What Could Be Wrong with My Calculations in the Infinite Potential Well Problem?

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SUMMARY

The discussion centers on calculating the energy levels and photon wavelength for an electron in an infinite potential well of width 0.120 nm. The ground state energy, E1, is calculated using the formula E_1 = (π² (ħ c)²) / (2M_e C² a²), resulting in 26.02 eV. The energy difference between the n=3 and n=2 states is computed as E_γ = E_3 - E_2, yielding 130.13 eV. The wavelength of the emitted photon is calculated using λ = hc / E_λ, resulting in 9.53 nm, which raises concerns about the expected wavelength being larger.

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  • Knowledge of fundamental constants such as Planck's constant (h), reduced Planck's constant (ħ), and the speed of light (c).
  • Ability to perform unit conversions and dimensional analysis in physics calculations.
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Students and professionals in physics, particularly those studying quantum mechanics, as well as educators seeking to clarify concepts related to energy levels and photon emissions in quantum systems.

Meekay
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three parts to this one, I can't seem to justify my values, units cancel, but the numbers don't seem right. I think I may have used a wrong equation for part B but I don't know what else to use.

Problem: An electron is confined to an infinitely deep potential well of width 0.120 nm.
a.) Calculate its ground state energy, E1
b.)If the electron makes a transition from the n=3 state to the n=2 state, how much energy is carried away by the emitted photon?
c.)What is the wavelength of this photon?equations:

a.) E_1 = \frac{pi^2 (hbar c)^2}{2M_e C^2 a^2}

b.) E_\gamma = E_3 - E_2

c.) \lambda = \frac{hc}{E_\lambda}My attempt:

a.) using hbar*c = 197 ev*nm and MeC^2 = 511000 ev i get a value of 26.02ev for E1

b.) using the same equation as above for the n=3 and n=2 states and subtracting I get 130.13ev

c.) using hc = 1240 ev*nm I get an answer of 9.53 nm which doesn't seem right to me. I feel like the photon should have a larger wavelength.
 
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Why do you think the photon should have a bigger wavelength?

Lets see - using ##h= 2\pi \hbar## ;$$E_n=\frac{n^2(h c)^2}{8mc^2a^2}=n^2E_1$$ (when you use LaTeX, put a backslash in front of the symbol name so \hbar renders as ##\hbar## etc.)

hc=1239.8 eV.nm
mc^2=511000 eV
a=0.120nm

Looks good to me:
Go through the arithmetic one step at a time, make sure you have squared the correct terms.
 

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