## anonymous 5 years ago the rate at which a radioactive isotope disntegrates is porportional to the amount present. if 30g sample will contain only 20g after 10 minutes, what is the half life of the isotope, correct to the nearest 10th. y=yoe^kt. :)

1. anonymous

your formula is $y=y_0e^{kt}$

2. anonymous

$y_0=30$ and you know that when t=10, y = 20

3. anonymous

so set $20=30e^{10k}$

4. anonymous

and solve for k: $20=30e^{10k}$ $\frac{2}{3}=e^{10k}$ $ln(\frac{2}{3})=10k$ $k=\frac{ln(\frac{2}{3})}{10}$

5. anonymous

6. anonymous

thank you thank youi thank you

7. anonymous

so now you know your formula is $y=30e^{-4.055t}$and you want to know when you have half of what you started with. you started with 30 and half of that is 15 so if you want you can write $15=30e^{-4.055t}$ or just cut to the chase and write $\frac{1}{2}=e^{-4.055t}$

8. anonymous

and now solve for t: $ln(.5)=-4.055t$ $t=\frac{ln(.5)}{-4.055}=$

9. anonymous

wait wait wait

10. anonymous

big mistake on my part

11. anonymous

$\frac{ln(\frac{2}{3}}{10}=-.04055$

12. anonymous

so wrong from there on in. it is $\frac{ln(.5)}{-.04055}=15.32$\]

13. anonymous

very sorry i was off by two decimal places

14. anonymous

thats fine :)

15. anonymous

hope you are still here to see my correction

16. anonymous

oh good.