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Nucleus A has a half-life T and nucleus B has a half-life 2T. Initially the number of nuclei of type A equals the number of nuclei of type B After a certain time , ...

Question

Nucleus A has a half-life T and nucleus B has a half-life 2T. Initially the number of nuclei of type A equals the number of nuclei of type B After a certain time , 10% of the nuclei of type B remain: At this same time, what fraction of the nuclei of type A remains?Select one: a_ 5%b. 0.01%50%d. 1%

Nucleus A has a half-life T and nucleus B has a half-life 2T. Initially the number of nuclei of type A equals the number of nuclei of type B After a certain time , 10% of the nuclei of type B remain: At this same time, what fraction of the nuclei of type A remains? Select one: a_ 5% b. 0.01% 50% d. 1%



Answers

A particular nucleus in a large population of identical radioactive nuclei servise 10 half-lives of that isotope. The probability that this surviving nucleus will servie the next half-life is (a) $\frac{1}{10}$ (b) $\frac{2}{5}$ (c) $\frac{1}{2}$ (d) $\frac{1}{2^{10}}$

In this question, we will learn about some basic concept of nuclear and the first order radio activity. The question is about a particular nucleus In a large population of identical radioactive nuclear Survives. lives of that ice adopts. The probability that is surviving nucleus will survive the next half life is what we have to find. Like. So in a given question, the particular nucleus in the large population that survived for have So the 5/2 lives, right? Then it will either survey for the next life or not. Then the probability that this surviving nucleus will survive the next half life is Half that is one x 2. Like. So there are two possibilities that whether it will survive the next half life cycle. Okay. Yeah. Or it will not survive. Right? So they do. Possibly. It is therefore, The probability of surviving is one x 2. Okay. Or 50%. Thank you. Hence CS the correct solution.

Hello students in this question we have given a radioactive element A. D. Case with into be with half life Tv. This is equal to five. Sorry this is equal to 4.5. More clear by 10 to the power nine years. Okay and this is the half life of A. Actually so ta. Okay and the beat have half life tv. It is equal to 5000 years. Okay. And in the study state uh number of nuclear and A. It is equal to two molecular weight. And to the power 20 then the number of nuclear that is N. B. It is equals two. So we can write that for the this is speaking in to be so number of nuclear nuclear produced of B. Will be equal to the left out nuclear of the A. So hence we can right here and a manipulator linda. A lambda A. This is equals two and be more clear by lambda. Be from the law of decades. Okay so we can right here that lambda which is decay constant and linda it is given by natural log of two divided by T. Have. Okay so we can right here that N. A divided by T. A. This is equal to and be developed by Tv. So from here we get envy. This is equals two Tv divided by T. A. Molecular bio and A. So and we will be equals two T. B. T. Half of B. Which is 5000 years, the verb in half like of A. Which is this 4.5 particular bait and to the power nine years molecular bio and A. Which is this to molecular by 10 to the power 20. Okay so from here after solving the number of number of nuclear of B. It is equal to 2.2 marker by 10 to the power 14. Okay so this becomes the answer for this question. An option B. It is the correct answer for this problem. Okay thank you.

Prominent number 78 for creature number eight. Type a new client. The case a first rate, then type E. So type yeah, has a longer half life For questions being convince the half life of two sandals A and B after two days we're in is equal to 10 notes e power Negative London Tea in notes is each 12 n a and and be or initial numbers of nuclear All you cry. So in a is equal to four a note mhm and n b is equal to and knows. Okay, Yeah, let in a equal and be after two days Have to two days. So for in Newt E polar Negative London tea Well equal to in those e power. Mhm. Negative Lund of P t. So for is equal to e power. Sorry, this is London 80 e Power London A minus London Fee wanted. Tonight P. That's it.

The fraction of a radioactive isotope remaining after time. T can be calculated by raising one half to the time that has passed, divided by the half life. So if 10 years have passed Then the fraction remaining will be equal to 1/2 Raised to the 10, divided by The half life for carbon 14 is 5730 years, or .9988. After 10,000 years, We have a fraction remaining of .29983. And then after 100,000 years, many many have lives have passed and there's hardly anything remaining often such a small amount. We can't even measure it. If we can't measure it, we can't determine half life. We're looking at 557, 7 times 10 to the six is the fraction that remains. Yeah. So it turns out that will get more accurate results with part B than we will with A. And C. This is because radiocarbon dating is more reliable. If we have a measurable amount, if hardly anything has decayed, then it's difficult to measure the amount that has decayed, as in part A. Or if pretty much all of it has decayed, that it's very difficult to measure how much is left, making it difficult to determine half life.


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