Calculate the kinetic energy of the alpha particle released when Radium-226 unde

Question

Calculate the kinetic energy of the alpha particle released when Radium-226 undergoes alpha decay to become Radon-222. i.e. 226 88Ra decays to 222 86Rn + 4 2He. You may assume that the radium nucleus is at rest before the decay and that the radon nucleus is in its ground state. Begin by calculating the ratio of the kinetic energies of the alpha particle and the radon nucleus. Express your answer in MeV. a. 3.77 MeV b. 3.96 MeV c. 4.11 MeV d. 4.53 MeV e. 4.78 MeV Please show me how you get the correct answer. Thanks!

Explanation / Answer

the mass of the radium = 226.0254098 amu the mass of the radon = 222.0175777 amu the amss of the alpha particle = 4.0015061 amu The mass deffect ?m = mass Ra - (massRn+massHe) = 0.006326 amu the mass can converted into K.E of the alpha patricle. K.E (alpha) = 0.006326*931.5 MeV ˜ 5.89 MeV

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