plrated from each 8. Model a helium-3 nucleus as three hard solid spheres of cha

Question

plrated from each 8. Model a helium-3 nucleus as three hard solid spheres of charge at the three co corners ot an equilateral triangle with side length 1 fm. Two are protons, and one is a neutron. protèn (a) If the residual strong force between each pair of nucleons is the same (regardless of whether it is a proton or neutron), then what is the minimum required residual strong force to hold this nucleus together? nucleus breaks into a proton and a deuteron (proton-neutron pair), how fast will the proton travel when it has traveled a distance of 1 micron (1 um on the order (b) If suddenly the strong force that holds one of the protons fails, so that the of the size of a cell)?

Explanation / Answer

a) The maximum repulsive force in this case is going to be proron proton repulsion

so distance between centres of protons , r = 2fm = 2*10^-15 m

so Force, F = ke^2/r^2 = 8.98*10^9*(1.6*10^-19)^2/(2*10^-15)^2 = 57.472 N

this is the minimum required residual force to hold nucleus together

b) if the residual force fails

then, initial Potential energy = ke^2/r [ e = 1.6*10^-19 C, charge on proton]

final Potential energy = ke^2/d

where d = 10^-6 m

so, PE lost = ke^2[1/r - 1/d] = gain in KE = 0.5mv^2 + 0.5*Mu^2[ m is mass of proton = 1.6*10^-27 KG, M is mass deutron = 2*1.6*10^-27 kg, v is proton velocity, u is deutron velocity]

from conservation of momentum

Mu = mv

u = 1.6*10^-27 *v/2*1.6*10^-27 = v/2

8.98*10^9*(1.6*10^-19)^2[10^15/2 - 10^6] = 0.5*1.6*10^-27*v^2 + 0.5*2*1.6*10^-27*v^2/4

v = 9.787*10^6 m/s

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