Half-Life of Silicon 32 - Kilani High School

[karush]

$\tiny{6.1.01}$
this was a homework problem from Kilani High School
The half life of silicon 32 is $710$ years. If $10g$ are present now
how much will be present in $600$ yrs?
find out $k$ using
\$\begin{array}{rlll}
&A=A_0 e^{kt}
&
&(1)\\
\textsf{make }A=5
&5=10e^{k \cdot 710}
&
&(2)\\
&k=\dfrac{\ln \left(2\right)}{710}=0.00097626&
&(4)\\
\end{array}$
hence A=$10e^{-.00097626 \cdot 600}=5.56685=5.57g$

ok did this 3 times but :unsure:
possible typos
suggestions ?

 

[topsquark]

$\tiny{6.1.01}$
this was a homework problem from Kilani High School
The half life of silicon 32 is $710$ years. If $10g$ are present now
how much will be present in $600$ yrs?
find out $k$ using
\$\begin{array}{rlll}
&A=A_0 e^{kt}
&
&(1)\\
\textsf{make }A=5
&5=10e^{k \cdot 710}
&
&(2)\\
&k=\dfrac{\ln \left(2\right)}{710}=0.00097626&
&(4)\\
\end{array}$
hence A=$10e^{-.00097626 \cdot 600}=5.56685=5.57g$

ok did this 3 times but :unsure:
possible typos
suggestions ?

The only problem I see is that \(\displaystyle A = A_0 e^{-kt}\) (Check the negative sign.) Other than that I see no problem.

-Dan

 

[skeeter]

$5 = 10e^{710k}$

$\dfrac{1}{2} = e^{710k}$

$\ln\left(\dfrac{1}{2}\right) = 710k$

$\dfrac{-\ln(2)}{710} = k$

 

[HOI]

I have no idea where the "A= 5"came from!

"The half life of silicon 32 is 710 years."
So $X(t)= A\frac{1}{2^{t/710}}$.

"If 10g are present now
how much will be present in 600 yrs?"

So A= 10 g and t= 600.
$X(600)= 10\frac{1}{2^{600/710}}$
$= 6.999$ grams.

 

[karush]

5 is half of 10

 

[topsquark]

5 is half of 10

I think what Country Boy was saying is not that you did it wrong, you just used numbers from the problem that you didn't need to. In general you have an amount A and after one half-life you have an amount A/2. In that way you can find k. You did this with A = 10 g and you did it correctly so no harm no foul. But you should recognize that you didn't have to use 10 g.

-Dan

 

[HOI]

Youy DO "have to use 10 g in this problem, because the problem SAID "10 gram are present now" when t= 0. Using 5= half of 10 means starting after the first "halving" when t= 710 years when the question was only asking about 600 year from "now".

 

[topsquark]

Youy DO "have to use 10 g in this problem, because the problem SAID "10 gram are present now" when t= 0. Using 5= half of 10 means starting after the first "halving" when t= 710 years when the question was only asking about 600 year from "now".

You need the 10 g in the second part of the problem, after you get the value of k.

Hey, I'm fine with using 10 g and 5 g to get k. But you don't need to.

-Dan