Photon emitted when proton changes state

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SUMMARY

The discussion focuses on calculating the energy and wavelength of a photon emitted when a proton transitions from the n=2 state to the ground state in an infinitely high square well of length 10 femtometers (fm). The energy calculations utilize the formula E = (h^2 n^2) / (8mL^2), yielding E_1 = 1.31E-12 J for n=2 and E_0 = 3.286E-13 J for the ground state. The energy of the emitted photon is calculated as E = 9.859E-13 J, resulting in a wavelength of λ = 2.016E-13 m, confirming that the photon is a gamma ray.

PREREQUISITES
  • Quantum mechanics principles, specifically energy quantization in potential wells.
  • Understanding of the Planck constant (h) and its application in energy calculations.
  • Familiarity with the concept of photon emission and wavelength calculations.
  • Basic knowledge of particle physics, particularly properties of protons.
NEXT STEPS
  • Explore quantum mechanics and the implications of infinite potential wells.
  • Learn about the properties and applications of gamma rays in physics.
  • Investigate the use of computational tools like WolframAlpha for verifying physics calculations.
  • Study the relationship between energy and wavelength in electromagnetic radiation.
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Students studying quantum mechanics, physicists interested in particle behavior, and educators teaching concepts related to photon emission and energy transitions in atomic systems.

wahaj
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Homework Statement


a proton is confined in an infinitely high square well of length 10 fm. If the proton transitions from n=2 to ground state determine the energy and wavelength of the photon emitted


Homework Equations



[tex]E = \frac{h^2 n^2}{8mL^2}[/tex]
[tex]E = \frac{hc}{\lambda} \ \ or \ \ \lambda = \frac{hc}{E}[/tex]

The Attempt at a Solution


I need some one to tell me if I did this right.
energy at n = 2
[tex]E_1 = \frac{(6.626E-34)^2( 2^2) }{8(1.67E-27)(1E-14)^2}[/tex]
[tex]E_1 = 1.31E-12[/tex]

energy at ground state
[tex]E_0 = \frac { (6.626E-34)^2}{8(1.67E-27)(1E-14)^2 }[/tex]
[tex]E_0 = 3.286E-13[/tex]

energy of photon released
[tex]E = E_1 - E_0[/tex]
[tex]E = 9.859E-13 \ J[/tex]

wavelength of photon
[tex]\lambda = \frac{(6.626E-34)(3E8)}{9.859E-13}[/tex]
[tex]\lambda = 2.016E-13 m = 0.2016 pm[/tex]
this would be a gamma ray.

So did I do this question right?
 
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There are some units missing, but a gamma ray is the right order of magnitude for such a photon and the approach looks fine. You can check the calculations with WolframAlpha, for example.
 
thanks.
 

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