Calculus.. help with questions (1 Viewer)

excellence1 · Mar 17, 2016

The half life of radium is 1600 years
a) find the percentage of radium that will be decayed after 500 years
b) find the number of years it will take for 75% radium to decay

InteGrand · Mar 17, 2016

excellence1 said:
The half life of radium is 1600 years
a) find the percentage of radium that will be decayed after 500 years
b) find the number of years it will take for 75% radium to decay

$\noindent Let time $t$ be measured in years. The half-life is given to be $\tau = 1600$. Let $A(t)$ be the amount of radium at time $t\geq 0$. Since we are only asked about percentages, we can just assume $A(0) = 1$. The standard exponential decay formula is then $A(t) = 2^{-\frac{t}{\tau}}$ $\Big{(}$in general, it is $A(t) = A(0)\cdot 2^{-\frac{t}{\tau}}$, but we let $A(0)=1$$\Big{)}$.$

$\noindent a) We have $A(500) = 2^{-\frac{500}{\tau}}= 2^{-\frac{500}{1600}}=2^{-\frac{5}{16}}=0.8052\ldots$ (this is the proportion that remains after 500 years). So the percentage that is decayed is found by subtracting this from 1, i.e. $1-2^{-\frac{5}{16}}\approx 0.1948 = 19.48 \%$.$

$\noindent (b) We are after the value of $t$ such that $A(t) = 0.25$ ($25 \%$ remaining). After one half-life, there is $50\%$ remaining. After two half-lives, there is $25\%$ remaining. But two half-lives is 3200 years, so this is the answer.$

Silly Sausage · Mar 17, 2016

There are many ways to do this question, this is typically what they expect for an HSC response (except the annotations)
The generic decay equation:
$N=N_0e^{-kt}$
Where $N_0$ is the initial amount of the element when $t=0$ and if you plug it into the equation you get $N=N_0$ . We can assume $N_0=1$
$k$ is the constant of proportionality.

a)Half life $t_{\frac{1}{2}}=1600$
Hence, since half of the original amount remains, $N=0.5$ and plugging $1600$ into $t$ in the decay equation we get:
$0.5=e^{-1600k}$
Solving for $k$
$ln(0.5)=ln(e^{-1600k})$
$ln(0.5)=-1600k$
$k=\frac{-ln(0.5)}{1600}$

Now we use the decay equation to find $N$ when $t=500$ .
$N=e^{-500k}$
$N=0.805$
$1-0.805=0.195$
Hence $19.5\%$ remains.

b) $75%$ decayed means $N=0.25$ .
$ln(0.25)=ln(e^{-kt})$
Just solve for $t$ .

There are many other methods such as Integrand's method which are simpler but they usually expect you to do like this.

InteGrand · Mar 17, 2016

Yeah, I intentionally used the base 2 form because it's more convenient when given the half-life. If they had instead said something like a "decay constant", we would have done it Silly Sausage's way with base e (the k there is known as the decay constant).

Silly Sausage · Mar 17, 2016

InteGrand said:
Yeah, I intentionally used the base 2 form because it's more convenient when given the half-life. If they had instead said something like a "decay constant", we would have done it Silly Sausage's way with base e (the k there is known as the decay constant).

Or if you're feeling extra lazy (NOT RECOMMENDED FOR HSC STUDENTS)...
$t_{\frac{1}{2}}=\frac{ln(2)}{k}$
$ln(\frac{N_0}{N_t})=kt$

InteGrand · Mar 17, 2016

Silly Sausage said:
Or if you're feeling extra lazy (NOT RECOMMENDED FOR HSC STUDENTS)...
$t_{\frac{1}{2}}=\frac{ln(2)}{k}$
$ln(\frac{N_0}{N_t})=kt$

Haha, that's like the non-lazy way, isn't it? The lazy way is what I did (go straight to base 2

).

Silly Sausage · Mar 17, 2016

InteGrand said:
Haha, that's like the non-lazy way, isn't it? The lazy way is what I did (go straight to base 2 ).

true haha I remember going through so many ways to do this is pchem

leehuan · Mar 18, 2016

Calculus.. help with questions (1 Viewer)

excellence1

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