Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing

This Question is Closed

s3aBest ResponseYou've already chosen the best response.0
The solution says: "To consider an electron's motion in classical terms, the uncertainties in its position and momentum must be negligible when compared to r and p; in other words, Δx << r and Δp << p." Why is it the consideration of the motion in classical terms that is reponsible for requiring that the uncertainties in position and momentum be negligible when compared to the actual values? Also, is it the case that the uncertainties need not be negligible when the motion is considered to be nonclassical? If so, why is this the case? The solution says: "Eq. (1.7.1) is incompatible with Eq. (1.7.2), unless n >> 1." Is the reason why Eq. (1.7.1) is incompatible with Eq. (1.7.2) because we can analyze a situation at a specific energy level (and therefore hold n constant for that situation) such that n >> 1 does not hold (for example when n = 1)? Similarly, is it not the case that n >> 1 because, Eq. (1.7.2) does not hold for all values of n (even if it holds for the majority of the values), we can say that the assumption using classical theory is false and we must therefore reject the semiclassical Bohr model for the hydrogen atom? Even if I got things right, please confirm it for me. Any help in fully understanding this problem, would be greatly appreciated!
 one year ago

henpenBest ResponseYou've already chosen the best response.1
Why is thinking in classical terms reponsible for requiring that the uncertainties in position and momentum be negligible when compared to the actual values? Because classical mechanics assumes that the uncertainties are 0. Classical mechanics is never exactly correct for the real atom, but it's a good approximation when the uncertainties are essentially=0 compared with the values. Also, is it the case that the uncertainties need not be negligible when the motion is considered to be nonclassical? That's the point of non classical (i.e. quantum) mechanics a better approximation (we don't have better) for large uncertainty values (in proportion to the values for p and x themselves). Is the reason why Eq. (1.7.1) is incompatible with Eq. (1.7.2) because we can analyze a situation at a specific energy level (and therefore hold n constant for that situation) such that n >> 1 does not hold (for example when n = 1)? Basically. If you set n=1 and make the 2 equations equal, you get \[1 \le \Delta x \Delta p/xp <<1\]Obviously incorrect.
 one year ago
See more questions >>>
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.