Oxbridge Physics Prelims Revision

voila!

Ok guys v.quick question (this is Chloé btw on Simon's account!!!):

In the "penguins" notes, he said:

p = hbar/a

thus min KE = p²÷2m
therefore: KE = h² ÷ 2ma²

surely this is a mistake? shouldn't it be hbar in the second equation (or is it the first one that's wrong, ie. p=h/a)? Thanks :smile:

Reply 983

oxymoron

voila!

Ok guys v.quick question (this is Chloé btw on Simon's account!!!):

In the "penguins" notes, he said:

p = hbar/a

thus min KE = p²÷2m
therefore: KE = h² ÷ 2ma²

surely this is a mistake? shouldn't it be hbar in the second equation (or is it the first one that's wrong, ie. p=h/a)? Thanks :smile:

I think its one of these fudge things. In that h and hbar are at the moment pretty interchangeable. The real value is I think hbar/2a or similar but its all orders of magnitude and so on so they tend to get exchanged. obviously h > hbar and so it definitely meets the uncertainty prinicple. So yes it is correct - just fudged. Just learn it as it is I say.

Edit: remmeber he calculated the value in the optical microscope as h/a not hbar/a. thats what hes used -as I said all very fudged. mmm tasty :rolleyes:

Reply 984

LennonMcCartney

\hat{p}=-i\hbar\nabla

, so that the commutation relation rule is satisfied:

Unparseable latex formula:

[\hat{p},{\hat{x}]=i\hbar

. And since KE:

T=p^2/2m

from classical mechanics, we say the operator for kinetic energy in quantum mechanics is:

\hat{T}=\hat{p}^2/2m=\frac{-\hbar^2}{2m}\nabla^{2}

, which agree with what we've been told in one dimension.

Reply 985

LennonMcCartney

OK, what is a?

EDIT: We're clearly talking about fairly different things (although not *really* different) but I just got up, so don't blame me.

Reply 986

you're confusing me (still Chloé on simon's account)

Right, could someone help me with all these questions? thanks:

====
Q1. June 2003
To calculate the maximum energy of the ejected electrons don we actually need to work out the intensity of the light source or do w ejust work out the energy from the wavelength and the minimum energy required?

How do I do the next bit?
====

====
Q4. June 2003
I'm stuck on what to do for the bit where we have to show that the results for the infinite potential well are consistent with the uncertainty principle. I know it can't have zero kinetic energy but how am I supposed to show it and how do I use the other results for the wave function?
====

Thanks guys, i'd better go now as simon's moanin at me for taking so long to type on his silly laptop! Sorry for all the questions - i just don't get this topic!

Reply 987

oxymoron

you're confusing me (still Chloé on simon's account)

Right, could someone help me with all these questions? thanks:

====
Q1. June 2003
To calculate the maximum energy of the ejected electrons don we actually need to work out the intensity of the light source or do w ejust work out the energy from the wavelength and the minimum energy required?

How do I do the next bit?
====

====
Q4. June 2003
I'm stuck on what to do for the bit where we have to show that the results for the infinite potential well are consistent with the uncertainty principle. I know it can't have zero kinetic energy but how am I supposed to show it and how do I use the other results for the wave function?
====

Thanks guys, i'd better go now as simon's moanin at me for taking so long to type on his silly laptop! Sorry for all the questions - i just don't get this topic!

Q4 first -

for a start we know delta x = 2a
delta p = hbar k = hbar n pi/ (2a) from the schrodinger equation. Multiply the 2 together and you get h/2

hence meets uncertainty principle.

Also KE = hbar^2 pi^2 /( 8ma^2) for ground state.

ie n must be an integer >= to 1.
hence there is a zero point energy - a minimum energy of the system

For Q1 you do 2 calculations - 1 quantum 1 classical

for quantum use E=hf and take difference.ie energy of photon is 2.81 eV - 2.3 eV of energy level difference = 0.51 eV of KE.

for classical - light is emitted spherically.
Intensity is power/ Area for the light emitted at the source

also power = W/t

so I = W/ta4pi where a is radius of the atom and w is the energy needed to eject it

then P/4pir^2 = W/ta4pi equating intensity a distance r away with intensity needed to eject electron.

rearrange for t bang it in and it should be around 10 hours.

Reply 988

F1 fanatic

Q4 first -

for a start we know delta x = 2a
delta p = hbar k = hbar n pi/ (2a) from the schrodinger equation. Multiply the 2 together and you get h/2

Woah! Surely if u multiply

\Delta x

and

\Delta p

you'll get hbar*n*

\pi

?

hence meets uncertainty principle.

Also KE = hbar^2 pi^2 /( 8ma^2) for ground state.

How do we know there's an 8 on the bottom? usually we have a 2...sorry I know I'm being very dense!!!

Thanks for all your help Stu :biggrin:

Reply 989

oxymoron

Woah! Surely if u multiply

\Delta x

and

\Delta p

you'll get hbar*n*

\pi

?

hence meets uncertainty principle.

How do we know there's an 8 on the bottom? usually we have a 2...sorry I know I'm being very dense!!!

Thanks for all your help Stu :biggrin:

well yes it is but the minimum is when n=1 so if it applies then it applies at all other. and hbar = h/2pi ==> hbar n pi = h/2 which may or may not have been what I said before.

8 at bottom - *cos its in his notes* no erm... cos its (2a)^2 which is 4 *2 that is usually there. hence 8.

Right - Im now going to try the unheard of trick of cramming quantum mechanics in 6 hours having done practically none up to this point. Thankfully its my speciality and so I already kind of get it. Wish me luck :rolleyes:

Reply 990

F1 fanatic

Right - Im now going to try the unheard of trick of cramming quantum mechanics in 6 hours having done practically none up to this point. Thankfully its my speciality and so I already kind of get it. Wish me luck :rolleyes:

Thanks ever so much Stu...I don't know why you're worried about the exam...you've been helping me!!! Thanks ever so much.

Reply 991

Why is Schrodinger equation compatible with the experimental results for matter waves?

Reply 992

oxymoron

Why is Schrodinger equation compatible with the experimental results for matter waves?

Well it has quantities that are both matter like and wave like. ie mass and wavelength. Also it is a wave equation for particles. Thats what it is. It works on probability waves. It has been tested by experiment and matches pretty much perfectly. Hence its compatible.

The probability function gives the probability wave for a particle - ie the chance of finding it in a particular place. the particle is defined by a wave function. :smile:

Hence all compatible.

Im gonna keep off here for the rest of the afternoon so if you have any more questions that no one else will answer MSN erise me. :smile:

Reply 993

Why should anular momentum be an integral number of h bar ???

Randomly put it in why don't you!!

Reply 994

Bezza

2

Nikking

Why should anular momentum be an integral number of h bar ???

Randomly put it in why don't you!!

otherwise it doesn't make sense i suppose

Reply 995

LennonMcCartney

Cos of units and a good guess.

It's like asking, "Why should F be proportional to d(mv)/dt? Randomly put it in, why don't you!" That was a guess, and it worked. So did L=n\hbar. As did the Schrodinger equation, choices for operators etc

Reply 996

Bezza

otherwise it doesn't make sense i suppose

lol. why wouldn't it? and why should an electron emit radiation when being accelerated? It's all good telling us it does, but WHY??

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Reply 997

Bezza

2

Nikking

lol. why wouldn't it?

because then you wouldn't get the discrete energy levels that you need for emission spectra

and why should an electron emit radiation when being accelerated? It's all good telling us it does, but WHY??

that's just how you make an em wave - don't know a better explanation i'm afraid. in a transmitter you just have an ac current and the accn of the electrons leads to em waves.

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I use cds

Reply 998

LennonMcCartney

Cos of units and a good guess.

It's like asking, "Why should F be proportional to d(mv)/dt? Randomly put it in, why don't you!" That was a guess, and it worked. So did L=n\hbar. As did the Schrodinger equation, choices for operators etc

Hmmm, why should F = ma? You've made me wonder that now!! Out the window goes quantum!
Do you never wonder why any of the stuff they teach us!!?

Reply 999