Nitrogen-16 is formed in the cooling water flowing through a hot reactor core in a nuclear power plant. It is formed when oxygen captures a neutron and then emits a b par-ticle. Determine the activity of 50.0 mg of 16N in units of Bq and Ci.

Video transcript

Welcome back, everyone. We're told that plutonium to 39 with an atomic weight of 2 39. to 2 a.m. use and a half life of 24,110 years is produced in a fast breeder reactor. Uranium 2 38 absorbs a neutron and then undergoes two consecutive beta decays to yield plutonium to 39. What is the activity of 245 mg of plutonium to 39 in Becquerel and curie units? Alright, let's begin by recalling that whenever we have radioactive decay, we will be following our first order rate law. We don't need the rate law in this case. But what we do need is our decay constant K, which is within our rate law. And recall that we can calculate our decay constant K by taking the Ln of two divided by R given half life for our radio isotope. And so we're going to begin by finding our decay constant for Plutonium 239. So we would take the Ln of two, which is the result being 0.693 divided by its given half life in the prompt as 24,110 years. And so we want to make note of the fact before we continue to solve this, That because we need to give our activity of our radio isotope plutonium to 39 in becquerels. We want to recall that one becquerel is equivalent to one disintegration per second. So note that in our denominator, we have units of seconds, but note that our decay constant or rather the half life given is in years. So we're going to need to convert this two years. So we're going to actually continue and multiply in a stock geometry step to go from One year, which has an equivalent of 365 days. This allows us to cancel out years. And now we're going to go from days where we have one day in our numerator equivalent to 24 hours, canceling out days were now going from hours to minutes. So we're called that one hour has an equivalence of 60 minutes. Canceling out ours, we can go now from our final conversion factor one minute equivalent to 60 seconds. And so canceling out minutes were left with seconds in our denominator. And so this will result in a decay constant equal to a value of 9.114428 times 10 To the negative 13th power inverse seconds since we're no longer in a fraction. Now, because again, we need to find the activity in Becquerel and Curie units, we want to recall that the activity of our radioisotope in Becquerels Typically is the same thing as our rate of decay of our plutonium to 39, which typically has units of time or rather inverse time that could be days, hours, seconds, whatever unit of time. And because we understand that our activity is the same as our rate of decay for our isotope. We can say that therefore, we would recall the law of radioactive decay where our rate of decay is equal to our decay constant K times the variable N Recall that this law states that the rate of decay of our radioactive material. In this case, plutonium 239 is going to be proportional to our number of radioactive Adams present N and R D K constant. Of course. And so we need to figure out because we know okay, we now need to figure out N for Adams of plutonium to 39. So that's our next step. We're going to begin by referring back to the prompt and noting down the sample size of our plutonium to 39, given us 245 mg of plutonium to 39. So we're going to want Adams to be our final unit. So we're going to go through psyche on a tree where in the denominator, we go from one mg where the prefix milli tells us that we have 10 to the negative third power of our base unit gram. So canceling out milligrams, we're going to now go from grams to moles and noting back from the prompt, the given mass of plutonium to 39 we would use it as a conversion factor where in our denominator, we have 239. g of plutonium 2 equivalent to one mole of plutonium to 39. So now canceling out g will go from Moles of Plutonium 2 39 to now Adams Of Plutonium 2 39. And so we're going to utilize avocados number here. So let me just make some room. So we're calling the number for avocados number. We have six point oh 22 times 10 to the 23rd power atoms of plutonium to 39 equivalent to one mole. And now canceling out most were left with atoms, which is what we want for our variable N, which is equal to a value of 6.17183192 times 10 to the 20th power atoms of plutonium to 39. So now we're gonna go back to our rate of decay equation where we say that the rate of decay for plutonium to 39 again is equal to or directly proportional to R D K constant, which we solved for as 9.1144 to 8 times 10 to the negative 13th power inverse seconds, which is multiplied by our atoms of our radio isotope plutonium 2 39 which we just calculated to be 6.17183192 times 10 to the 20th power atoms. And so, because we're multiplying our decay constant K times and our atoms of our radio isotope, our units are now going to be disintegration per second for our rate of decay. And so we can say that our rate, when we take the product is equal to 5.6 to 5 to seven times 10 to the eighth power disintegration per second. Again, we still have inverse time being seconds in the denominator. And so, as we stated above the activity of our radio isotope is the same as the rate of decay. And so therefore, we can say that the activity of plutonium 2 39 represented by a is equal to, we want three sig figs because that was our minimum number of sig figs in the prompt 5.62 times 10 to the eighth power Becquerel. Since we recalled above that one Becquerel is equivalent to one disintegration per second. And so this is this would be our first answer here as our activity of plutonium 2 39 in Becquerel. And now we need this inquiry. And so we're going to recall that one Curie is equivalent to 3.7 times 10 to the 10th Power Becquerels. And so beginning with finding again, the Activity of Plutonium 2 39, we would find that by taking our activity in Becquerels, which we found above as 5.6 to 5 to seven times 10 to the eighth Power Becquerels, which is multiplied by our conversion factor, which we just recalled as 3.7 times 10 to the 10th Power Becquerels in the denominator equivalent to one curie. And so canceling out Becquerels, we get our activity of plutonium 2 39 equal to a value of 0.152. Curious which also represents three sig figs as our second final answer for the activity of plutonium to 39. So it's highlighted in yellow are our two final answers corresponding to choice D and the multiple choice I hope that everything I went through is clear if you have any questions, leave them below and I'll see everyone in the next practice video.