Quantum Well with Infinite Barriers

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SUMMARY

The discussion focuses on calculating the wavelengths of photons emitted during electronic transitions in a one-dimensional rectangular potential well with infinite barriers, specifically for quantum numbers n = 2 and n = 3 transitioning to n = 1. The calculated wavelengths are λ21 ≈ 1.15 µm and λ31 ≈ 0.43 µm. The participant initially misapplied the energy difference formula ΔE = En+1 − En, which is valid only for neighboring states, leading to incorrect wavelength calculations. The correct approach involves calculating the energies of each state separately before finding the differences.

PREREQUISITES
  • Quantum mechanics fundamentals, specifically the concept of quantum wells.
  • Understanding of energy quantization in potential wells.
  • Familiarity with Planck's equation ε = hf for photon energy calculations.
  • Basic knowledge of the Schrödinger equation and its application to one-dimensional systems.
NEXT STEPS
  • Study the derivation of energy levels in quantum wells using the Schrödinger equation.
  • Learn about the implications of infinite potential barriers in quantum mechanics.
  • Explore the application of the energy difference formula for non-neighboring quantum states.
  • Investigate the relationship between energy transitions and photon emission in quantum systems.
USEFUL FOR

This discussion is beneficial for physics students, educators, and researchers focusing on quantum mechanics, particularly those studying quantum wells and electronic transitions in confined systems.

beastman
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Homework Statement


An electron is in a one-dimensional rectangular potential well
with barriers of infinite height. The width of the well is equal to L = 5 nm.
Find the wavelengths of photons emitted during electronic transitions from the
excited states with quantum numbers n = 2, λ21, and n = 3, λ31, to the ground
state with n = 1. (Answer: λ21 ≈ 1.15 µm and λ31 ≈ 0.43 µm.)

Homework Equations



E1 = (∏^2)*h^2/2meL^2 = 0.3737/L^2 eV

ΔE = En+1 − En = (2n + 1)E1

ε = hf

The Attempt at a Solution



I found the ground state energy to be 0.0149 eV. Then using the ΔE equation for n=2,3 I found the energies of the emitted photons to be 0.0745 eV and 0.1043 eV, respectively.
Using these energies in Plancks formula is getting me the wrong wavelengths, what am I doing wrong?

Please help!
 
Last edited:
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beastman said:
ΔE = En+1 − En = (2n + 1)E1

This formula will only work for jumping between two neighboring states. So, it should work for E2-E1. But what value should you use for n in the formula for this case? (It's not n = 2.)

For the jump from the n = 3 to the n = 1 case you might just want to calculate the energies of each state separately and then subtract.
 

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