Problem 31.4 In Rutherfords scattering experiments, alpha particles (charge - +2

Question

Problem 31.4 In Rutherfords scattering experiments, alpha particles (charge - +2e) were fired at a gold foil. Consider an alpha particle with an initial kinetic energy K heading directly for the nucleus of a gold atom (charge +79e) The alpha particle will come to rest when all its initial kinetic energy has been converted to electrical potential energy Part A Find the distance of closest approach between the alpha particle and the gold nucleus for the case K- 35 MeV Express your answer using two significant figures. In Submit Provide Feedback

Explanation / Answer

By using energy conservation,

d(K.E.) = d(P.E.)

K - 0 = U - 0

K = U = k*q1*q2/d

here, K = 3.5 MeV = 3.5*10^6*1.6*10^-19 J = 5.6*10^-13 J

q1 = 2e = charge of alpha atom

q2 = 79e = charge of nucleus of gold atom

d = distance of closest approach = ??

k = 9*10^9 N.m^2/C^2

So, 5.6*10^-13 = 9*10^9*2*e*79*e/d

d =9*10^9*2*79*(1.6*10^-19)^2/(5.6*10^-13) = 6.5*10^-14 = 65 fm

d = 65 fm

please upvote.

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