Question

A hospital uses cobalt-60 in its radiotherapy treatments for cancer patients. Cobalt-60 has a half-life of 7

Answer 1

Answer:

We have 197 g of Co-60 after 18 months.

Step-by-step explanation:

We can use the decay equation.

$M_{f}=M_{i}e^{-\lambda t}$

Where:

• M(f) and M(i) are the final and initial mass respectively
• λ is the decay constant (ln(2)/t(1/2))
• t(1/2) is the half-life of Co
• t is the time at the final amount of m

$M_{f}=228e^{-\frac{ln(2)}{7} 1.5}$    

$M_{f}=197\: g$    

Therefore, we have 197 g of Co-60 after 18 months.

I hope it helps you!