In Rutherford’s famous scattering experiments (which led to the planetary model

Question

In Rutherford’s famous scattering experiments (which led to the planetary model of the atom), alpha particles (having charges of +2 e and masses of 6.64×1027 kg) were fired toward a gold nucleus with charge +79 e. An alpha particle, initially very far from the gold nucleus, is fired at 1.61 × 107 m/s directly toward the gold nucleus.

How close does the alpha particle get to the gold nucleus before turning around? Assume the gold nucleus remains stationary. The fundamental charge is 1.602 × 1019 C and the Coulomb constant is 8.98755 × 109 N · m2 /C 2 . Answer in units of m.

Explanation / Answer

Here, velocity of the alfa - particle, v = 1.61 x 10^7 m/s.

Now -

KE of the approaching alpha = ½mv² = ½(6.64*10^-27)(1.61 x 10^7)² = 9.26 x 10^-13 J

Again, when the alpha comes to rest (about to be turned) when all it's KE is transferred into electrical potential energy (due to the electric field created by gold nucleus)

Elec. pot. energy = elec potential (V as J/C) x Q2(C) =1/4(Q1)/R x Q2 .. in J
Q1 = gold charge =79e Coulomb
Q2 = alpha charge = 2e Coulomb
e = 1.6*10^-19 C
1/4 = 9*10^9Nm²/C²
R = separation of the two (point) charges (m)

E.pot.energy = (9*10^9)(79e*2e) / R = (9*10^9)*158(1.6*10^-19)² /R = (3.64*10^-26) /R

Now, put the values from above -

9.26 x 10^-13 J = (3.64x10^-26) /R

R = (3.64x10^-26) / (9.26 x 10^-13) = 3.93 x 10^-14 m

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