Solving for the speed of a recoiling H atom

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To solve for the speed of a recoiling hydrogen atom after it emits a photon during a transition from the n=2 to the ground state, the energy difference between the two states is calculated using E = -13.6/n^2. The user initially attempted to equate this energy difference to (1/2)mv^2 but was unsuccessful. A more effective approach is to apply conservation of momentum, which correctly leads to the answer of 3.4 m/s for the recoil speed. This method accounts for the momentum of the emitted photon and the recoiling atom. The discussion emphasizes the importance of using conservation principles in such quantum mechanics problems.
silex
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Hey everyone,
I have a practice problem that I can't figure out:

A hydrogen atom is initially at rest and in the n=2 state. Subsequently, it decays to the ground state by emitting a photon. What is the speed of the recoiling atom afterwards? (Mass of H atom = 1.67 * 10^-27 kg)

My approach is to find the difference in energy between the two states by using E = -13.6/n^2 for each state (n=1 and n=2), then finding the energy difference, and setting that equal to (1/2)mv^2 and solving for v, but I'm not getting the right answer. (The correct answer is 3.4 m/s)
Can someone help me with this?

Thanks in advance

-s
 
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use 4 in work reverse?
 
I would use conservation of momentum.
 
Using conservation of momentum solved the problem, thanks.
 

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