## anonymous 3 years ago What is the rate of decay of awesome-10 which has a half-life of 650 days?

1. anonymous

I can't figure out how to solve these rate of decay questions, so I just made up my own as a way to help me learn this. How would I solve this?

2. anonymous

if it has a half life of 650 time units it means every 650 time units 1/2 of the samle dissapears.

3. anonymous

Yep. :) How would I find the rate of decay, though?

4. anonymous

halflife= -ln2/k that is the half life equation

5. anonymous

I have a formula $N=N _{0}e ^{kt}$ How/where would I plug everything in?

6. anonymous

I'm not sure I understand @Spectrum

7. anonymous

I know spectrum

8. anonymous

i gave him all right answers

9. anonymous

for his teset he did good to

10. anonymous

I think they are asking for k you got N=1/2No t=650 so you plug all numbers in and find k

11. anonymous

I mean, I don't understand what he meant. I'm having a lot of trouble with logarithms, so I'm not quick on my feet. :P

12. anonymous

$N=\frac{ 1 }{ 2 }N _{0}^{650k}$ Okay, so what do I do next?

13. anonymous

Would I put some test number in for No? Like 100 or something?

14. anonymous

no,|dw:1354966748637:dw| you plug 1/2No for Nthen you can simplify for No

15. anonymous

|dw:1354967053153:dw|

16. anonymous

|dw:1354952485674:dw|

17. anonymous

|dw:1354952600983:dw|

18. anonymous

|dw:1354952743876:dw|

19. anonymous

|dw:1354953110349:dw|

20. anonymous

I think now it makes sense. Doesn't it?

21. anonymous

KIS means keep it simple.

22. anonymous

Derp still need help understanding =/

23. anonymous

The easiest way I believe is to derive it from calculus (if you've taken Diff EQ , you should know how to). Doing this you get an equation much like yours only N_0 is denoted c $A=ce^{kt}$

24. anonymous

now we're given two values to find what c is you need to know that $N(0)=N_0$ $N_0=ce^0=c$ $N=N_0e^{kt}$

25. anonymous

same as your equation now we know that after 650 days , we will have half of the original amount soo $N(650 days)=.5N_0$ $.5N_0=N_0e^{650k}$ solve for k, divide by $N_0$

26. anonymous

$\frac{.5N_0}{N_0}=e^{650k}$ the initial N's cancel so you get $.5=e^{650k}$ to get rid of e take the ln of both sides $ln(.5)=650k$ $\frac{ln(.5)}{650}=k$

27. anonymous

your rate of decay is $\frac{ln(.5)}{650}=k$ your equation will be $N=N_0e^{\frac{ln(.5)t}{650}}$

28. anonymous

@Outkast3r09 : same thing I did. That is why I was confused with @mahmit2012 had done

29. anonymous

$\frac{dP}{dt}=kP$ $\frac{dP}{dt}-kP=0$ $e^{\int{kdt}}=e^{kt+c}=ce^{kt}$

30. anonymous

you integrate the left side you'll just get that and the right side will be P