Question

The radioactive isotope of lead, Pb-209, decays at a rate proportional to the amount present at time t and has a half life of 3.3 hours. if 1 gram of this isotope is present initially, how long will it take for 90% to decay

Answer 1

\[\frac{dP}{dt}=kP\]

given

P(0)=1g
P(3.3)=.5g
P(x)=.1g

Answer 2

doing the integration again you get

\[P=ce^{kt}\]

Answer 3

\[1=ce^0\]
c=1

Answer 4

\[P=e^{kt}\]

Answer 5

i think it should be
dP/dt = -kP as this is decay

Answer 6

does it matter as k is a constant

Answer 7

i'm looking at an example now and the book uses kP

Answer 8

right - good point - its just that in all problems like this this is the way the textbooks do them

Answer 9

ooh - ok

Answer 10

like i understand how to do it but i feel i messed up in simplifying because i get a negative time

Answer 11

\[.5=e^{3.3k}\]

Answer 12

\[ln(.5)=ln(e^{3.3k})\]
\[ln(.5)=3.3k\]

\[k=\frac{ln(.5)}{3.3}\]

Answer 13

try .5 = e ^(-3.3t)

Answer 14

\[P=e^{\frac{ln(.5)t}{3.3}}\]

Answer 15

-3.3k i mean

Answer 16

oh wait i found a simplifying error but i don't think it'll do anything

Answer 17

\[P=e^{\frac{ln(.5)}{3.3}}e^t\]

Answer 18

wups i meant this

\[e^{\frac{ln(.5)}{3.3}}e^{\frac{t}{3.3}}\]

Answer 19

\[{.5^{\frac{1}{3.3}}}e^\frac{t}{3.3}=p\]

Answer 20

i guess i'm pretty rusty at these....

Answer 21

but lets see hwat i get with this

Answer 22

\[\frac{.1}{.5^{\frac{1}{3.3}}}=e^{\frac{t}{3.3}}\]

Answer 23

it's not even close to the answer lol

Answer 24

@asnaseer

Answer 25

you have it correct up to this point:\[P=e^{\frac{ln(.5)t}{3.3}}\]you can then say the following:\[P=e^{\frac{ln(.5)t}{3.3}}=(e^{ln(.5)})^{\frac{t}{3.3}}=(0.5)^{\frac{t}{3.3}}\]

Answer 26

which can also be written as:\[P=\frac{1}{2^{\frac{t}{3.3}}}=2^{-\frac{t}{3.3}}\]

Answer 27

still not gettin the right answer

Answer 28

ok... this is how to do it

\[0.5 = 1\times e^{3.3k}\]

solve for k... you'll need logs

\[\ln(0.5) = 3.3k........ k = \frac{\ln(0.5)}{3.3} ..... k = -0.210045\]

so your model is

\[P = 1 \times e^{-0.210045t}\]

Answer 29

next find when P = 0.1

\[01. = 1\times e^{-0.210045t}\]

take the log of both sides

\[l(0.1) = -0.210045t....... t = \frac{\ln(0.1)}{-0.210045}.... t=10.9623\]

Answer 30

the time taken is 10.9623 hours for the isotope to decay to 0.1 grams from its original of 1 gram