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A 200.0 - mCi sample of a radioactive isotope is purchased by a medical supply house. If the sample has a half - life of 14.0 days, how long will it keep before its activity is reduced to 20.0 mCi?

46.5 d

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University of Washington

Hope College

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McMaster University

okay for number 53 we have radioactive material that has 1/2 life of 14 days, and it starts with an activity 200 military. And we wanna help Ling, is it? Take until doesn't activity generally 20 um, liqueur. I like to use this equation for that. Even though this end is the number of nuclear, the number of nuclear end with the number of nuclear, you start with, um, 1/2 raised to the number of half lives. I just think it's equations easy to deal with. I don't need to convert units, because if I just multiply both sides of this by the decay constant now instead of number of nuclear, this is activity. So I can think of this is activity, um or this is activity Original are, um lungs are in the same unit works, so I wanted to be 20 but it started it 200 and this is 1/2 raised to the number of half Life's. If I wonder how long it takes its some multiple of of half lives that some number I'm gonna call X, but divided by 14. So this will tell me because 14 is 1/2 life. So x will be how many days it is. And then that makes this the number of half lives. So I just need to solve this. I'm going to divide both sides by 200. So now I have 0.1 equals 1/2 raised to the exit 14. So to solve this one, you take little log of both sides. Um, so this will be log of 0.1 over here. This will be log of 1/2 would call it 0.5 now raised to that. Remember how logs work the exponents. I can just bring that out in front. So this is gonna be X over 14 in front. So now, to get the exploit itself, I'm gonna on the side of have longer 0.1. I'm gonna divide by the local 0.5. I'm gonna multiply by the 14 on this vehicle X and the way this was set up that 14 risen days for my exhibition days and I get 46.5 days

University of Virginia