Colliding hydrogen atoms find velocity to raise to N=2

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Homework Help Overview

The problem involves two hydrogen atoms in the ground state colliding to achieve excitation to the n = 2 energy level. Participants are exploring the minimum speed required for this collision, focusing on energy conservation principles and the conversion of units.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply energy conservation principles, questioning their calculations related to kinetic energy and the mass of hydrogen. Some participants suggest clarifying the use of the center of mass frame and ensuring correct unit conversions, particularly from eV to joules.

Discussion Status

Participants are actively engaging with the problem, with some providing corrections and alternative approaches. There is an acknowledgment of errors in calculations and unit conversions, but no consensus has been reached regarding the final solution.

Contextual Notes

Participants note the importance of using the correct mass for hydrogen and the necessity of working in MKS units. There is also a mention of the energy balance equation needing to account for the masses of the hydrogen atoms post-collision.

Bill Headrick
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Homework Statement



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?
Answer in units of m/s

1 mol of hydrogen is 1.008g

Homework Equations



En = K + Uelec

For a hydrogen atom: En= 13.6/N^2

The Attempt at a Solution



Okay, so I know that there is a change in energy that excites the atoms to N=2. This change in energy must be from kinetic energy.

So,
Two atoms collide into each other. And after the collision they are raised to N=2.

A. 2(Ei+K)=2(Ef) ;note, i used 2 because it is two atoms colliding
The twos cancel?
B. Ei+K=Ef
-13.6/(1*1)+1/2(mh)v^2=-13.6/(2*2) ;mh is mass of hydrogen
-13.6+1/2(mh)V^2=-3.4
C. 10.2=1/2(mh)v^2 ;add 13.6 to both sides

Am I correct up until here? (Part C)
To find mass of hydrogen
D. 1.008/6.02*10^23=1.67*10^-24

E. 10.2=1/2(1.67*10^-24)*(v^2)
20.4=(1.67*10^-24)*(V^2)
1.22155 E 25=V^2

Where did I mess up?
 
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Okay, I figured it out:

10.2=1/2(mh)V^2
20.4=(mh)V^2

mh is equivalent to the mass of a proton which is 938E6 MeV/C^2

so, 20.4/938E6=v^2

so V^2 = 2.17484E-8
v= 1.474E-4

You have to multiply by c, I am not exactly sure the reasoning but
1.474E-4 * 299792458 = 44211.4107m/s

Hope if you stumble upon this with the same problem it helps.

Id still appreciate an explanation if anyone wants to give one.
 
First, be clear that you are using the center of mass frame. In that frame, the situation with respect to the two H is symmetric, so you can just look at one of them.

Second, use the right units. Since you need your answer to be in MKS units, you need to convert the 13.6 eV to joules.
 
Bill Headrick said:
mh is equivalent to the mass of a proton which is 938E6 MeV/C^2
You are off by a factor of a million.

Also, don't forget that in your energy balance equation the Hs still exist after the collision, so their masses are still there.
 

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