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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.

Any help is greatly appreciated!

~Courtney

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# Hydrogen atom collision

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