need help with this question pleas 2-8. In 1906, in the course of a historic exp

Question

need help with this question pleas

2-8. In 1906, in the course of a historic experiment which demonstrated the small size of the atomic nucleus, Rutherford observed that an alpha particle (Q = 2 X 1.6 X 10-19 coulomb) with a kinetic energy of 7.68 × 106 electron-volts (7.68 × 100 X 1.6 X 10-19 joule) making a head-on collision with a gold nucleus (Q-79 1.6 X 10-19 coulomb) is repelled. (a) What is the distance of closest approach at which the electrostatic poten- tial energy is equal to the initial kinetic energy? Express your result in fem- tometers (10-15 meter). (b) What is the maximum force of repulsion? (c) What is the maximum acceleration in g's? The mass of the alpha particle is about 4 times that of a proton, or 4 X 1.7 X 10-27 kilogram

Explanation / Answer

A)

All the KE is converted to PE stored in the electric field.

So, 0.5*mv^2 = k*q1*q2/r

So, 7.68*10^6*1.6*10^-19 = 9*10^9*(3.2*10^-19)*(79*1.6*10^-19)/r^2

So, r = 2.96*10^-14 m = 29.6 fm <------ closest approach

B)

Max force of repulsion = k*q1*q2/r^2

= 9*10^9*(3.2*10^-19)*(79*1.6*10^-19)/(2.96*10^-14)^2

= 41.55 N

C)

Max acceleration = F/m

= 41.55/(4*1.7*10^-27)

= 6.11*10^27 m/s2

= 6.11*10^27/9.81

= 6.22*10^26 g <------ in terms of g

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