Alpha particles of kinetic energy 470 MeV are scattered at 9

Alpha particles of kinetic energy 4.70 MeV are scattered at 90° by a gold foil. (a) What is the impact parameter?
[Correct: 24.2] fm

(b) What is the minimum distance between alpha particles and gold nucleus?
[Correct: 58.2] fm

(c) Find the kinetic and potential energies at that minimum distance.

U= ______MeV
K= _______MeV

(IT IS NOT 0 & 4.70 OR VICE VERSA)

Solution

C) minimum distance is r = 58.2 fm = 58.2 x 10^-15 m

Potential energy = kq1q2 / r

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

charge on Gold nucleus = 79 x 1.6 x 10^-19 C

PE = (9 x 10^9 x 2 x 1.6 x 10^-19 x 79 x 1.6 x 10^-19 ) / (58.2 x 10^-15)

PE = 6.255 x 10^-13 J

PE (in eV ) = 6.255 x 10^-13 / (1.6 x 10^-19)

= 3.91 x 10^6 eV = 3.91 MeV

So U = 3.91 MeV

Using energy conservation,

Ui + Ki = Uf + Kf

0 + 4.70 = 3.91 + K

K= 0.79 MeV