1118_Physics ProblemsTechnical Physics

# 1118_Physics ProblemsTechnical Physics - Chapter 39 P39.75...

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Chapter 39 459 P39.75 Since the total momentum is zero before decay, it is necessary that after the decay pp E cc nucleus photon keV == = γ 14 0 . . Also, for the recoiling nucleus, Ep c m c 22 2 2 2 =+ ej with mc 29 860 10 538 = .. J G e V . Thus, mc K mc 2 2 2 2 2 14 0 += + af . keV or 1 14 0 1 2 2 2 2 + F H G I K J = F H G I K J + K mc mc . k eV . So 11 14 0 1 1 2 14 0 2 2 2 + F H G I K J ≈+ F H G I K J K mc mc mc keV keV (Binomial Theorem) and K mc ≈= = × × 14 0 2 14 0 2538 10 182 10 2 2 2 9 3 . . . . keV 10 eV eV eV 3 . P39.76 Take m = 100 . k g . The classical kinetic energy is Km um c u c u c c F H G I K J F H G I K J 1 2 1 2 450 10 2 16 2 J and the actual kinetic energy is K uc mc r = F H G G I K J J F H G G I K J J 1 1 1 9 00 10 1 1 1 2 21 6 2 bg J . u c KK cr JJ 0.000 0.000 0.000 0.100 0.045 0.0453 0.200 0.180 0.186 0.300 0.405 0.435 0.400 0.720 0.820 0.500 1.13 1.39 0.600 1.62 2.25 0.700 2.21 3.60 0.800 2.88 6.00
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