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Challence problem on electric potential

  • Thread starter mrjoe2
  • Start date
  • #1
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Edit to the title: CHALLENGE problem on electric potential energy*

Homework Statement


A proton and an alpha particle (q = +2e, m = 4u) are fired directly toward each other from far away, each with an initial speed of 0.010c. what is their distance of closes approach, as measured between their centers?


Homework Equations


u = proton mass = 1.67262158 × 10-27 kilograms
e = elementary charge = 1.60217646 × 10-19 coulombs
Kinetic energy Ke= 1/2 mv^2
U = Kq1q2/r
K = 9.0exp9
deltaKe + deltaU = 0


The Attempt at a Solution


(deltaKe1 + deltaU1) + (deltaKe2 + deltaU2) = 0 where 1 represents the u mass proton and 2 represents the alpha particle.
((0-.5u(.01c)^2) + (Ke*2e/r)) + ((0-2u(0.01c)^2) + (Ke^2e/r)) = 0
solve for r =2.45 *10^-14

the answer in the textbook is 1.93 * 10 ^-14

Iv done this over and over and over and am getting this 2.45 answer. please help!!!!! it is very frustrating! thanks!!!
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,832
249
Hi mrjoe2! :smile:
A proton and an alpha particle (q = +2e, m = 4u) are fired directly toward each other from far away, each with an initial speed of 0.010c. what is their distance of closes approach, as measured between their centers?

((0-.5u(.01c)^2) + (Ke*2e/r)) + ((0-2u(0.01c)^2) + (Ke^2e/r)) = 0
ah … but they're not stationary at closest approach, are they? :wink:
 
  • #3
rl.bhat
Homework Helper
4,433
5
If you consider a proton and an alpha particle as a system, at the closest approach the total kinetic energy of the system is converted to the potential energy.
Hence
.5u(.01c)^2) + + 2u(0.01c)^2) = (Ke*2e/r))
 
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