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A proton is located at the origin, and a second proton is located on the x-axis

ID: 1580990 • Letter: A

Question

A proton is located at the origin, and a second proton is located on the x-axis at x -6.40 fm (1 fm - 10-15 m) (a) Calculate the electric potential energy associated with this configuration 359-14 2e, mass 6.64 x 10-27 kg) is now placed at (x, y)-(3.20, 3.20) fm. Calculate the electric potential energy associated with this configuration (b) An alpha particle (charge 23.95e-14 (c) Starting with the three particle system, find the change in electric potential energy if the alpha particle is allowed to escape to infinity while the two protons remain fixed in place. (Throughout, neglect any radiation effects.) 1.677e-14X Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. (d) Use conservation of energy to calculate the speed of the alpha particle at infinity 7.106e6 X Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. m/s (e) If the two protons are released from rest and the alpha particle remains fixed, calculate the speed of the protons at infinity. m/s

Explanation / Answer

(c)If alpha particle is allowed to escape to infinity it will not contribute anything to the total p.e.
So the change in potential will be the difference between result of (a) and (b).

20.36*10^-14 J

(d) Change of electrostatic energy (answer of (c)) = 1/2 * mass of alpha * v^2
All are known except v, put the values.
(20.36*10^-14 ) = 1/2 * 6.64*10^-27 * v^2

v^2 = 6.13*10^13

v = 7.83*10^6 m/s

(e) total potential energy of the system (answer of (b)) = 2 * 1/2 * mass of proton * v^2

(23.95*10^-14) = 2 * 1/2 * 1.6726219 × 10-27* v^2

v^2 = 1.43*10^14

v = 1.2*10^7 m/s

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