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

Question

A proton is located at the origin, and a second proton is located on the x axis at x1 = 5.66 fm (1 fm = 10-15 m). (a) Calculate the electric potential energy associated with this configuration. 4.06 e 14 ) (b) An alpha particle (charge = 2e, mass = 6.64 x 10-27 kg) is now placed at (x2, y2) = (2.83, 2.83) fm. Calculate the electric potential energy associated with this configuration. 2.280-13 Your response differs from the correct answer by more than 10%. Double check your calculations.) (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.) -1.870-13 Your response differs from the correct answer by more than 10%. Double check your calculations.) (d) Use conservation of energy to calculate the speed of the alpha particle at infinity. 17.5206 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. 1.16e7 Your response is within 10% of the correct value. This may be due to rouncloff 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

Explanation / Answer

The electric potential energy can be calculated as UE = K Qq/r

K=9*109 which is the coulumbs constant

b)The potential energy of the system of charges can be calculated as U=K(q1q2/r12 + q2q3/r23+q1q3/r13)

= 9*109(1.602*10-19* 1.602*10-19/5.66 +1.602*10-19*2*1.602*10-19/4.0022 + 1.602*10-19*2*1.602*10-19/4.022)1015

= 2.704*10-13

c)The work done to move the alpha particle to infinity is QV

=2*2.704*10-13 =5.408*10-13

Therfore change in potential energy is 5.408*10-13-2.704*10-13 =2.704*10-13

d) Potential energy at infinity is equal to kinectic energy of alpha particle

kinectic energy =1/2mv2 = 2.704*10-13

v2=2.704*10-13*2/6.64*10-27

v= 0.902*107 =9.02*106m/s

e)Speed of protons

v2= 2.704*10-13*2/1.672*10-27

v = 17.98*106m/s

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