he 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

A) F = 21.134 N

B) a = 3180.76 × 10^(24) m/s²

Explanation:

A) We are given;

Mass of alpha particle; m = 4.0026 u

Now, 1u = 1.66 × 10^(-27) kg

Thus; m = 4.0026 × 1.66 × 10^(-27)

Distance apart; r = 6.60 × 10^(−15) m

Charge on the alpha particle is;

q = 2e = 2 × 1.6 × 10^(-19) C

Formula for the force between the two alpha particles is;

F = kq1.q2/r²

k = 8.99 × 10^(9) N.m²/C²

q1 = q2 = 2 × 1.6 × 10^(-19) C

F = 8.99 × 10^(9) × (2 × 1.6 × 10^(-19))²/(6.60 × 10^(−15))²

F = 21.134 N

B) acceleration is given by;

a = F/m

Thus; a = 21.134/(4.0026 × 1.66 × 10^(-27))

a = 3180.76 × 10^(24) m/s²

the working equation for the angular size is the product of the physical size and 360 degrees divided by the product of two pi and distance from earth. the minimum size would be 0.559 degrees while the maximum size would be 0.49 degrees.