Let’s consider what we learned in the last lecture. 238U 234Th + 4He + QConservation of energy (and equivalence with mass) yields: Q= Δ(238U) – (Δ(234Th) + Δ(4He)) Q= (47.305) – (40.610) – (2.425) = 4.27 MeV Q> 0. What does this tell us ? 238U 237U+ nQn = Δ(238U) – (Δ(237U) + Δ(n)) Qn = (47.305) – (45.385) – (8.071) = -6.151 MeV Qn < 0. What does this tell us ? 238U 237Pa + 1H Qp = (47.305) – (47.6) – (7.289) = -7.584 MeV Qp < 0. What does this tell us ?
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Facts from the last lecture: Q> 0 for all A > 140 nuclei; They range from 1.5 to 12 MeVAlpha decay half-lives range from 1016y > t 1/2> 10-21sWHY ?Draw a picture to explain why Qfalls in such a small range while the range for t1/2is so large. Clearly label all important aspects of your picture. Then explain your picture in a short paragraph. In explaining your picture you should reference the concepts of thermodynamic stability and kinetic stability and make use of the discussion in class about Q values.