Question

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?

Answer 1

`color(blue)(" The half life is 15 hours.")`

Explanation:

We need to find an equation of the form:

`A(t)=A(0)e^(kt)`

Where:

`bb(A(t))=` the amount after time t.

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

`bbk=` the growth/decay factor.

`bbe=` Euler's number.

`bbt=` time, in this case hours.

We are given:

`A(0)=160`

`A(45)=20`

We need to solve for `bbk`:

`20=160e^(45k)`

Divide by 160:

`1/8=e^(45k)`

Taking natural logarithms of both sides:

`ln(1/8)=45kln(e)`

`ln(e)=1`

Hence:

`ln(1/8)=45k`

Dividing by 45:

`ln(1/8)/45=k`

`:.`

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

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

`A(t)=160(1/8)^(t/45)`

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

`A(t)=80`

So we need to solve for t in:

`80=160*(1/8)^(t/45)`

`80/160=(1/8)^(t/45)`

`1/2=(1/8)^(t/45)`

Taking natural logarithms:

`ln(1/2)=t/45ln(1/8)`

`45*(ln(1/2))/(ln(1/8))=t=15`

The half life is 15 hours.

Answer 2

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 * 15`

So the half-life is 15 years.

• More formal way

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

`X(t) = X_o (1/2)^(t/tau) qquad square`

So:

`X(0) = X_o, X(tau) = X_o/2, X(2tau) = X_o/4,...`

Plugging the values that are given into `square` :

`20 = 160 * (1/2)^(45/tau)`

`implies (1/2)^(45/tau) = 1/8 qquad qquad [= (1/2)^3]`

`implies 45/tau = 3 implies tau = 15`