Question

A radio active substance has a half life of 20-days

how much time is required so that only 1/32 of the original amount remains?

Find the rate of decay at this time..

Answer 1

100 days because
1/32=3.125%
100/2=50 1st half life
50/2=25 2nd half life
25/2=12.5 3rd half life
12.5/2=6.25 4th half life
6.25/2=3.125% 5th half life
and since each half life is 20 days
20*5=100 days

Answer 2

In other words,
\[\frac{1}{32}=\left(\frac{1}{2}\right)^5\]
5 half-lives = 5×20days.

Answer 3

yeah better..

Answer 4

what about the rate of decay/

Answer 5

That you need to solve for using logarithms. How are you with taking logs of stuff?

Answer 6

am good

Answer 7

Sweet. Here's how it goes:
Half-life is 20days, so we'll start with t=20.
\[A(t)=A_0e^{kt}.\]
(General exponential growth/decay model)
Half-life means that A(20)=0.5A_0 ->
\[\frac{1}{2}=e^{20k}.\]
Now take natural log of both sides.

Answer 8

After solving for k, you can put it back in the equation for
\[\frac{1}{32}=e^{kt}\]
and solve for t to verify that it is 100 days.

Answer 9

Clear so far, @khally92 ?

Answer 10

yeah sure thanks...

Answer 11

OK, you should get something around -0.035 for the decay rate.