Sodium-24 has a half-life of 15 hours. How much sodium-24 will remain in an 18.0 g sample after 60 hours?

jeverett1784
Mar 23, 2015

#1**+5 **

Using

A = 18e^(kt)... where A is the amount after t hours, we have

9 = 18e^(k*15) divide both sides by 18

1/2 = e^(15*k) take the ln of both sides

ln(1/2) = ln e^(15* k) and by a log property, we can write

ln(1/2) = (15k) lne and lne = 1 so we can drop that part

ln(1/2) = 15k divide both sides by 15

k = ln(1/2) / 15 = -.0462

So after 60 hours, we have

A = 18e^(60* -0462) = about 1.126 g

Here's the graph of the situation.....where x = t ....the relevant values are where x ≥ 0

https://www.desmos.com/calculator/wvzugktxmp

CPhill
Mar 23, 2015

#1**+5 **

Best Answer

Using

A = 18e^(kt)... where A is the amount after t hours, we have

9 = 18e^(k*15) divide both sides by 18

1/2 = e^(15*k) take the ln of both sides

ln(1/2) = ln e^(15* k) and by a log property, we can write

ln(1/2) = (15k) lne and lne = 1 so we can drop that part

ln(1/2) = 15k divide both sides by 15

k = ln(1/2) / 15 = -.0462

So after 60 hours, we have

A = 18e^(60* -0462) = about 1.126 g

Here's the graph of the situation.....where x = t ....the relevant values are where x ≥ 0

https://www.desmos.com/calculator/wvzugktxmp

CPhill
Mar 23, 2015