Quantum Mechanics / Uncertainty Principle Question

Click For Summary

Homework Help Overview

The discussion revolves around a question related to quantum mechanics, specifically the probability of finding an electron in a 1s orbital within a specified volume. The context involves the Heisenberg Uncertainty Principle and the interpretation of wave functions in quantum chemistry.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relevance of the initial statement about wave functions and question the meaning of "centred at" in the context of the problem. There is a discussion about the absence of a specific wave function and whether the question is merely theoretical.

Discussion Status

Some participants have offered insights into the nature of the probability calculation and suggested looking up the wave function for the ground state. However, there remains a lack of consensus on how to approach the problem without a given wave function.

Contextual Notes

Participants note that the problem may not be straightforward due to the absence of a specific wave function, raising questions about the assumptions underlying the problem setup.

fazer2014
Messages
8
Reaction score
0

Homework Statement



The square of a wave function gives the probability of finding a particle at a given point. What is the probability of finding an electron in a 1s orbital within a volume of 1pm^3, centred at:
a) the nucleus
b) 50pm away from the nucleus?

Homework Equations



Heisenberg Uncertainty Principle

The Attempt at a Solution



I sense that this is actually a straightforward question, but I just can't get my head around what it's asking. I feel like the first sentence is not actually relevant to solving the question, just a little ditty of information? I'm also thrown by the 'centred at' thing. If anyone can offer an explanation for how to think about this problem it would be greatly appreciated, thanks.
(Sorry, this is actually for a chemistry class, but I searched the forums and there are a few questions on this topic, though none I could find that answered this specific question).
 
Physics news on Phys.org
fazer2014 said:

Homework Statement



The square of a wave function gives the probability of finding a particle at a given point. What is the probability of finding an electron in a 1s orbital within a volume of 1pm^3, centred at:
a) the nucleus
b) 50pm away from the nucleus?

Homework Equations



Heisenberg Uncertainty Principle

The Attempt at a Solution



I sense that this is actually a straightforward question, but I just can't get my head around what it's asking. I feel like the first sentence is not actually relevant to solving the question, just a little ditty of information? I'm also thrown by the 'centred at' thing. If anyone can offer an explanation for how to think about this problem it would be greatly appreciated, thanks.
(Sorry, this is actually for a chemistry class, but I searched the forums and there are a few questions on this topic, though none I could find that answered this specific question).

It is not really an uncertainty question.

First, note that the probability is not given by the square of the wave function! The probability of finding the particle in a small volume dV is actually given by

\bigl| \psi (r, \theta, \phi) \bigr|^2 \, dV

So just square the wave function at the values of r given in the questions and multiply by the small volume.
 
I see... thanks, I understand in theory. But we weren't actually given a wave function. So is it just a thought experiment or something?
 
You were given the state (1s orbital). I suggest looking up the wave function for the ground state in a hydrogen-like potential.
 

Similar threads

  • · Replies 23 ·
Replies
23
Views
5K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 47 ·
2
Replies
47
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
6K