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 3.40 10-15 m apart? Your response differs from the correct answer by more than 10%. Double check your calculations. 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) The force between the two alpha particles can be given as

F = 4k( q² / r² )

Here, k is an coulombs force constant = 8.988 * 10^-9 N

r = diatance between the alpha particles = 3.4 * 10^-15 m ( given )

q = proton charge = 1.602 * 10^-19 C

Now, from the above formula 'F' can be calculated as

F = 4 * ( 8.988 * 10^-9 ) * ( (1.602 * 10^-19)²) / ( ( 3.4 * 10^-15 )² )

F = ( 9.2267 * 10^-46 ) / ( 1.156 * 10^-29 )

=> F = 7.9816 * 10^-17 N

b) We know that,

Force, F = mass ( m ) * acceleration ( a )

Given that,

the mass of alpha particle m = 4.0026 u

Now, the relation between the atomic mass units and kilogram is given as

m_kg = m_u * 1.66053886 * 10^-27

That means, 1 u = 1.66053886 * 10^-27 kg

Then, the mass of alpha particle, m = 4.0026 * 1.66053886 * 10^-27 kg

=> m = 6.6464 * 10^-27 kg

Here,

from the above formula force and mass are known then the acceleration can be calculated as

acceleration of alpha particles, a = F / m

=> a = ( 7.9816 * 10^-17 ) / ( 6.6464 * 10^-27 )

=> a = 1.2 * 10¹⁰ m/s²