# How to Calculate the decay constant, half-life and the mean life for a radioisotope which activity is found to decrease by 25% in one week??

##### 1 Answer
May 21, 2018

$\lambda \approx 0.288 \textcolor{w h i t e}{l} {\text{week}}^{- 1}$
${t}_{\frac{1}{2}} \approx 2.41 \textcolor{w h i t e}{l} \text{weeks}$
$\tau \approx 3.48 \textcolor{w h i t e}{l} \text{weeks}$

#### Explanation:

The first-order decay constant $\lambda$ comprises the expression for the decay activity at a particular time $A \left(t\right)$.

$A \left(t\right) = {A}_{0} \cdot {e}^{- \lambda \cdot t}$
${e}^{- \lambda \cdot t} = \frac{A \left(t\right)}{A} _ 0 = \frac{1}{2}$

Where ${A}_{0}$ the activity at time zero. The question suggests that A(1color(white)(l)"week")=(1-25%)*A_0, thus

e^(-lambda*1color(white)(l)"week")=(A(1color(white)(l)"week"))/(A_0)=0.75

Solve for $\lambda$:

lambda=-ln(3/4)/(1color(white)(l)"week")~~0.288color(white)(l)"week"^(-1)

By the (self-explanatory) definition of decay half-life

${e}^{- \lambda \cdot {t}_{\frac{1}{2}}} = \frac{A \left({t}_{\frac{1}{2}}\right)}{A} _ 0 = \frac{1}{2}$
$- \lambda \cdot {t}_{\frac{1}{2}} = \ln \left(\frac{1}{2}\right)$
${t}_{\frac{1}{2}} = \ln \frac{2}{\lambda} \approx 2.41 \textcolor{w h i t e}{l} \text{weeks}$

Mean life $\tau$ represents the arithmetic mean of all individual lifetimes and is equal to the reciprocal of the decay constant.

$\tau = \frac{1}{\lambda} = 3.48 \textcolor{w h i t e}{l} \text{weeks}$