A proton with an initial speed of 4.7 times 10^6 m/s is projected straight at th
ID: 1600539 • Letter: A
Question
A proton with an initial speed of 4.7 times 10^6 m/s is projected straight at the nucleus of a gold atom. Neglecting the motion of the gold nucleus, what is the closes the proton will get to the nucleus before being repelled? a. A proton is made out of three quarks. Two up quarks, each with a charge +2/3 e, and one down quark with a charge -1/3 e. (Total charge +2/3 e + 2/3 e - 1/3 e = +e). Assuming the charges are at the comers of an equilateral triangle with sides of 1 times 10^-15 m (= 1 fm, also called a Fermi), calculate the total electric potential energy of these three quarks. b. The neutron is made out of two down quarks and one up quark, total charge -1/3 e - 1/3 e + 2/3 e = 0. Calculate the total electric potential energy of these three quarks.Explanation / Answer
6) we know, atomic number of proton, z = 79
let
Q = 79e = 79*1.6*10^-19 C
q = 1.6*10^-19 C
let d is the closest approch.
Apply conservation of energy
initial kinetic energy of proton = final potential energy of proton and gold nucleus
(1/2)*m*v^2 = k*Q*q/d
d = 2*k*Q*q/(m*v^2)
= 2*9*10^9*79*1.6*10^-19*1.6*10^-19/(9.1*10^-31*(4.7*10^6)^2)
= 1.81*10^-9 m
7)
a) let q1 = q2 = +2*e/3, q3 = -e/3
d = 1*10^-15 m
total potential energy
U = k*q1*q2/d + k*q2*q3/d + k*q3*q1/d
= (k/d)*(4*e^2/9 - 2*e^2/9 - 2*e^2/9)
= 0
b)
let q1 = q2 = -e/3, q3 = +2e/3
d = 1*10^-15 m
total potential energy
U = k*q1*q2/d + k*q2*q3/d + k*q3*q1/d
= (k/d)*(e^2/9 - 2*e^2/9 - 2*e^2/9)
= (k/d)*(-e^2/3)
= (9*10^9/(1*10^-15))*(-(1.6*10^-19)^2/3)
= -7.68*10^-14 J
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