# What is the half-life of (Na^24) if a research assistant made 160 mg of radioactive sodium (Na^24) and found that there was only 20 mg left 45 h later?

Jul 7, 2018

$\textcolor{b l u e}{\text{ The half life is 15 hours.}}$

#### Explanation:

We need to find an equation of the form:

$A \left(t\right) = A \left(0\right) {e}^{k t}$

Where:

$\boldsymbol{A \left(t\right)} =$ the amount after time t.

bb(A(0)= the amount at the start. i.e. t = 0.

$\boldsymbol{k} =$ the growth/decay factor.

$\boldsymbol{e} =$ Euler's number.

$\boldsymbol{t} =$ time, in this case hours.

We are given:

$A \left(0\right) = 160$

$A \left(45\right) = 20$

We need to solve for $\boldsymbol{k}$:

$20 = 160 {e}^{45 k}$

Divide by 160:

$\frac{1}{8} = {e}^{45 k}$

Taking natural logarithms of both sides:

$\ln \left(\frac{1}{8}\right) = 45 k \ln \left(e\right)$

$\ln \left(e\right) = 1$

Hence:

$\ln \left(\frac{1}{8}\right) = 45 k$

Dividing by 45:

$\ln \frac{\frac{1}{8}}{45} = k$

$\therefore$

$A \left(t\right) = 160 {e}^{t \left(\ln \frac{\frac{1}{8}}{45}\right)}$

A(t)=160e^(t/45(ln(1/8))

$A \left(t\right) = 160 {\left(\frac{1}{8}\right)}^{\frac{t}{45}}$

Since by definition the half life is the time period when we have half of the starting amount:

$A \left(t\right) = 80$

So we need to solve for t in:

$80 = 160 \cdot {\left(\frac{1}{8}\right)}^{\frac{t}{45}}$

$\frac{80}{160} = {\left(\frac{1}{8}\right)}^{\frac{t}{45}}$

$\frac{1}{2} = {\left(\frac{1}{8}\right)}^{\frac{t}{45}}$

Taking natural logarithms:

$\ln \left(\frac{1}{2}\right) = \frac{t}{45} \ln \left(\frac{1}{8}\right)$

$45 \cdot \frac{\ln \left(\frac{1}{2}\right)}{\ln \left(\frac{1}{8}\right)} = t = 15$

The half life is 15 hours.

Jul 7, 2018

15 hours

#### Explanation:

• Quick Way

As the amount of a decaying substance halves over each half-life (hence the name), halving the amount in steps requires 3 steps to get from 160 to 20:

• $160 \to 80 \to 40 \to 20$

And $45 = 3 \cdot 15$

So the half-life is 15 years.

• More formal way

For half-life $\tau$, where $X \left(t\right)$ is the amount (mass/ number of particles/ etc) remaining at time t:

$X \left(t\right) = {X}_{o} {\left(\frac{1}{2}\right)}^{\frac{t}{\tau}} q \quad \square$

So:

$X \left(0\right) = {X}_{o} , X \left(\tau\right) = {X}_{o} / 2 , X \left(2 \tau\right) = {X}_{o} / 4 , \ldots$

Plugging the values that are given into $\square$ :

$20 = 160 \cdot {\left(\frac{1}{2}\right)}^{\frac{45}{\tau}}$

$\implies {\left(\frac{1}{2}\right)}^{\frac{45}{\tau}} = \frac{1}{8} q \quad q \quad \left[= {\left(\frac{1}{2}\right)}^{3}\right]$

$\implies \frac{45}{\tau} = 3 \implies \tau = 15$