p44_027

# p44_027 - 27(a Let vni be the initial velocity of the...

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Unformatted text preview: 27. (a) Let vni be the initial velocity of the neutron, vnf be its ﬁnal velocity, and vf be the ﬁnal velocity of the target nucleus. Then, since the target nucleus is initially at rest, conservation of momentum 2 2 2 yields mn vni = mn vnf + mvf and conservation of energy yields 1 mn vni = 1 mn vnf + 1 mvf . We 2 2 2 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 vnf in terms of vf and substituting the expression into the conservation of energy equation. We solve the resulting equation for vf . We obtain vf = 2mn vni /(m + mn ). The energy lost by the neutron is the same as the energy gained by the target nucleus, so 1 1 4m 2 m n 2 ∆K = mvf = v2 . 2 2 (m + mn )2 ni 2 The initial kinetic energy of the neutron is K = 1 mn vni , so 2 4m n m ∆K . = K (m + m n )2 (b) The mass of a neutron is 1.0 u and the mass of a hydrogen atom is also 1.0 u. (Atomic masses can be found in Appendix G.) Thus, ∆K 4(1.0 u)(1.0 u) = 1. 0 . = K (1.0 u + 1.0 u)2 Similarly, the mass of a deuterium atom is 2.0 u, so (∆K )/K = 4(1.0 u)(2.0 u)/(2.0 u+1.0 u)2 = 0.89. The mass of a carbon atom is 12 u, so (∆K )/K = 4(1.0 u)(12 u)/(12 u + 1.0 u)2 = 0.28. The mass of a lead atom is 207 u, so (∆K )/K = 4(1.0 u)(207 u)/(207 u + 1.0 u)2 = 0.019. (c) During each collision, the energy of the neutron is reduced by the factor 1 − 0.89 = 0.11. If Ei is the initial energy, then the energy after n collisions is given by E = (0.11)n Ei . We take the natural logarithm of both sides and solve for n. The result is n= ln(E/Ei ) ln(0.025 eV/1.00 eV) = = 7. 9 . ln 0.11 ln 0.11 The energy ﬁrst falls below 0.025 eV on the eighth collision. ...
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## This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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