Question

a) Calculate the electric force Fe between a stationary proton and uranium nucleus...separated by a distance of 3.6 nm: (consider the uranium a point charge)

b) is this force repulsive or attractive? Why?

c) determine the intial acceleration of the proton:

Answer 1

a)

Q_p = charge on proton = 1.6 x 10^-19 C

Q_n = charge on uranium nucleus = 92 protons = 92 x 1.6 x 10^-19 C = 147.2 x 10^-19 C

r = distance = 3.6 nm = 3.6 x 10^-9 m

Electric force is given as ::

$F_{e} = \frac{kQ_{p}Q_{n}}{r^2}$

F_e = (9 x 10⁹) (1.6 x 10^-19) (147.2 x 10^-19) / (3.6 x 10^-9)²

F_e = 1.64 x 10^-9 N

b)

the force is repulsive since charge possesed by both the proton and the uranium nucleus is Positive. and we know that like charges repul each other.

c)

mass of proton = m = 1.67 x 10^-27 kg

acceleration is given as ::

a = F_e/m

a = (1.64 x 10^-9) / (1.67 x 10^-27 )

a = 9.82 x 10¹⁷ m/s²