Hydrogen atom emitting a photon

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SUMMARY

The discussion focuses on the emission of a photon by a hydrogen atom transitioning from the n = 4 state to the n = 3 state, resulting in a photon wavelength of 1.86 x 103 nm. The recoil speed of the hydrogen atom upon photon emission is calculated using the conservation of momentum, yielding a speed of 0.22 m/s. The momentum of the emitted photon is determined using the formula p = h/λ, where h is Planck's constant. The calculations demonstrate the relationship between photon emission and atomic recoil in quantum mechanics.

PREREQUISITES
  • Understanding of quantum mechanics principles, specifically atomic transitions.
  • Familiarity with the equation λ = hc/ΔE for photon wavelength calculations.
  • Knowledge of momentum conservation in physics.
  • Ability to apply kinetic energy equations, such as KE = 1/2 mv².
NEXT STEPS
  • Study the derivation and applications of the equation λ = hc/ΔE in quantum mechanics.
  • Learn about the conservation of momentum in photon emission scenarios.
  • Explore the implications of recoil in atomic physics and its effects on measurements.
  • Investigate the relationship between energy levels in hydrogen and photon emission wavelengths.
USEFUL FOR

Students and educators in physics, particularly those focusing on quantum mechanics, atomic transitions, and photon interactions. This discussion is beneficial for anyone studying the principles of momentum and energy conservation in atomic systems.

STJ
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Homework Statement



Consider the following.
(a) For a hydrogen atom making a transition from the n = 4 state to the n = 3 state, determine the wavelength of the photon created in the process. (Already solved this, 1.86x10^3 nm)



(b) Assuming that the atom was initially at rest, determine the recoil speed of the hydrogen atom when it emits this photon.


Homework Equations



λ=hc/ΔE

The Attempt at a Solution



I've done one like this previously, but we were given the energy of the photon colliding with the atom. Found the Kinetic Energy, set it equal to 1/2mv^2 and solved for v. I don't think I can do this for part b. I think this has something to do with momentum perhaps? I'm kind of lost on this one.
 
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Momentum of photon = ## \frac{h}{ \lambda} ##

Conservation of momentum says that this momentum must be in the opposite direction of the hydrogen atom.

## p = \frac{h}{\lambda} = \frac{6.63*10^{-34}}{1.83*10^{-6}} = 3.62*10^{-28} kg \cdot ms^{-1} ##

## V_{H-atom} = \frac{p}{m} = \frac{3.62*10^{-28}}{1.67*10^{-27}} = 0.22 ms^{-1} ##
 
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