Question

Most of the energy released in nuclear fission is electrostatic. Suppose a uranium-236 nucleus (92 protons and 236-92=144 neutrons), which is initially at rest, splits into two identical “daughter” nuclei each having 46 protons and 72 neutrons. (Protons and neutrons have the same mass, namely 1.67×10-27 kg.)(a) [15 pts.] Initially the two daughter nuclei are separated by 13×10-15 m. How fast will they be moving when they get very far apart?(b) [5 pts.] Fission products usually are not the same size, so when they get far apart they will have different speeds. To solve for those two speeds you would need two equations, and those two equations would come from two conservation laws. What are the two quantities that are conserved in this situation, that would give you the two equations you would need?

Answer 1

a)
let M is the mass of the Uranium nucleus.

mass of daughter nuclie, m = M/2

= 118*1.67*10^-27 kg

d = 13*10^-15 m

charge of each nuclei, q = 46*e

let v is the speed of each daughter nuclei when they move far away

Apply conservation of energy

final kinetic energy of daughter nuclei = initial electrostatic potential energy

2*(1/2)*m*v^2 = k*q^2/d

m*v^2 = k*q^2/d

v^2 = k*q^2/(m*d)

v = q*sqrt(k/(m*d))

= 46*1.6*10^-19*sqrt(9*10^9/(118*1.67*10^-27*13*10^-15))

= 1.38*10^7 m/s <<<<<<<-------------Answer

b) let m1 and m2 are masses of daughter nuclei

let q1 and q2 are the charges of daughter nuclei

let v1 and v2 are the velocities of duaghter nuclei.

Apply conservation of momentum

final momentum = initial momentum

m1*v1 + m2*v2 = 0 -----------------(1)

Apply conservation of energy

final kinetic energy = initial electrostatic potential energy

(1/2)*m1*v1^2 + (1/2)*m2*v2^2 = k*q1*q2/d -----------(2)

The two quantities which are conserevd are momentum and energy.