A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing

This Question is Closed

CliffSedge
 2 years ago
Best ResponseYou've already chosen the best response.4Is this an angular momentum thing?

SheldonEinstein
 2 years ago
Best ResponseYou've already chosen the best response.0^ mvr = angular momentum.

CliffSedge
 2 years ago
Best ResponseYou've already chosen the best response.4Would you like to do this via photon energy and Coulomb's Law?

CliffSedge
 2 years ago
Best ResponseYou've already chosen the best response.4(I'm assuming h = Planck's constant. Is this a reasonable assumption?)

CliffSedge
 2 years ago
Best ResponseYou've already chosen the best response.4And you're comfortable with basics like Newton's second law and conservation of energy and all that?

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.0Physic ! ask in right section.

SheldonEinstein
 2 years ago
Best ResponseYou've already chosen the best response.0Yes... F=ma and ...energy can neither be created nor be destroyed.

CliffSedge
 2 years ago
Best ResponseYou've already chosen the best response.4Ok... Let's start with \[\large f=\frac{c}{\lambda}\]

CliffSedge
 2 years ago
Best ResponseYou've already chosen the best response.4and \[\large \Delta E=hf\]

CliffSedge
 2 years ago
Best ResponseYou've already chosen the best response.4\[\large \rightarrow \frac{1}{hc} \Delta E = \frac{1}{\lambda}\] Cool so far?

CliffSedge
 2 years ago
Best ResponseYou've already chosen the best response.4So we'll now express DeltaE in terms of kinetic and potential energy differences. \[\large \Delta E =0.5m(\Delta v)^2ke^2(r_1^{1}r_2^{1})\] Where m, v, e, and r are the mass, velocity, charge, and orbital radius of the electron. Let me know when you're ready to continue.

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.0ΔE=0.5m(Δv)2−ke2(r−11−r−12) I guess sshould be +ke2(...)

CliffSedge
 2 years ago
Best ResponseYou've already chosen the best response.4@mahmit2012 , this is assuming an electron moving from an upper level to a lower level, so potential energy is decreasing. I should have been more clear that r_1 is the upper and r_2 is the lower. That is confusing now that I look at it like that.

CliffSedge
 2 years ago
Best ResponseYou've already chosen the best response.4Anyway, from here, we can introduce L=mvr > v=L/mr, and make some substitutions.

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.0so it shows the potential is negative, then delta pot=U2U1 ke2/r2(ke2/r1)

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.0And most important is how do you want reach to angle momentom!

CliffSedge
 2 years ago
Best ResponseYou've already chosen the best response.4Where I was going to go with this, but I don't think @SheldonEinstein is still interested, was: \[\large \Delta E =\frac{L^2}{2m}(r_1^{2}r_2^{2})ke^2(r_1^{1}r_2^{1})\] Putting together F=ke^2/r^2 and a=v^2/r with F=ma, \[\large \frac{ke^2}{r^2}=\frac{mv^2}{r}\] \[\large \rightarrow r=\frac{L^2}{kme^2}\]

CliffSedge
 2 years ago
Best ResponseYou've already chosen the best response.4Now the energy equation can be written as \[\large \Delta E = \frac{k^2e^4m}{2}(L_1^{2}L_2^{2})\] Now this can be put into \[\large \frac{\Delta E}{hc}=\frac{1}{\lambda}\] and then set equal to \[\large \frac{1}{\lambda} =R(2^{2}n^{2})\] This is the Balmer Series, where R = Rydberg constant. (This part might take more explanation . . .) Anyway, to make a long story shorter, L=mvr=nh/2π is the only solution for L that works.

SheldonEinstein
 2 years ago
Best ResponseYou've already chosen the best response.0sorry for my absence here @CliffSedge but I have a small question.

SheldonEinstein
 2 years ago
Best ResponseYou've already chosen the best response.0May I ask it? (but it is stupid one, but please allow)

CliffSedge
 2 years ago
Best ResponseYou've already chosen the best response.4Alright, as long as it isn't *too* stupid. ;) j/k. Ask anything.

SheldonEinstein
 2 years ago
Best ResponseYou've already chosen the best response.0have you proved that : mvr = nh/2 pi the above you wrote is to be read by me but after getting your response to my above short question...

CliffSedge
 2 years ago
Best ResponseYou've already chosen the best response.4Let me check . . . Yes. I can show some of the intermediate steps. I assumed you could do a lot of the algebra yourself.

SheldonEinstein
 2 years ago
Best ResponseYou've already chosen the best response.0Yes, but let me tell you that this is "bohrs second postulate" and postulates can't be proved. Majesty, you proved it! Any comments by you on this?

SheldonEinstein
 2 years ago
Best ResponseYou've already chosen the best response.0I assume that you might have taken anything unproven... that makes your work as not exactly a proof. I hope you're getting my point.

CliffSedge
 2 years ago
Best ResponseYou've already chosen the best response.4No, I'm not. Postulates are assumed to true for the purpose of proving a statement given certain assumptions.

SheldonEinstein
 2 years ago
Best ResponseYou've already chosen the best response.0Right so without assumptions, we have not be able to prove that ?

CliffSedge
 2 years ago
Best ResponseYou've already chosen the best response.4And postulates can be proven. You are mistaken on that point.

CliffSedge
 2 years ago
Best ResponseYou've already chosen the best response.4Nothing can be deduced, except tautologies, without assumptions.

CliffSedge
 2 years ago
Best ResponseYou've already chosen the best response.4If you're just wasting my time by trying to be cute playing with semantics, then be warned that I have plenty of time to waste.

SheldonEinstein
 2 years ago
Best ResponseYou've already chosen the best response.0I am controlling myself at present, be nice to others (at least to the one who has not done any thing offensive ..) The people interested to leave have an exit door also ...

SheldonEinstein
 2 years ago
Best ResponseYou've already chosen the best response.0http://mathworld.wolfram.com/Postulate.html http://answers.yahoo.com/question/index?qid=20101027183239AAcXeJa http://en.wikipedia.org/wiki/Axiom http://openstudy.com/study#/updates/4fea1f19e4b0bbec5cfa34ae Hope it justifies my answer as no to "postulates can be proven" ...

CliffSedge
 2 years ago
Best ResponseYou've already chosen the best response.4Not really . . . Ah, so you are just being cute playing with semantics. That's nice, dear. Regarding link 1: I understand that modern mathemeticians use the words "axiom" and "postulate" interchangeably, but they are not exactly the same thing in all contexts. Regarding link 2: Some level7 neckbeard on Yahoo! Answers doesn't qualify as a reliable source to me. Regarding link 3: That is a wikipedia page on "Axiom" not "Postulate," though I assume you are using it as reference to define 'postulate' as nonlogical axiom. This is all irrelevant, because  Regarding link 4: Even if we do accept the narrow definition of postulate to be synonymous with axiom, this statement is important, "Postulate is the starting point of any argument, the basis of any reasoning and proof. " Proofs follow from axioms and postulates (and other intermediate theorems). If you are trying to assert that nothing based on an assumption can be proven, then you are stuck in a Cartesian 'theonlythingthatcannotbedoubtedisthefactthatIamdoubting' / 'cogito ergo sum' bit of philosphical nonsense. With that out of the way, I suppose we can come to this point, viz. Bohr's second postulate is "Each emission or absorption of radiation energy represents the electron transition from one stationery orbit to another. The radiation emitted during such transition is homogeneous." So your remark that I was proving Bohr's second postulate is false. Lastly, if mvr=nh/2π is a postulate then why was I able to derive it from basic principles? Your claims are without merit or basis. Your reply @SheldonEinstein ?

CliffSedge
 2 years ago
Best ResponseYou've already chosen the best response.4So it seems that you are merely trying to play a little logic games that goes like Premise one: Postulates can't be proven. Premise two: Here' a postulate, try to prove it. Premise three: Haha, you tried to prove it; but, by premise one, postulates can't be proven, nanynanybooboo! Conclusion: You're a dummy. Look at how smart I must be in comparison. That is very childish, and all the more embarrassing for you because your premises were false. I hope you are more successful at playing your childish games elsewhere. I'm sure you derive much pleasure from them.

SheldonEinstein
 2 years ago
Best ResponseYou've already chosen the best response.0@CliffSedge I respect your knowledge and opinion but the way you talked here is quite bad to me at least, I respect your work done here but I didn't do the way of your talking here. The way you lost your temper itself shows that you are a "child" not me, your this behavior proved a lot. Thanks!

CliffSedge
 2 years ago
Best ResponseYou've already chosen the best response.4I'm sorry you feel that way. What was your intention for posting this question?

SheldonEinstein
 2 years ago
Best ResponseYou've already chosen the best response.0It happened like this sir: I was watching a lecture, of Atomic Structure and the teacher said "It is still a mystery for me that how did Bohr came to this formula > mvr = nh/2 pi and how can we prove it? " This produced just a willingness to see that can experts here on OS help to prove? I posted it first in chemistry section, they said 'postulates can't be proven" and my stupidness I agreed them some how , I came to the maths section then , asked here and got the solution from you but unfortunately I was not able to understand anything, I studied the lecture of the teacher more and came to know something about Rydberg's constant and other constants, variables used by you here.

CliffSedge
 2 years ago
Best ResponseYou've already chosen the best response.4Then you would have been better served by asking more questions instead of boldly claiming that I had attempted to prove a postulate that could not be proved.

SheldonEinstein
 2 years ago
Best ResponseYou've already chosen the best response.0I lost my language and the way I talked was of course have felt bad to you, sorry about that sir.

SheldonEinstein
 2 years ago
Best ResponseYou've already chosen the best response.0Yes sir, I agree with you that I must have asked quest. rather than claiming with overconfidence, I am guilty for that :( , sorry , forgive me sir....

CliffSedge
 2 years ago
Best ResponseYou've already chosen the best response.4You asked me to do a certain task and then I did it, then you claimed that I had not without even understanding what I had done. You jumped to a conclusion. I forgive you for that, but take that as a lesson to be more patient in the future; because please know: that when it comes to arguing points of logic, I am *very* patient, and very ruthless at attacking and destroying bad arguments. It may seem sometimes that I am attacking the person, but that is not so, I only brutally attack and dismantle bad ideas, not people.

CliffSedge
 2 years ago
Best ResponseYou've already chosen the best response.4If you are actually interested in learning the physics, I would be happy to discuss this further.

SheldonEinstein
 2 years ago
Best ResponseYou've already chosen the best response.0Agreed sir, thanks for forgiving, the lesson saved permanently in my mind. Yes sir, why not? Can I post this as a new question here in maths section but my first question will be : "CCan postulates be proved?"

SheldonEinstein
 2 years ago
Best ResponseYou've already chosen the best response.0Should I go for it? sir.
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.