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

  • Thread starter Thread starter sss1
  • Start date Start date
  • Tags Tags
    Diagram
AI Thread 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 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top