manipulating the expressions but I'm glad I at least remember this much basic mechanics.