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Kinetic energy of electrons/protons

by thisisfudd
Tags: electrons or protons, energy, kinetic
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thisisfudd
#1
Feb15-05, 01:57 PM
P: 56
Hi, I'm working on the following:

An electron starting from rest acquires 6.3 keV of KE in moving from point A to point B. (a) How much KE would a proton acquire, starting from rest and moving to point A? (b) Determine the ratio of their speeds at the end of their respective trajectories.

So I understand that the proton will also have a KE of 6.3 keV because it has an equal but opposite charge to the electron (part A). But could someone give me a hint about the whole ratio of speeds part? Would I use some equation for velocity, and if so, how would I use the keV in it? Or do I use the respective masses of electrons and protons? Thx.
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Gamma
#2
Feb15-05, 02:20 PM
Gamma's Avatar
P: 333
Use the relativistic equation

[tex] E = \gamma m c^2[/tex]

where m is the rest mass to find the velocities of e and p.
MathStudent
#3
Feb15-05, 02:23 PM
P: 281
Think about energy conservation.

There talking about kinetic energy,
The equation for kinetic energy is

[tex] KE = \frac{1}{2} m v^2 [/tex]

using conservation of energy, how would the KE of the electron compare to the KE of the proton?

edit: This is assuming your looking at it from the "classical" sense. If not, look to the post above

dextercioby
#4
Feb15-05, 02:26 PM
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P: 11,927
Kinetic energy of electrons/protons

There's no need for relativistic formula,the ratio is approx 1/100 for the electron and 1/180000 for the proton (if the latter has the same 6.3 KeV).

Daniel.
thisisfudd
#5
Feb15-05, 02:54 PM
P: 56
I'm sorry Dexter, I don't think I understand what you're saying?? The answer, according to my friend, is something like 1:43. So I am looking for the ratio of their two velocities. This is dependent on mass, correct? Thanks.
Andrew Mason
#6
Feb15-05, 06:04 PM
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Quote Quote by thisisfudd
An electron starting from rest acquires 6.3 keV of KE in moving from point A to point B.
Are we to assume that it acquires this KE by passing through an electric field? There are other ways it can acquire kinetic energy.

AM
dextercioby
#7
Feb15-05, 06:14 PM
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Quote Quote by thisisfudd
I'm sorry Dexter, I don't think I understand what you're saying?? The answer, according to my friend, is something like 1:43. So I am looking for the ratio of their two velocities. This is dependent on mass, correct? Thanks.
I was making the ratio:
[tex] (\frac{KE}{Rest \ mass \times c^{2}})_{electron} \sim \frac{6.3KeV}{511KeV}\sim \frac{1}{85} [/tex]

Well,i approximated (not too accurately,though) to 1/100 and called the use of relativistic energy formula a mere complication...

Daniel.
xanthym
#8
Feb15-05, 07:50 PM
Sci Advisor
P: 412
Quote Quote by thisisfudd
Hi, I'm working on the following:

An electron starting from rest acquires 6.3 keV of KE in moving from point A to point B. (a) How much KE would a proton acquire, starting from rest and moving to point A? (b) Determine the ratio of their speeds at the end of their respective trajectories.

So I understand that the proton will also have a KE of 6.3 keV because it has an equal but opposite charge to the electron (part A). But could someone give me a hint about the whole ratio of speeds part? Would I use some equation for velocity, and if so, how would I use the keV in it? Or do I use the respective masses of electrons and protons? Thx.
SUMMARY OF ESTABLISHED PROBLEM ITEMS:

[tex] :(1): \ \ \ \ (ElectronKineticEnergy) = (ProtonKineticEnergy) = (6.3 keV) [/tex]

[tex] :(2): \ \ \ \ \frac { M_{electron} V^{2}_{electron} } {2} = \frac { M_{proton} V^{2}_{proton} } {2} = (6.3 keV) [/tex]

{Electron Mass} = (9.1093897e-31 kg)
{Proton Mass} = (1.6726231e-27 kg)


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