The half-life for radioactive decay (a first-order process) of
p...

Question

The half-life for radioactive decay (a first-order process) of 
plutonium- 
239 is 24,000 years. How many years does it take for 
one mole of this radioactive material to decay until just one 
atom remains?

Accepted Answer

everyone in this example we have americium 2 43 with a half life of over 7500 years and decays in first order kinetics. We need to calculate the time that it will take for one mole of americium 2 43 to decay only until 25 atoms remain. So what we should recall is that for a first order reaction we would calculate by the following formula where we take the natural log of our concentration of our atom at a given time and set that equal to negative one times our equilibrium constant. Or sorry, our rate constant multiplied by time, which is then added to the natural log of our concentration of our our atom at its initial time. And so now we want to recall the half life for a first order reaction. And we would recall that that is calculated by taking t one half or half life and setting that equal to the natural log of two divided by our rate constant. K. So our first objective is to calculate our value for our rate constant. So we would reorganize the formula above for half life and say that the rate constant is equal to the natural log of two divided by our time. To the one half or rather our half lifetime. And so we would say that our rate constant is equal to The Ln of two divided by or half life given in the prompt as years. And so what we would get for our rate constant is a value equal to 9.242 times 10 to the negative fifth power inverse years. So now we want to go back to that first order reaction equation and we will plug in our rate constant value. So right now we would have that the natural log of our 25 atoms according to the prompt that should remain after the half life Is equal to negative one times our rate constant which we just found above as 9.24, 2 times 10 to the negative fifth power in verse years. We're going to multiply this by time, which is what we're solving for. And then at the natural log of our initial concentration of our um americium 2 43 which we would assume is equal to six point oh 22 times 10 to the 23rd power atoms initially. Because we're going to recall avocados number here. And so what we would do is sell for time here. So we would say that t is equal to 5. times 10 to the 5th power years. And this would be our final answer for the amount of time that it will take for one mole of americium 2 43 to decay until 25 atoms remain. So I hope that everything I explained was clear. If you have any questions, just leave them down below and I will see everyone in the next practice video

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