Hydrogen atom emitting a photon

AI Thread Summary
For a hydrogen atom transitioning from the n = 4 state to the n = 3 state, the emitted photon has a wavelength of 1.86 x 10^3 nm. To determine the recoil speed of the hydrogen atom after photon emission, the conservation of momentum principle is applied. The momentum of the emitted photon is calculated as 3.62 x 10^-28 kg·m/s, which leads to the hydrogen atom's recoil speed being approximately 0.22 m/s. The calculations involve using the relationship between wavelength, energy, and momentum. This analysis highlights the interplay between photon emission and atomic recoil in quantum mechanics.
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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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