UAM & Electrostatics: Does It Apply At Atomic Scale?

[metalmagik]

Do UAM equations apply on an atomic scale when dealing with protons and electrons etc?

 

[OlderDan]

Do UAM equations apply on an atomic scale when dealing with protons and electrons etc?

You mean the The Urban Airshed Model - UAM-IV?

http://www.ccl.rutgers.edu/~ssi/thesis/thesis-node56.html

Or maybe you are talking about uniformly accelerated motion?

The classical notions of particle motion do not hold up on the atomic scale. The theory of quantum mechanics is used to analayze such problems.

 

[metalmagik]

I am talking about Uniformly Accelerated Motion. hm, I see they cannot be used. How do I then find a final velocity for a proton when I am given the initial velocity, electric field magnitude, and distance?

 

[OlderDan]

I am talking about Uniformly Accelerated Motion. hm, I see they cannot be used. How do I then find a final velocity for a proton when I am given the initial velocity, electric field magnitude, and distance?

They can be used. See the other thread where the context of your problem is stated. You need to change your understanding of what atomic scale means. It is not about the size of the particle. It is about the distances involved in the motion.

 

[metalmagik]

I see, I understand this now, thank you very much.

 

[metalmagik]

If you're still around. Here is my work for the problem I was confused about using UAM equations with:

A uniform electric field has a magnitude of 3.0 x 10^3 N/C. In a vacuum, a proton begins with a speed of 2.4 x 10^4 m/s and moves in the direction of this field. Find the speed of the proton after it has moved a distance of 1.0 mm.

$$F = qE$$

$$F=(1.6 x 10^-19)(3 x 10^3)$$

$$F = 4.8 x 10^-16 N$$

$$F = ma$$

$$4.8 x 10^-16 N = (1.67 x 10^-27 kg) a$$

$$a = 2.87 x 10^(11) m/s^2$$

$$Vf^2 = Vi^2 + 2ad$$

$$Vf^2 = (2.4 x 10^4)^2 + 2(2.87 x 10^11)(.001)$$

$$Vf = 3.39 x 10^4 m/s$$

eh some of the exponents in the latex got skewed, but I am sure you can figure it out. If you can verify this answer for me, that'd be great.

 

[OlderDan]

If you can verify this answer for me, that'd be great.

Looks OK . . . .