Lecture11RadioactiveDecayModes-Betadecay_001

Lecture11RadioactiveDecayModes-Betadecay_001 - Let's...

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Let’s consider what we learned in the last lecture. 238 U 234 Th + 4 He + Q Conservation of energy (and equivalence with mass) yields: Q = Δ ( 238 U) – ( Δ ( 234 Th) + Δ ( 4 He)) Q = (47.305) – (40.610) – (2.425) = 4.27 MeV Q > 0. What does this tell us ? 238 U 237 U + n Q n = Δ ( 238 U) – ( Δ ( 237 U) + Δ (n)) Q n = (47.305) – (45.385) – (8.071) = -6.151 MeV Q n < 0. What does this tell us ? 238 U 237 Pa + 1 H Q p = (47.305) – (47.6) – (7.289) = -7.584 MeV Q p < 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 MeV Alpha decay half-lives range from 10 16 y > t 1/2 > 10 -21 s WHY ? Draw a picture to explain why Q falls in such a small range while the range for t 1/2 is 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.
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Beta-Decay: Neutron-Proton Transformations Connects Isobars Z A = most probable charge NEUTRON EXCESS Negatron Decay n 1 H + + PROTON EXCESS Positron/electron capture Decay 1 H 1 n + + + 1 H + e 1 n +
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