Question

The nucleus of 8Be, which consists of 4 protons and 4 neutrons, is very unstable and spontaneously breaks into two alpha particles (helium nuclei, each consisting of 2 protons and 2 neutrons).

(a) What is the force between the two alpha particles when they are 6.60 ✕ 10−15 m apart? N
(b) What is the initial magnitude of the acceleration of the alpha particles due to this force? Note that the mass of an alpha particle is 4.0026 u. m/s2

Answer 1

a) F = 21.16 N, b) a = 3.17 10²⁸ m / s

Step-by-step explanation:

a) The outside between the alpha particles is the electric force, given by Coulomb's law

F =
`k ( q_1 q_2)/(r^2)`

in that case the two charges are of equal magnitude

q₁ = q₂ = 2q

let's calculate

F =
`9 \ 10^9 \ ( (2 \ 1.6 \ 10^(-19) )^2 )/( (6.60 \ 10^(-15) )^2 )`

F = 21.16 N

this force is repulsive because the charges are of the same sign

b) what is the initial acceleration

F = ma

a = F / m

a =
`(21.16)/(4.0026 \ 1.67 \ 10^(-27) )`21.16 / 4.0025 1.67 10-27

a = 3.17 10²⁸ m / s

this acceleration is in the direction of moving away the alpha particles