(f) During each collision, the energy of the neutron is reduced by the factor 1 – 0.89 = 0.11. If Eiis the initial energy, then the energy after ncollisions is given by E= (0.11)nEi. We take the natural logarithm of both sides and solve for n. The result is ln( /)ln(0.025 eV/1.00 eV)7.98.ln 0.11ln 0.11iEEn===≈The energy first falls below 0.025 eV on the eighth collision. 23. (a) Let vnibe the initial velocity of the neutron, vnfbe its final velocity, and vfbe the final velocity of the target nucleus. Then, since the target nucleus is initially at rest, conservation of momentum yields mnvni= mnvnf+ mvfand conservation of energy yields 122122122mvnniff=+.We solve these two equations simultaneously for vf. This can be done, for example, by using the conservation of momentum equation to obtain an expression for vnfin terms of vfand substituting the expression into the conservation of energy equation. We solve the resulting equation for
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