ch43-p023

# ch43-p023 - 23(a Let vni be the initial velocity of the...

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(f) During each collision, the energy of the neutron is reduced by the factor 1 – 0.89 = 0.11. If E i is the initial energy, then the energy after n collisions is given by E = (0.11) n E i . We take the natural logarithm of both sides and solve for n . The result is ln( / ) ln(0.025 eV/1.00 eV) 7.9 8. ln 0.11 ln 0.11 i EE n == = The energy first falls below 0.025 eV on the eighth collision. 23. (a) Let v ni be the initial velocity of the neutron, v nf be its final velocity, and v f be the final velocity of the target nucleus. Then, since the target nucleus is initially at rest, conservation of momentum yields m n v ni = m n v nf + mv f and conservation of energy yields 1 2 2 1 2 2 1 2 2 mv nn i f f =+ . We solve these two equations simultaneously for v f . This can be done, for example, by using the conservation of momentum equation to obtain an expression for v nf in terms of v f and substituting the expression into the conservation of energy equation. We solve the resulting equation for
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## This note was uploaded on 05/19/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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