Question

A sample of Iron-59 contains 50 mg. One hundred days later, the sample contains 10.75 mg. What is the half-life of iron-59, in days? Round to the nearest tenth.

Answer 1

`t_(1/2)=45.09 d`

Explanation:

We can use the decay equation:

`M=M_(0)e^(-\lambda t)`

• M(0) is the initial mass
• M is the mass after t (1000 days) time
• λ is the decay constant

But:

`\lambda = ln(2)/t_(1/2)`

• t(1/2) is the half-life.

So, we can rewrite the initial equation:

`M=M_(0)e^{-(ln(2))/(t_(1/2))t}`

Now, we just need to solve it for t(1/2):

`ln((M)/(M_(0)))=-(ln(2))/(t_(1/2))t`

`t_(1/2)=-(ln(2))/(ln((M)/(M_(0))))t`

`t_(1/2)=-(ln(2))/(ln((10.75)/(50)))100`

`t_(1/2)=45.09 d`

I hope it helps you!