# Kinetic energy of electrons/protons

by thisisfudd
Tags: electrons or protons, energy, kinetic
 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.
 PF Gold P: 329 Use the relativistic equation $$E = \gamma m c^2$$ where m is the rest mass to find the velocities of e and p.
 P: 281 Think about energy conservation. There talking about kinetic energy, The equation for kinetic energy is $$KE = \frac{1}{2} m v^2$$ 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
HW Helper
P: 11,834

## 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.
 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.
HW Helper
P: 6,510
 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
HW Helper
P: 11,834
 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:
$$(\frac{KE}{Rest \ mass \times c^{2}})_{electron} \sim \frac{6.3KeV}{511KeV}\sim \frac{1}{85}$$

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

Daniel.
P: 412
 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:

$$:(1): \ \ \ \ (ElectronKineticEnergy) = (ProtonKineticEnergy) = (6.3 keV)$$

$$:(2): \ \ \ \ \frac { M_{electron} V^{2}_{electron} } {2} = \frac { M_{proton} V^{2}_{proton} } {2} = (6.3 keV)$$

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

~~

 Related Discussions High Energy, Nuclear, Particle Physics 30 Introductory Physics Homework 1 High Energy, Nuclear, Particle Physics 1 General Physics 8 General Physics 5