11. lodine decays with the emission of beta particles according to the equation

Question

11. lodine decays with the emission of beta particles according to the equation IJ'153 X + oa, + v-(neutrino, no mass) Find the nucleons and charge for X. 12. Calculate the N/Z ratio for the "Cs atom. A. 1. 567 B. 2.345 1.333 D. 1.666E. 1.886 13. Calculate is the value of "atomic mass unit" 2 1.660540 x 107 B. 1.660540 x 1024. g C. 1.660540 x 1024 kg D. 1.87960 x 1025 gE 1 689609 x 1025 g. 14. Calculate the total binding energy in kJ per mole of Calcium 40Ca2o. A. 1.450 x 1010 kJ mole1 B3.890x 1010 kJ mole-1 C. 3.300 x 1010 kJ mo D. 1.686 x 1010 kJ mole E. 2.646 x 1010 kJ mole-1 15. Calculate the total binding energy in MeV per atom of Rubidium 87Rbn A. 342.1. MeV B. 8.711. MeV C. 1801.7. MeV D 757.9. MeV E. 665.2. MeV 16. Calculate the binding energy in kJ per mole of nucleons in Bes atom. 5.44. x 10° kJ/ mol nucleons C) 6.06 x 1010 kJ/mol nucleons E. 5.36x 10° kJ/mol nucleons B. 6.24 x 103 kJ /mol nucleons D. 3.46 x 10 kJ/mol nucleons

Explanation / Answer

11. Option C is correct.

131-I-53 ----> 131-Xe-54 + Beta particle + Neutrino

12. Option C is correct.

N = 8

Z = 6

N/Z = 1.33

13. Option B is correct.

1 amu = 1.66054 X 10^-24 g

14. Option D is correct.

Calculate the binding energy in KJ per mole of Calcium 40Ca20

Neutrons = 40-20 = 20

20 X 1.007825 = 20.1565

20 * 1.008665 = 20.1733

20.1733 + 20.1565 = 40.3298

MW = 39.963

TOTAL BINDING ENERGY = 342.051941 Mev = 1.60218 X 10^-16

BINDING ENERGY PER NUCLEON = 8.551298525 Mev

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