
#1
Feb1505, 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. 



#2
Feb1505, 02:20 PM

PF Gold
P: 330

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. 



#3
Feb1505, 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 



#4
Feb1505, 02:26 PM

Sci Advisor
HW Helper
P: 11,863

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. 



#5
Feb1505, 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.




#6
Feb1505, 06:04 PM

Sci Advisor
HW Helper
P: 6,573

AM 



#7
Feb1505, 06:14 PM

Sci Advisor
HW Helper
P: 11,863

[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. 



#8
Feb1505, 07:50 PM

Sci Advisor
P: 412

[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} = (9.1093897e31 kg) {Proton Mass} = (1.6726231e27 kg) ~~ 


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