Homework SourseThe radioactive decay of carbon11 is modelled into an equati
The radioactive decay of carbon-11 is modelled into an equation:
1. Determine the half life or carbon-11, where it is the amount of time it takes for the amount to decayto half of its initial mass.
2. The researchers of this experiment wanted there to be 11 picograms of carbon-11 in the patient\'d body 10 minutes after injection. Using the decay rate, how much should they inject at time 0?
3. How long will it take for the amount left in the body to be 12.5% of what was injected? Assuming that radioactive decay is the only way of removing carbon-11 from the body.
2tSolution
1. When carbon 11 decaya to half of its initial mass, we have N(t) = 1/2 (N0) so that ( 1/2 ) N0 = N0e-t . Also, = 0.03428792, so that (1/2)N0 = N0e-0.03428792t or, 1/2 = e-0.03428792t . On taking natural logarithms of both the sides, we have ln(1/2) = - 0.03428792t ( as ln e = 1). or, - 0.69314718 = - 0.3428792t so that t = 0.69314718 / 0.03428792 = 20.21549years = 20 years, 2 months, 18 days approximately.
2. . Let the initial quantity at t = 0 be N0 picogram. Here, t = 10 minutes . We know that 1 year = 365*24 *60 minutes = 525600 minutes. Therefore 10 minutes = 10/525600 year = 0.190258751 * 10-4 year. Also, N ( 10min) = 11 picogram, Then, we have N10m = N0 e - ( 0.03428792* 0.190258751* 10-4). or, 11= No e- (0.03428792* 0.190258751* 10-4 ) or, Therefore, N0 = 11*e(0.03428792* 0.190258751* 10-4 ) picogram.= 11* e(6.523576834*10-7 ) picogram.
3. Let years be the time when the amount left in the body is 12.5 % of the initial amount. Then Nt = 12.5 % (N0) = 0.125N0 . Then 0.125N0 = N0 e- 0.03428792t or, 0.125 = e- 0.03428792t . On taking natural logarithms of both the sides, we have ln ( 0.125) = - 0.03428792 t or, t = ( - 2.079441542) / (- 0.03428792) = 60.6467671 years = 60 years, 7 months23 days approximately.
