(a) What is the energy release in a decay of 218Po, given the mass of 218Po is 2
ID: 2066504 • Letter: #
Question
(a) What is the energy release in a decay of 218Po, given the mass of 218Po is 218.008965 u? 84 84(b) This isotope can also decay by ß- emission. What is the energy release in this case? What is the maximum kinetic energy of the emitted ß- particle?
Explanation / Answer
(a)1 po-218 ----> 1 Pb-214 + 1 He-4 + mass lost mass lost = 218.008965 - 213.9997981 - 4.002604 = 0.006563 amu 0.006563 amu x (1.660x10^-27 kg / 1 amu) = 10.895x10^-30 kg E = mc^2 = 10.895x10^-30 kg x (2.998x10^8 m/s)^2 = 9.792x10^-13 kJ / atom (b)1 po-218 ----> 1 At-218 + -1 e-0 + mass lost mass lost = 218.008965 - 210.000000 - 0.0005486 = 8.008416 amu 8.008416amu x (1.660x10^-27 kg / 1 amu) = 13.294x10^-27 kg E = mc^2 = 13.294x10^-27 kg x (2.998x10^8 m/s)^2 = 1.195x10^-9 kJ / atom You can calculate the total mass of the nuclei before and after the beta emission, which will give you (by subtraction) the total energy available. But a half MEV of this is the rest mass of the beta particle, which can't appear as part of its kinetic energy -- so you have to subtract that off.
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