Question

In separate experiments, four different particles each start from far away with the same speed and impinge directly on a gold nucleus. The masses and charges of the particles are particle 1: mass m0, charge q0 particle 2: mass 2m0, charge 2q0 particle 3: mass 2m0, charge q0/2 particle 4: mass m0/2, charge 2q0 Rank the particles according to the distance of closest approach to the gold nucleus, from smallest to largest.
A. 1, 2, 3, 4
B. 4, 3, 2,1
C. 3, 1 and 2 tie, then 4
D. 4, 1 and 2 tie, then 1

Answer 1

Answer:

Potential energy at the beginning: Ui =Qq/R*k ≈ 0 (particles are far apart)

Kinetic energy at the beginning: KEi =$mv^{2}$/2

Potential energy at the end: Uf =(Qq/r)*k

Kinetic energy at the beginning: KEf = 0

From energy conservation:

r = 2*Qk/$v^{2}$*q/m

Ranking of the closest approach to the gold nucleous comes from q/m:

- particle 3 has the smallest q/m = 1/4 ⋅q0/m0  and, hence, comes the closest

- particle 4 has the largest q/m = 4⋅q0/m0  and, hence, comes the last in ranking

- particles 1 and 2 have equal q/m=q0/m0 and, hence, come tie in between particles 3 and 4

Explanation:

Potential energy at the beginning: Ui =Qq/R*k ≈ 0 (particles are far apart)

Kinetic energy at the beginning: KEi =$mv^{2}$/2

Potential energy at the end: Uf =(Qq/r)*k

Kinetic energy at the beginning: KEf = 0

From energy conservation:

r = 2*Qk/$v^{2}$*q/m

Ranking of the closest approach to the gold nucleous comes from q/m:

- particle 3 has the smallest q/m = 1/4 ⋅q0/m0  and, hence, comes the closest

- particle 4 has the largest q/m = 4⋅q0/m0  and, hence, comes the last in ranking

- particles 1 and 2 have equal q/m=q0/m0 and, hence, come tie in between particles 3 and 4