The common isotope of uranium, 238U, has a halflife of 4.47x10^9 ...

Question

The common isotope of uranium, 238U, has a halflife of 4.47x10^9 years, decaying to 234Th by alpha emission. (a) What is the decay constant? (b) What mass of uranium is required for an activity of 1.00 curie? (c) How many alpha particles are emitted per second by 10.0 g of uranium?

Accepted Answer

Hi, everyone in this practice problem, we're being asked to calculate the decay constant, the mass of radium and the rate of emission of alpha particles where we'll have a bracket therapy, a sealed source of radium. 226 used to deliver a radiation dose to a certain tumor radium. Tw 2 26 has a half life of 1600 years and decays by producing rain. Uh 222 and an alpha particle, we are being asked to calculate the decay constant of radium 2 where a patient need a dose of 20 millicu to treat his cancer. We are being asked to also calculate the mass of radium designed to be swallowed by the patient and lastly calculate the rate of emission of alpha particles for a one microgram sample of radium. 2 26. The options given listed the decay constant of radium. 2 26 the mass of radium designed to be swallowed by the patient and the rate of emission of alpha particles. So in order for us to calculate or tackle this problem, let's tackle this problem uh one part at a time. So let's start with the first part where we're being asked to calculate the decay constant of radium. 2 26 we want to recall that the relationship between the half-life, the half and the decay constant. Lambda is given by a lambda equals to L N of two divided by the half, the half life of radium 2 26 or T half is given to be 1600 years, which will have to be converted into seconds in order for us to get the decay constant. But we can definitely substitute everything into our Lambda equation where lambda will then be equals to L N of two divided by the T half, which is 1600 years. And we wanna multiply this with um seconds per year, which is going to be 3.1557 times 10 to the power of seven seconds for every year, close parenthesis. And in calculating this, we will get the decay constant lambda to then be equal to 1.373 times 10 to the power of negative seconds to the power of negative one. So that will be the first part of this problem. And then moving on to the second part which is calculating the mass of radium designed to be swallowed by the patient. We have given the dose of the radium which is going to be millicu, which means that we should have 20 mili query or 20 multiplied by 10 to the power of negative three query multiply that by um the decays per seconds divided by one query because we want to convert the query into decays per seconds. So this will then be equals to three po uh or multiplied by 3.7 times to the power of 10 decays per second for every one C I or one query. And they all give us uh the decays per second to be 7.4 times 10 to the power of eight decays per seconds just like. So, so the activity of the radioactive source is going to be described by the formula of D N divided by D T equals to negative lambda N. And rearranging this formula, we can get a formula for N which will then be equals to negative open parenthesis D N divided by D T. All of that divided by lambda. The N divided by D the represent represents the number of disintegrations per seconds or the number of nuclei emitting alpha particles per seconds. So the number of nu needed to produce 20 Meli can then be calculated by using this formula where N will be equal to the negative of the N divided by D T is going to be equal to the decays per second, which is 7.4 times 10 to the power of eight decays per seconds to fight that with the decay constant or LAMBDA, which is going to be 1.373 times 10 to the power of negative 11 seconds. Calculating this the N or the number of nuclei needed is going to then be equal to 5.390 times 10 to the power of 19 nuclei. And then what we need to do next is by this number of nuclei, we just have to calculate the mass of radium designed to be swallowed by the patient. So the mass of one nucleus of radium or M is going to equals to approximately 226 U where the U is going to be equal to 1.67 times 10 to the power of negative kg. So the mass will then be equal to 226 times multiplied by or 226 multiplied by 1.67 times 10 to the power of negative kg. And we can then calculate the total number or the total mass of the radium needed in one dose for a patient. So then we can then uh calculate that by um total to then be equal to 226 times 1.67 times 10 to the power of negative 27 kg multiply that with the number of nuclei in one dose, which is 5.39 times 10 to the power of 19 nuclei. And then that will give us the total mass or M total to then be equal to two point oh three times 10 to the power of negative five kg just like. So, so the mass of radium designed to be swallowed by the patient is two point oh three times to the power of negative five kg. And that will uh give the answer to the second part. And then lastly moving on to the third part, we can uh recall that the number of nuclei for one mil microgram of radium 226 can be calculated by using N equal to one microgram or essentially in kilogram. That will be one times 10 to the power of negative nine kg. So one microgram divided by the mass of individual nuclei which is going to be dismissed right here, which is 226 times 1.67 times 10 to the power of negative 27 kilograms. And in calculating this, we will get the number of nuclei in one microgram of radium 226 to be 2.65 times 10 to the power of 15 nuclei. And then we can calculate the rate of emitted alpha particles which will be equal to the number of disintegrations per second, which will then be the absolute value of D N divided by D T equals to lambda, multiplied by N. And the D N divided by D T can then be calculated by multiplying the lambda which is 1.373 times 10 to the power of negative 11 seconds to the power of negative one multiply that with the end that we just calculated, which is 2.65 times 10 to the power of 15 nuclei are just uh disregard this nuclei here. And then this will then give us the D N divided by D T value to then be equal to 3.64 times 10 to the power of four alpha particles per seconds. So there we have the rate of emission of alpha particles for one microgram sample of radium to 26 which will be 3.64 times 10 to the power of four alpha particles per second. For the second part, we get the total a mass of radium designed to be swallowed by the patient to be two point oh 22 point oh three, sorry times 10 to the power of negative five kg. And then for the first part, the decay constant is 1.373 times 10 to the power of negative 11 seconds to the power of negative one. And all of this answer will correspond to option B in our answer choices. So option B 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 lesson 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 common isotope of uranium, 238U, has a halflife of 4.47x10^9 years, decaying to 234Th by alpha emission. (a) What is the decay constant? (b) What mass of uranium is required for an activity of 1.00 curie? (c) How many alpha particles are emitted per second by 10.0 g of uranium?

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