Question

An investigator collects a sample of a radioactive isotope with an activity of 490,000 Bq.48 hours later, the activity is 110,000 Bq. Part A For the steps and strategies involved in solving a similar problem, you may view a Video Tutor Solution What is the half-life of the sample?

Answer 1

Answer:

The correct answer is "22.27 hours".

Explanation:

Given that:

Radioactive isotope activity,

= 490,000 Bq

Activity,

= 110,000 Bq

Time,

= 48 hours

As we know,

⇒ $A = A_0 e^{- \lambda t}$

or,

⇒ $\frac{A}{A_0}=e^{-\lambda t}$

By taking "ln", we get

⇒ $ln \frac{A}{A_0}=- \lambda t$

By substituting the values, we get

⇒ $-ln \frac{110000}{490000} = -48 \lambda$

⇒    $-1.4939=-48 \lambda$

                 $\lambda = 0.031122$

As,

⇒ $\lambda = \frac{ln_2}{\frac{T}{2} }$

then,

⇒ $\frac{ln_2}{T_ \frac{1}{2} } =0.031122$

⇒ $T_\frac{1}{2}=\frac{ln_2}{0.031122}$

         $=22.27 \ hours$