Hcc.18 the half life of silicon-32 is 710 years.

[karush]

oops this is a pre calc question

$\tiny{hcc.18}$
the half life of silicon-32 is $710$ years.
If $10g$ are present now
how much will be present in 600 yrs?
to find out $k$ using
$$A=A_0 e^{kt}$$
$$\frac{1}{2}=e^{k \cdot 710}$$
$$\ln\left[\frac{1}{2}\right]=k\cdot710$$
$$\frac{\ln(1/2)}{710}=k=-0.00097626$$
I continued with this but the answer was ?

 

[MarkFL]

Given that we know the half-life $H$, we may write:

\(\displaystyle A(t)=A_0\left(\frac{1}{2}\right)^{\frac{t}{H}}\)

Plug in the given data:

\(\displaystyle A(t)=10\left(\frac{1}{2}\right)^{\frac{t}{710}}\)

And so:

\(\displaystyle A(600)=?\)

 

[karush]

Re: hcc.18 the half life of silicon-32 is \$710\$ years.

where is $e$ ??

 

[MarkFL]

Re: hcc.18 the half life of silicon-32 is \$710\$ years.

where is $e$ ??

Between 2 and 3...hehehe.

Seriously though...we know:

\(\displaystyle A(t)=A_0e^{-kt}\)

And we know:

\(\displaystyle A(710)=\frac{1}{2}A_0=A_0e^{-710k}\implies \frac{1}{2}=e^{-710k}\implies e^{-k}=\left(\frac{1}{2}\right)^{\frac{1}{710}}\)

And so we have:

\(\displaystyle A(t)=A_0e^{-kt}=A_0\left(e^{-k}\right)^t=A_0\left(\frac{1}{2}\right)^{\frac{t}{710}}\)

 

[karush]

ok
I was pacing the floor wondering:D

great help and insight
again from MHB

 

[MarkFL]

ok
I was pacing the floor wondering:D

great help and insight
again from MHB

Mentally, I didn't go through all that I posted...I simply reasoned that for every 710 years the amount of substance is cut in half, which leads directly to the relation I posted. However, I felt it should be mathematically justified. (Yes)