Question

element x decays into element y with a half life of 20 years. suppose that a sample of element x was found to contain 70% of element y. estimate the age of this sample.I don't know how to start this, Thanks!

Answer 1

I think it should be a good start to put the relation,
\[N=N_0e^{-\lambda t}\]

Answer 2

Then, it should be interesting to see that there is still 30% of element x, so
\[N=0.3N_0\]

Answer 3

is this simply the decay formula? p=poe^-rt?

Answer 4

Yes, I would say it.

Answer 5

ok. but this is a differential equations class.. i dont understand.

Answer 6

Oh, for sure. Differential equations are coneccted with the decay process,
\[dN=-\lambda N dt\]

Answer 7

what does the differentiation mean?

Answer 8

Also you can start from this formula, integrate and obtain the decay formula, and then solve.

Answer 9

This is the "differential", so this is an exact differential equation that can be integrated, easily,
\[\frac{dN}{N}=-\lambda dt\Rightarrow \int_{N_0}^N\frac{dN}{N}=-\lambda \int_0^t dt\]

Answer 10

aha. i understand that i think. i'm sorry would you mind going over this problem with me in steps?

Answer 11

Yes, when you integrate the last formula you obtain,
\[\ln(N/N_0)=-\lambda t\Rightarrow N=N_0e^{-\lambda t}\]

Answer 12

Now you need the conditions of the problem.

Answer 13

can i just start the solution with N=N0e^rt
?

Answer 14

I think yes.

Answer 15

and the conditions would it be N0=.3 and r=20 yearS?

Answer 16

Better,
\[N=0.3N_0\]And for the halflife
\[r=\lambda=\frac{\ln2}{T}\]with T=20 years.

Answer 17

r or λ is the constant of desintegration, that is related to the halflife. You can demonstrate this relation easily. But I give you a link where you can read it quietly.
http://www.math24.net/radioactive-decay.html

Answer 18

thanks so much. i will read over this and hopefully i will be able to get an answer

Answer 19

Yes, you will be able to do it, and in case you have any more doubts, tell me.

Answer 20

ok so i got lamda to equal (1/20)*ln 2
and the equation to end up looking like this: 0.3 =e^(1/2ln2)t
is that right?

Answer 21

\[0.3 =e^{(ln2/20)t}\]

Answer 22

with the minus sign in the exponent I forgot.

Answer 23

why is it ln 2/20?

Answer 24

Because,
\[\lambda=\frac{\ln 2}{T}=\frac{\ln 2}{20}\]

Answer 25

I would then apply logarithms,
\[0.3=e^{-\frac{\ln2}{20}t}\Rightarrow \ln(0.3)=-\frac{\ln2}{20}t\Rightarrow t=-\frac{20\ln(0.3)}{\ln 2}\]

Answer 26

i got it thanks so much!