**Question:**How many half-lives would have to elapse for a sample of a radioactive isotope to decrease from an activity of 48 cpm to an activity of 3 cpm?---**Explanation:**This question involves calculating the number of half-lives required for a radioactive substance to decrease in activity from 48 counts per minute (cpm) to 3 cpm. The activity of a radioactive isotope decreases by half with each half-life. Therefore, you can solve this by using the formula:\[ \text{Final Activity} = \text{Initial Activity} \times \left(\frac{1}{2}\right)^n \]Where:- Final Activity = 3 cpm- Initial Activity = 48 cpm- \( n \) = number of half-livesYou can rearrange and solve this equation to find \( n \):\[ 3 = 48 \times \left(\frac{1}{2}\right)^n \]By dividing both sides by 48 and solving for \( n \), you can find how many half-lives have elapsed.

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Answered: How many halE lives would hae to elapse…

How many halE lives would hae to elapse for a Sample of a radioachtve isotope to decraase Of 3cpm? fram an activity of 48 cpm to an achyity o! acthuily

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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