Anonymous
Hi could you please help me with this? The activity of a radioactive source falls from 600 to 100 in 6 hours. What is the half-life of the isotope? I'm completely stuck :(

Hi!
You can solve your problem using the equation $N(t)=N_{0}e^{-\lambda t}$ where N(t) is the number of atoms at a given time t, N_0 is the number of atoms at t = 0, and lambda is the decay constant.

The activity (A) can be written as $A = \lambda N$ so we can multiply both factors for lambda and have the equation that will solve our problem:
$A(t)=A_{0}e^{-\lambda t}$  since we know A(t), A_0 and t from the data of our problem we can find lambda.
$100 = 600 e^{-6 \lambda }$ , $\lambda=-\frac{ln(\frac{1}{6})}{6}$, $\lambda=0.299$

Since $t_{1/2}=\frac{ln2}{\lambda}$ the half-life of our isotope is t_(1/2) = 2.32 hours.

I really hoped this helped you! Let us know if you need any more clarification on the matter of radioactive decay!
-Alex

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