# A researcher made160mg of radioactive sodium(Na24) and found that there was only 20 mg left after 45 hours. a). What is the half-life of Na24?

### 1 Answer | Add Yours

Since the amount of NA24 is decreasing, let's apply the formula of exponential decay, which is:

`y = ae^(kt)`

where y- amount of substance at time t, a - amount of substance at the start and k - rate of decay.

To solve for the half-life of NA24, we have to determine k first. So, substitute y=20, a=160 and t=45 to the formula.

`20=160e^(45t)`

To simplify the equation, divide both sides by 160.

`20/160=(160e^(45t))/160`

`0.125=e^(45t)`

To bring down k, take the natural logarithm of both sides.

`ln0.125 = lne^(45k)`

Then at the right side of the equation, apply the power property of logarithm which is ` ln x^m= m*ln x ` .

`ln0.125=45k*lne`

Note that ln e= 1.

`ln 0.125=45k*1`

`ln0.125=45k`

And divide both sides by 45.

`(ln0.125)/45=(45k)/45`

`k=-0.046`

Now that we know the value of k, let's solve proceed to solve for half-life of NA24.

Note that half-life means the time when y = a/2. So to determine t, substitute y = a/2 and k=-0.046 to the formula of exponential decay and follow the steps above.

`y =ae^(kt)`

`a/2=ae^(-0.046t)`

Divide both sides by a.

`(a/2)/a=(ae^(-0.046t))/a`

`1/2=e^(-0.046t)`

Take the natural logarithm of both sides.

`ln(1/2)=ln e^(-0.046t)`

Then, apply the power property of logarithm.

`ln(1/2)=-0.046t*lne`

`ln(1/2)=-0.046t*1`

`ln(1/2)=-0.046t`

And, divide both sides by -0.046.

`(ln(1/2))/(-0.046)=(-0.046t)/(-0.046)`

`t=15.068`

**Hence, the half-life of NA24 is 15.068 hours.**