Half-Life of Silicon 32 - Kilani High School

• MHB
• karush
In summary, 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 $A=A_0 e^{kt}$. A=$10e^{-.00097626 \cdot 600}=5.56685=5.57g$.
karush
Gold Member
MHB
$\tiny{6.1.01}$
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 ?

Last edited:
karush said:
$\tiny{6.1.01}$
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

$5 = 10e^{710k}$

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

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

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

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.

5 is half of 10

karush said:
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

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".

Country Boy said:
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

What is the half-life of Silicon 32?

The half-life of Silicon 32 is 170 years. This means that it takes 170 years for half of the original amount of Silicon 32 to decay into other elements.

Why is Silicon 32 important?

Silicon 32 is important because it is a radioactive isotope that is used in various scientific and medical applications, including cancer treatment and radiocarbon dating.

How is Silicon 32 produced?

Silicon 32 is produced through the process of neutron activation, where stable silicon atoms are bombarded with neutrons to create the radioactive isotope.

What are the dangers of Silicon 32?

Silicon 32 is a low-energy beta emitter, which means that it poses little danger to humans. However, it can be harmful if ingested or inhaled, so proper safety precautions should be taken when handling it.

How is the half-life of Silicon 32 measured?

The half-life of Silicon 32 is measured using a technique called radiometric dating, which involves measuring the rate of decay of the isotope over time.

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