The unstable isotope 40K is used for dating rock samples. Its hal...

Question

The unstable isotope 40K is used for dating rock samples. Its half-life is 1.28x10^9 y. (a) How many decays occur per second in a sample containing 1.63x10^-6 g of 40K? (b) What is the activity of the sample in curies?

Accepted Answer

Hi, everyone. In this practice problem, we are being asked to calculate the number of radioactive disy the patients per sample and also expressing this activity in the sample. And Q E, we will have a decay of rubidium. 87 R B employed to estimate the age of a very old mineral sample containing 3.25 g of rubidium. We're being asked to first calculate the number of radioactive disintegrations per second within the sample. If R B has a half life of 4.92 times 10 to the power of 10 years and second express this activity of the sample in query. The options given listed the number of radioactive disintegrations per seconds in disintegrations per seconds and also this in Q. So in order for us to calculate this first, we want to recall that the relationship between the health life, the health and the decay constant lambda is given by a lambda to then be equal to L N of two divided by the half, the rubidium has a half life of or T half of 4.92 times 10 to the power of years. And essentially, we can then calculate the lambda or the um decay constant by substituting our half life value. So then our lambda will then be equal to L N of two divided by 4.92 times 10 to the power of 10 years. We wanna multiply this and uh convert this into secondss by multiplying the years with 3.1557 times 10 to the power of seven seconds for every one year. So then we can cross out the years and then we can then calculate this to get the decay constant lambda to be equal to 4.464 times 10 to the power of negative 19 2nd to the power of negative one. So then we wanna next recall that the activity of the radioactive source rubidium is going to be described by the equation of D N divided by D T equals to negative Lambda N N is the number of rubidium nuclei in the sample. And lambda is the decay constant. The mass of one nucleus of R B 87 is going to be approximately 87 U where one U is going to equal to 1.66 times 10 to the power of negative 27 kg. Uh In this case, we can then calculate the mass of one nucleus of rubidium 87 where mass will then be equal to 87 multiplied by 1.66 times 10 to the power of negative 27 kg, which will then come out to B 1.44 times 10 to the power of negative 25 kg. The number of rubidium nuclei in 3.25 g of rubidium can then be calculated or then the end value can be calculated by taking 3.25 g or 3.25 times 10 to the power of negative three kg. And dividing that with uh the mass of each nuclei uh or the mass of one nucleus of rubidium 87 which will be 1.44 times 10 to the power of negative kg. In calculating this, we will get the number of R B or rubidium nuclei in our sample, which is going to be 2.257 times 10 to the power of 22 nuclei. We can then calculate the uh activity of the sample, which is going to be the D N divided by D T which will be equals to negative lambda multiplied by N the negative lambda. We just calculated earlier which is 4.464 times 10 to the power of negative 19 seconds to the power of negative one multiplied that with the N or the number of uh rubidium 87 nuclei which is 2.257 times 10 to the power of 22. And that will give us our D N divided by the T or the activity of the sample to then be equals to 1.0 times 10 to the power of four, this in the gras per seconds. So then the activity of the sample is one times 10 to the power of four disintegrations per second. And we can convert this activity into query. So for the second part, that is exactly what we're gonna do. So we know that one query or one C I is going to be 3.7 times 10 to the power of 10 disintegrations per second. So the activity of this sample in query or D N divided by D T, which will be equal to 1.0 times 10 to the power of four disintegrations per seconds. We can multiply this with one query and divide that with 3.7 times 10 to the power of disintegrations per seconds to then obtain our um activity of the sample in query, which will come out to a value of D N divided by D T equals to 2.70 times 10 to the power of negative seven Q E or C I. So there we have it the activity of our sample. The N divided by D T is going to be uh 10 to the power of four disintegrations per second or 2.7 times 10 to the power of negative seven Q E which will both uh correspond to option A and R answer choices. So option A will be the answer to this particular practice problem and that will be it for this video. If you guys still have any sort of confusion, please feel free to check out our other videos on similar topics available on our website. But other than that, that will be it for this one. Thank you.

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The unstable isotope 40K is used for dating rock samples. Its half-life is 1.28x10^9 y. (a) How many decays occur per second in a sample containing 1.63x10^-6 g of 40K? (b) What is the activity of the sample in curies?

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