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anonymous
 one year ago
If we start with 8000 atoms of radium226, which has a half life of 1600 years, how much would remain after 3200 years?
anonymous
 one year ago
If we start with 8000 atoms of radium226, which has a half life of 1600 years, how much would remain after 3200 years?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@arw1998 Hello, did u try this? it was one of the incredible questions so simple.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I tried and Got The amount that would remain is 2.0345e+3 atoms of Radium., before to copynpaste it to your homework, check it please.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well im not really even sure how to do it. my teacher showed us today but I didn't quite understand and im trying to figure out how to do it for the test tomorrow.

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.1hey! what does it mean when someone says halflife?

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1The half life is how much time it takes for half your sample to decay. in the statement it says 1600 years, so after 1600 years half of our sample will be gone. 4,000 atoms would be left, because half of 8,000 is 4,000. so after 1600 years you're left with 4000. But the question asks for how much is left after 3200 years. Remember half life refers to half of our sample that we (have currently) so now you have 4,000 atoms left how many would you have left after another 1,600 years?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Photon 336 explained it well. If you need the formula it is: \[N_{(t)}=N_{(0)}(\frac{ 1 }{ 2 })^{\frac{ t }{ t (1/2) }}\] which means: \[Amount_{(Final)}=Amount_{(Initial)}(\frac{ 1 }{ 2 })^{\frac{ time }{ time of (1/2)life }}\] You can plug in your values: N(0) = 8000 t = 3200 t(1/2) = 1600 \[N_{(t)}=8000(\frac{ 1 }{ 2 })^{\frac{3200 }{1600}}\] For the example given using the formula is overkill but for different values the formula can be helpful.
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