Problem set 8 Crib 2011

# Problem set 8 Crib 2011 - 7.0581346 7.01600 = .0421346 amu....

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CHM 24100 Spring 2011 Practice Problem Set Crib A. Exactly half of the nuclides in the grouping below are unstable. 64 Fe 64 Co 64 Ni 64 Cu 64 Zn 64 Ga 1) Identify those most likely to decay, and explain your reasoning. 64 Co, 64 Cu, and 64 Ga because Z and N are odd 2) One member of the group decays strictly by electron capture. Predict which and explain your reasoning. Write a balanced equation for the electron-capture process in question. 64 27 Co + 0 1 -1 e 0 64 26 Fe + νe Process converts a proton into a neutron and is favored when N or N/Z is too low. B. Calculate the BEPN (binding energy per nucleon) of 7 Li if it has a mass of 7.01600 amu. ( 1 1 H mass = 1.007825 amu; 1 0 n mass = 1.0086649 amu) First, must find the mass deficit in the nucleus. There are 3 protons and 4 neutrons in the nucleus. 3 (1.007825)+ 4 (1.0086649) 7.0581346 amu mass deficit = mass of nucleons - Li atomic mass
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Unformatted text preview: 7.0581346 7.01600 = .0421346 amu. E= m(931.49)/total number of nucleons = (.0421346)(931.49)/7 = 5.606 MeV/nucleon C. Xenon has a number of stable isotopes; in fact, only two in the following string are radioactive. One decays by electron capture with a half life of 36.4 d, and the other decays by electron (beta) emission with a half life of 5.2 d. Identify the two isotopes from the list most likely to decay, predict the pathway of decay for each, and write a balanced equation. Briefly, explain your reasoning. 126 Xe 127 Xe 128 Xe 129 Xe 130 Xe 131 Xe 132 Xe 133 Xe 134 Xe Expect highest and lowest N/Z ratios with N=odd number to be radioactive. Therefore, 127 and 133 are radioactive. -1 e + 127 Xe 127 I + 133 Xe 133 Cs + -1 e +...
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