Question

An unknown radioactive sample is observed to decrease in activity by a factor of two in a one hour period. What is its half-life?

Answer 1

Answer:

The half-life is $t_{1/2} = 1.005 h$

Explanation:

Using the decay equation we have:

$A=A_{0}e^{-\lambda t}$

Where:

• λ is the decay constant
• A(0) the initial activity
• A is the activity at time t

We know the activity decrease by a factor of two in a one hour period (t = 1 h), it means that $A = \frac{A_{0}}{2}$

$\frac{A_{0}}{2}=A_{0}e^{-\lambda*1 h}$

$0.5=e^{-\lambda*1 h}$

Taking the natural logarithm on each side we have:

$ln(0.5)=-\lambda$

$\lambda=0.69 h^{-1}$

Now, the relationship between the decay constant λ and the half-life t(1/2) is:

$\lambda = \frac{ln(2)}{t_{1/2}}$

$t_{1/2} = \frac{ln(2)}{\lambda}$

$t_{1/2} = \frac{ln(2)}{0.69}$

$t_{1/2} = 1.005 h$

I hope it helps you!