The half-life of a radioactive isotope represents the average time it would take
ID: 1788417 • Letter: T
Question
The half-life of a radioactive isotope represents the average time it would take half of a collection of this type of nucleus to decay. For example, you start with a sample of 1000 Oxygen-15 (15O) nuclei, which has a half-life of 122 seconds. After 122 seconds, half of the 15O nuclei will have decayed into Nitrogen-15 (15N) nuclei. After another 122s, half of the remaining Oxygen nuclei will have also decayed, and so on. Suppose you start with 2.10×10^3 15O nuclei and zero 15N nuclei. How many 15O nuclei remain after 122 s has passed? The half-life of a radioactive isotope represents the average time it would take half of a collection of this type of nucleus to decay. For example, you start with a sample of 1000 Oxygen-15 (15O) nuclei, which has a half-life of 122 seconds. After 122 seconds, half of the 15O nuclei will have decayed into Nitrogen-15 (15N) nuclei. After another 122s, half of the remaining Oxygen nuclei will have also decayed, and so on. Suppose you start with 2.10×10^3 15O nuclei and zero 15N nuclei. How many 15O nuclei remain after 122 s has passed? You are correct. Your receipt no. is 157-9832 How many 15N nuclei are there after 122 s has passed? 1050 You are correct. Your receipt no. is 157-2919 How many 150 nuclei remain after 244 s has passed? 525 You are correct. Your receipt no. is 157-72086) How many 15N nuclei are there after 244 s has passed? 1575 You are correct. Your receipt no. is 157-8497 Suppose you start with 5.01x103 Carbon-14(14c) nuclei. 14C has a half-life of 5730 years and decays into Nitrogen-14(14N) via a beta decay. How much time has passed if you are left with 2.50×103 14C nuclei? (The units for years is 'yr'.) 2.50 10A3 yr Submit Answer Incorrect. Tries 2/10 Previous Tries How much time has passed if you are left with 1.25x103 14C nuclei? Submit AnswerTries 0/10Explanation / Answer
1. N = N(0)2^(-t/t_half life)
N = 2.5 X 10^3
N(0) = 5.01 x 10^3
t_half life = 5730 years
This gives: t = 5730 yr
2. N = 2.5 X 10^3
t = 2 x 5730 = 11460 yr
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