Half-Life Calculation
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# Half-Life Calculation

[From: ] [author: ] [Date: 11-05-25] [Hit: ]
75and find the vale for 24.Remeber to subtract this from the original 30 g.-ln(A0/A)=kt,Now t(1/2)=0.or k=0.693/t(1/2)=0.......
"Sodium-24 has a half-life of 15 hours. How much of a 30g sample will remain undecayed after one day?"

I know it's really easy but have completely forgotten how to do these, working out would be appreciated :D cheers

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Simplest is to graph since you are only dealing with 2 significant figures.

0.30 15,15 30, 7.5 45, 3.75 and find the vale for 24. Remeber to subtract this from the original 30 g.

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ln(A0/A)=kt, where A0 is the initial amount and A is the current amount remaining at time t
Now t(1/2)=0.693/k
or k=0.693/t(1/2)=0.693/15=0.0462
now no. of moles in 30g sample=30/24=5/4=1.25 moles
So
ln(1.25/A)=0.0462*24 (because 1 day=24 hours)
so ln(1.25/A)=1.1088
hence 1.25/A=3.031
and A=1.25/3.031=0.412 moles
so amount remaining=24*0.412=9.89g

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The relevant time equation is

Amount = Amount at start x exp( - k * time) where k is a specific constant.

To find k we need the fact that

15 = 30*exp(-15k) which gives a k of ln(1/2)/-15

Thus one day = 24 hours ==> A = 30*exp( -( ln(1/2)/-15 )*24)

==> A = 9.896g
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