How Do Electron Transitions Correspond to Energy Levels in Quantum Wells?

  • Thread starter Thread starter sss1
  • Start date Start date
  • Tags Tags
    Diagram
Click For Summary
The discussion focuses on electron transitions between energy levels in quantum wells, specifically analyzing transitions from states n=4 to n=2, n=5 to n=4, n=5 to n=1, and n=4 to n=1. It establishes that these transitions correspond to specific states in different wells, such as well C and well D. The second problem involves calculating node separation, leading to the conclusion that n=10, with an energy difference formula derived from this state. There is a correction regarding the denominator in the energy calculation, which requires squaring, and a clarification on the correct measurement of 200 pm instead of 300 pm. The conversation emphasizes the importance of accurate parameters in quantum well calculations.
sss1
Messages
50
Reaction score
2
Homework Statement
Can someone check whether I've done the problems in the pictures properly? not sure if i've done them correctly.
Relevant Equations
n^2h^2/8(m_e)a^2
Screen Shot 2023-10-09 at 22.18.32.png
Screen Shot 2023-10-09 at 22.19.23.png
Screen Shot 2023-10-09 at 22.19.28.png

For the problem with the diagram, I'm getting from n=4 to n=2, n=5 to n=4, n=5 to n=1 and n=4 to n=1. n=4 to n=2 corresponds to n=1 to n=2 in well C; n=5 to n=4 corresponds to n=1 to n=2 in well D; n=5 to n=1 corresponds to n=1 to n=3 in well D and n=4 to n=1 corresponds to n=1 to n=2 in well E.
For the second problem, since the node separation is given by L/n where n is the state, and given that there is no nodes between 60pm and 80pm, i worked out that n=10. So the difference in energy is (10^2-9^2)h^2/(8(m_e)(300pm))?
 
Physics news on Phys.org
sss1 said:
For the problem with the diagram, I'm getting from n=4 to n=2, n=5 to n=4, n=5 to n=1 and n=4 to n=1. n=4 to n=2 corresponds to n=1 to n=2 in well C; n=5 to n=4 corresponds to n=1 to n=2 in well D; n=5 to n=1 corresponds to n=1 to n=3 in well D and n=4 to n=1 corresponds to n=1 to n=2 in well E.
For the second problem, since the node separation is given by L/n where n is the state, and given that there is no nodes between 60pm and 80pm, i worked out that n=10. So the difference in energy is (10^2-9^2)h^2/(8(m_e)(300pm))?
That appears correct, except the denominator in the energy at the end.
 
DrClaude said:
That appears correct, except the denominator in the energy at the end.
Needs squaring?
 
sss1 said:
Needs squaring?
Yes, and why 300 pm?
 
DrClaude said:
Yes, and why 300 pm?
Oh right, misread it. 200pm?
 
sss1 said:
Oh right, misread it. 200pm?
Yes
 
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

Similar threads

  • · Replies 30 ·
Replies
30
Views
8K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 3 ·
Replies
3
Views
882
Replies
12
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 24 ·
Replies
24
Views
2K
Replies
12
Views
900