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

Answer:

$t_{1/2}=45.09 d$    

Step-by-step 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:

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

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

$ln(\frac{M}{M_{0}})=-\frac{ln(2)}{t_{1/2}}t$

$t_{1/2}=-\frac{ln(2)}{ln(\frac{M}{M_{0}})}t$

$t_{1/2}=-\frac{ln(2)}{ln(\frac{10.75}{50})}100$                

$t_{1/2}=45.09 d$    

I hope it helps you!