Answer :
phosphorus-30, which has a half-life of 150 seconds, decays by positron emission. it will take t = 1146.82 seconds for 4.6% of this nuclide to decay.
phosphorus-30, which has a half-life of 150 seconds, decays by positron emission. It takes 1146.82 seconds long for 4.6% of this nuclide to decay. The rate constant can be solved using the equation for first-order kinetic half life. Basically: when concentration becomes half of initial concentration then time become half life ln (.5 [A]0/[A]0) = -kt1/2 --> ln(1/2) = -kt1/2 ,t1/2 = ln(1/2)/k ,t1/2 = 0.693/k . So, solve the equation for half-life using 150 seconds, i.e.: 150 = 0.693/k I get an answer of k = 0.00462 .Now, you can use the equation ln (Nt/N0) = -kt, since k is given. Since concentrations are given in percentages, it would be easy to assume you start of with 100 nuclei (the N0 value). For example, for the first part of the question, you are trying to figure out how long it takes for 2% to decay, so you just assume a loss of 2 from 100 and set up your equation to find t: ln (98/100) = -0.00462*t ,I get t = 4.37 seconds .Similarly, set up the rate equation for 99.5% decay by assuming a loss of 99.5 from 100. A quick set-up and calculation: ln (0.5/100) = -0.00462*t
t = 1146.82 seconds
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