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Proton accelerated thru potential difference - find final speed

  • #1

Homework Statement


A proton is accelerated from rest through a potential difference of 107 V. Calculate the final speed of this proton.


Homework Equations


I'm not really sure which equations to use, but I figured I give all I tried...
V=-Ed
V=PE/q
PE=-qEd
and I think this applies...v(f)^2= v(i)^2 + 2ad

The Attempt at a Solution



I really can't figure out what equation i should use. What I did was..
1)used the equation V = Ed
2) becomes V= Kc(q/d^2)d
3) becomes V=Kc(q/d)
4)107= 8.99x10^9 (1.602x10^-19)/d
5)Solved for d and found it was 1.34598 x 10^-11
6) used the equation v(f)^2= v(i)^2 + 2ad
7) became v = sqrt (2a (1.34598x10^-11)
....but I don't know what "a" is so I can't solve. Any help would be greatly appreciated. I gave this a shot and really couldn't figure it out. Thanks, everyone.
 

Answers and Replies

  • #2
378
2

Homework Statement


A proton is accelerated from rest through a potential difference of 107 V. Calculate the final speed of this proton.


Homework Equations


I'm not really sure which equations to use, but I figured I give all I tried...
V=-Ed
V=PE/q
PE=-qEd
and I think this applies...v(f)^2= v(i)^2 + 2ad

The Attempt at a Solution



I really can't figure out what equation i should use. What I did was..
1)used the equation V = Ed
2) becomes V= Kc(q/d^2)d
3) becomes V=Kc(q/d)
4)107= 8.99x10^9 (1.602x10^-19)/d
5)Solved for d and found it was 1.34598 x 10^-11
6) used the equation v(f)^2= v(i)^2 + 2ad
7) became v = sqrt (2a (1.34598x10^-11)
....but I don't know what "a" is so I can't solve. Any help would be greatly appreciated. I gave this a shot and really couldn't figure it out. Thanks, everyone.
Try energy method.
Find relationship between initial energy, and final energy.
What is the initial energy?, and what's the final energy?
 
  • #3
529
1
Yes, by far the easiest method is to just use energy conservation.

Proton KE = potential difference
 
  • #4
Don't forget that 1eV = 1.60E-19 J

Since the potential difference is given (107 V), you have to find the kinetic energy that it would take for that to happen. Set up a ratio like this: (1 eV)/(1.60E-19 J)=(107 V)/(x). Solve for x. This value, x, is the kinetic energy (in J) that this would take.

Now, use 1/2mv^2=KE. Since you have the KE to be x, solve for v.

v=sqrt(2KE/m)
 
  • #5
Now, use 1/2mv^2=KE. Since you have the KE to be x, solve for v.

v=sqrt(2KE/m)
what is "m"?
what is "sqrt"?
 
  • #6

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