Q7 of 106 Page 384

Find the energy liberated in the reaction

223Ra → 209Pb + 14C.


The atomic masses needed are as follows.


223Ra 209Pb 14C


223.018 u 208.981 u 14.003 u






MPb = mass of Pb209=208.981 u


MC = mass of C14 =14.003 u


MRa = mass of Ra223=223.018 u




The energy released is 31.671 MeV


More from this chapter

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3

Calculate the mass of an α-particle. Its binding energy is 28.2 MeV.

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Find the binding energy per nucleon of if its atomic mass is 196.96 u.

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32P beta-decays to 32S. Find the sum of the energy of the antineutrino and the kinetic energy of the β-particle. Neglect the recoil of the daughter nucleus. Atomic mass of 32P = 31.974 u and that of 32S = 31.972 u.

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Potassium -40 can decay in three modes. It can decay by β–-emission, β+-emission or electron capture.

(a) Write the equations showing the end products.


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