In Rutherford\'s scattering experiments, alpha particles (charge = +2 e ) were f

Question

In Rutherford's 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.

Find the distance of closest approach between the alpha particle and the gold nucleus for the case K= 2.6 MeV .

Express your answer using two significant figures.

Explanation / Answer

K = 2.6 MeV

1 eV = 1.6 x 10-19 J

K = 2.6 x 106 x 1.6 x 10-19 J

K = 4.16 x 10-13 J

d is the closest approach

charge on alpha particle q1 = 2e = 2 x 1.6 x 10-19 C

q1 = 3.2 x 10-19 C

Charge on gold foil q2 = 79e = 79 x 1.6 x 10-19 C

q2 = 126.4 x 10-19 C

initial Kinetic energy = final potential energy

4.16 x 10-13 = kq1q2 / d

4.16 x 10-13 = (9 x 109)(3.2 x 10-19)(126.4 x 10-19) / d

d =  (9 x 109)(3.2 x 10-19)(126.4 x 10-19) / 4.16 x 10-13

d = 8.8 x 10-14 m

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