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Emily778
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The halflife of a radioactive substance is the time it takes for half of the material to decay. Phosphorus32 is used to study a plant's use of fertilizer. It has a halflife of 14.3 days. Write the exponential decay function for a 50mg sample. Find the amount of phosphorus32 remaining after 84 days.
 7 months ago
 7 months ago
Emily778 Group Title
The halflife of a radioactive substance is the time it takes for half of the material to decay. Phosphorus32 is used to study a plant's use of fertilizer. It has a halflife of 14.3 days. Write the exponential decay function for a 50mg sample. Find the amount of phosphorus32 remaining after 84 days.
 7 months ago
 7 months ago

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coolsday Group TitleBest ResponseYou've already chosen the best response.0
The halflife formula is: dw:1389736938636:dw
 7 months ago

coolsday Group TitleBest ResponseYou've already chosen the best response.0
A is the final amount , Ao is the initial amount, h is the halflife of the substance, and t is the time
 7 months ago

coolsday Group TitleBest ResponseYou've already chosen the best response.0
just plug the variables into the formula to solve for A.
 7 months ago

Emily778 Group TitleBest ResponseYou've already chosen the best response.0
how do I do the rest?
 7 months ago

coolsday Group TitleBest ResponseYou've already chosen the best response.0
Ao is 50 mg, h is 14.3, t is 84 plug it into the formula and solve for A
 7 months ago

Emily778 Group TitleBest ResponseYou've already chosen the best response.0
wouldn't the answer be 146.85?
 7 months ago

Emily778 Group TitleBest ResponseYou've already chosen the best response.0
@coolsday
 7 months ago

Emily778 Group TitleBest ResponseYou've already chosen the best response.0
by plugging those in?
 7 months ago

dape Group TitleBest ResponseYou've already chosen the best response.1
So the halflife formula can also be written as \(2^{\lambda t}\), where \(\lambda\) is the 'decay constant', or just \(1/h\), where h is half life. So the exponent is \(84/14.3\approx5.874\), which says that the sample will have time to halve in size about 5.9 times in the 84 days (this is an easy way to remember the formula). Putting this in, we have that \(2^{84/14.3}\approx1.7\%\). So about 1.7% of the sample will remain. Starting with 50 mg this means that about \(50\times1.7\%=0.85\) mg of the sample will remain.
 7 months ago

dape Group TitleBest ResponseYou've already chosen the best response.1
Oh, and the 'exponential decay function' we just get by putting all this together, so \[P(t)=50\times2^{t/14.3}\] Where t is time in days.
 7 months ago

Emily778 Group TitleBest ResponseYou've already chosen the best response.0
So that's the answer?
 7 months ago

Emily778 Group TitleBest ResponseYou've already chosen the best response.0
hellloooo?
 7 months ago

coolsday Group TitleBest ResponseYou've already chosen the best response.0
0.85 mg will remain after 84 days if you use the formula.
 7 months ago
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