# Thread: Half-life.

1. ## Half-life.

A radioactive substance has a half-life of 20 days.

a) How much time is required so that only $\frac{1}{32}$ of the original amount remains?
b) Find the rate of decay at this time.

My attempts:
a) Since after 20 days the substance decays by half of its mass, after 40 days it decays to ${1}{4}$, and after 60 days it decays to $\frac{1}{8}$. After 80 days it decays to $\frac {1}{16}$ of its original mass and after 100 days it decays to ${1}{32}$ of its original mass. Therefore, it will take 100 days for the substance to decay to [tex]\frac{1}{32}/MATH] of its original mass.
b) $f(t) = \frac{1}{32}(\frac{1}{2})^{\frac{t}{20}$
$f'(t) =\frac{1}{32}(\frac{1}{2})^{\frac{t}{20}} × (ln\frac{1}{2}) × (\frac{1}{20})$
$f'(100) =\frac{1}{32}(\frac{1}{2})^{\frac{100}{20}} × (ln\frac{1}{2}) × (\frac{1}{20})$
$f'(100) = -3.384 × 10^{-5}$ is the rate of decay when $t = 100$ days.

I just wanted to know if my answers are correct? I checked with derivative calculators and they have given me very different answers for b. Thanks in advance.

2. Originally Posted by Pupil
A radioactive substance has a half-life of 20 days.

a) How much time is required so that only $\frac{1}{32}$ of the original amount remains?
b) Find the rate of decay at this time.

My attempts:
a) Since after 20 days the substance decays by half of its mass, after 40 days it decays to ${1}{4}$, and after 60 days it decays to $\frac{1}{8}$. After 80 days it decays to $\frac {1}{16}$ of its original mass and after 100 days it decays to ${1}{32}$ of its original mass. Therefore, it will take 100 days for the substance to decay to [tex]\frac{1}{32}/MATH] of its original mass.
b) $f(t) = \frac{1}{32}(\frac{1}{2})^{\frac{t}{20}$
$f'(t) =\frac{1}{32}(\frac{1}{2})^{\frac{t}{20}} × (ln\frac{1}{2}) × (\frac{1}{20})$
$f'(100) =\frac{1}{32}(\frac{1}{2})^{\frac{100}{20}} × (ln\frac{1}{2}) × (\frac{1}{20})$
$f'(100) = -3.384 × 10^{-5}$ is the rate of decay when $t = 100$ days.

I just wanted to know if my answers are correct? I checked with derivative calculators and they have given me very different answers for b. Thanks in advance.
The first part is correct However you could have solved it this way

$\displaystyle \left( \frac{1}{2}\right)^\frac{x}{20}=\frac{1}{32}=\left ( \frac{1}{2}\right)^5 \iff \frac{x}{20}=5 \implies x=100$

For the 2nd part all we know is

$\displaystyle f(t)=A_0\left( \frac{1}{2}\right)^{\frac{x}{20}}=A_0 e^{\frac{x}{20}\cdot \ln(\frac{1}{2})}$

Now taking the derivative gives

$\displaystyle f'(t)=\left( {\frac{1}{20}\cdot \ln(\frac{1}{2})}\right)A_0 e^{\frac{x}{20}\cdot \ln(\frac{1}{2})}=A_0\left( {\frac{1}{20}\cdot \ln(\frac{1}{2})}\right)\left( \frac{1}{2}\right)^{\frac{x}{20}}$

3. Ah, I forgot to take the log of both sides and solve it much easier.

Okay, so I derived the equation correctly? And the question asks me to find the rate of decay when $t = 100$ days. Knowing the derivative and the quantity at the time, wouldn't I just need to plug in 100 to find the rate of decay?

4. Yes, f'(100) should tell you the rate of decay on the 100th day.

5. Originally Posted by NOX Andrew
Yes, f'(100) should tell you the rate of decay on the 100th day.
So, is the rate of decay $-3.384 × 10^{-5}$ when $t = 100$?

6. What is the original amount? It should be the original amount times -log(2)/640.

7. Originally Posted by NOX Andrew
What is the original amount? It should be the original amount times -log(2)/640.
It was not given. I only have $\frac{1}{32}$ of the original substance to work with.