Through what voltage must an αα-particle, with its charge of +2e+2e, be accelerated so that it has just enough energy to reach a distance of 1.5×10−14 mm from the surface of a gold nucleus? (Assume the gold nucleus remains stationary and can be treated as a point charge.)

5101.35v

Explanation:

Radius of gold nucleus=7.3×10-15m and a charge of +79e

Q= 79e

e=1.6×10^-19

q= +2e

The nucleus is considered as the point charge where the potential energy between the charges are

U = 1/(4×3.142×Eo)×(qQ)/r

Where r is distance between the charges and the nucleus

r=R+d

V=U/q

U= 1/(4×3.142×Eo)×Q/r

V= 1/(4×3.142×Eo)×Q/(r+d)

9.0×10^9 ×(79×10^-19)/(7.3×10^-15)+(1.5×10^-14)

V= 9.0×10^9 ×(1.264×10^-17)/(2.23×10^-14)

V= 9×10^9×(5.67×10^-14)= 5101.35v

a.) travels faster in solids because the particles are closer together.

air is a gas water is a liquid

the answer is a. hope this .