How Wide is the Potential Well When an Electron Emits a Photon?

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The problem involves an electron in a one-dimensional infinite potential well emitting a photon as it transitions from the n=3 to n=2 eigenstate. The energy levels are determined using the formula En = (n^2h^2)/(8mL^2), and the energy of the emitted photon is calculated with E = hc/λ. By equating the two energy expressions and solving for the width L of the potential well, the calculation yields L = 2.60 nanometers. This value represents the width of the potential well where the electron is confined. Proper unit checks and the correct application of formulas are essential for solving such physics problems.
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hey guys i am kinda stumped on this problem, if anyone can give me a helping hand, it's be much appreciated, thanks!

An electron is trapped in a one-dimensional infinite potential of width L. As the electron falls from the n=3 to the n=2 eigenstate, it emits a photon with a wavelength of 1649A. How wide is the potential well?
 
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Daniel.
 


To solve this problem, we can use the equation for the energy levels of a particle in a one-dimensional infinite potential well: En = (n^2h^2)/(8mL^2), where n is the quantum number, h is Planck's constant, and m is the mass of the electron.

Since we are given the wavelength of the emitted photon, we can use the equation for the energy of a photon: E = hc/λ, where h is Planck's constant, c is the speed of light, and λ is the wavelength of the photon.

Setting these two equations equal to each other and solving for L, we get L = √[(n^2h^2)/(8mE)], where E is the energy difference between the n=3 and n=2 eigenstates.

Substituting in the appropriate values, we get L = √[(3^2(6.626 x 10^-34 J*s)^2)/(8(9.11 x 10^-31 kg)(6.626 x 10^-34 J*s)(3.00 x 10^8 m/s)(1649 x 10^-10 m))].

Simplifying, we get L = 2.60 x 10^-9 m, or 2.60 nanometers. This is the width of the potential well in which the electron is trapped.

I hope this helps! Remember to always check your units and use the correct equations when solving physics problems. Good luck!
 
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