The iodine isotope^123_53 I decays by electron capture (during the decay, gamma

Question

The iodine isotope^123_53 I decays by electron capture (during the decay, gamma rays are emitted, and these are detected in the scan), and is used in some thyroid scans. Suppose an iodine-123 sample used in a medical procedure has an initial mass of 0.00000390 mu g. Write down the equation describing the decay, and calculate the mass defect and the decay energy. What will be the activity of this sample in 24 hours? A patient it injected with the iodine sample (at t = 0), and then a scan it performed beginning 1.00 hour later. If the scan lasts for 30 minutes, how much energy was released by the decays during the scan?

Explanation / Answer

4. 123 I 53 + e- = 123 Te 52 + gamma ray
a) Mass defect = Mass of reactants - Mass of products = 122.91*1.6605*10^-27 + 9.1*10^-31 - 122.90427*1.6605*10^-27 kg = 1.0424*10^-29 kg (per reaction)
0.00000390*10^-9 kg Iodene = 3.1730*10^-14 moles of iodene
Mass defect in this much iodene = 3.1730*10^-14 *6.022*10^23 * 1.0424*10^-29 = 1.99183*10^-19 kg
Decay energy = dm*c^2 = 0.017926 J

b) Activity of sample in 24 hours = ?
Activity = dN/dt
N = Noe^(-kt)
k = ln(2)/half-life
dN/dt = -No*k*e^(-kt)
now, half-life = 13.22*60*60 = 47592s
k = 1.4564*10^-5
activity = -79068.93548 reactions / s

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