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

Answer:

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

Explanation:

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

          F = $k \frac{ q_1 q_2}{r^2}$

in that case the two charges are of equal magnitude

          q₁ = q₂ = 2q

let's calculate

         F = $9 \ 10^9 \ \frac{ (2 \ 1.6 \ 10^{-19} )^2 }{ (6.60 \ 10^{-15} )^2 }$

         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 = $\frac{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