Hydrogen atom collision

  • Thread starter cocomisk
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Two hydrogen atoms, both initially in the ground state, undergo a head on collision. If both atoms are to be excited to the n=2 level in this collision, what is the minimum speed each atom can have before the collision?
Ke = 8.99 *10^9
e = 1.602 *10^-19
ħ = 1.05 * 10 ^-34
Mass of electron = 9.11 * 10^-31
mass of proton = .672 * 10^-27



Mass of electron*v*r = nħ This is the angular momentum equation
Total Energy= 1/2 mv^2 - Ke *(e^2 / r)
v^2 = (n^2 * ħ)/(m^2*r^2) = (Ke*e^2)/(m*r)
Radius of n = n^2 * .0529 nm




I can't figure out how to use angular momentum to solve this and that's the only equation my book provides for momentum concerning hydrogen. The closest I have come is using the v^2 equation and using the proton mass rather than the electron mass and using the radius for n=1. This gives me 51043 m/s. The answer is supposed to be 44200 m/s. I assume I need to incorporate the changes in velocity somehow, but I can't figure out how to format an equation to do that.:confused:

Any help is greatly appreciated!


~Courtney
 

Kurdt

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I think this is a lot more simple than you're trying to make it. All you need to know is how much energy it takes to get an electron into the n=2 level and then make that equal to the kinetic energy of the atoms and solve for speed. Because the collision is head on you can assume all the energy goes into making the electron jump to the n=2 level.
 
E of the ground state = 13.6 eV
E of the n=2 state = 3.4 eV

So the energy required is 10.2 eV or about 6.367 *10^19 J

When I set this equal to 1/2 mv^2, I get a huge number for v...like numbers *10^23. That's incredibly far from the listed answer. What am I doing wrong?
 

Kurdt

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The energy should be 6.367x10-19 should get a more reasonable answer like 15 000 m/s.
 
Yeah I just realized I was converting the numbers wrong.

Thanks Kurdt.
 

Kurdt

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no problem :smile:
 

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