Question

consider the radioisotope $^{41}$ca (half life = 1.03e5 years). what is the first - order rate constant for $^{41}$ca?

Explanation:

Step1: Recall the half - life formula for first - order reactions

The formula for the half - life ($t_{1/2}$) of a first - order reaction is $t_{1/2}=\frac{\ln 2}{k}$, where $k$ is the first - order rate constant.

Step2: Rearrange the formula to solve for $k$

We can rewrite the formula as $k = \frac{\ln 2}{t_{1/2}}$. Given $t_{1/2}=1.03\times 10^{5}$ years and $\ln 2\approx0.693$.
$k=\frac{0.693}{1.03\times 10^{5}\text{ years}}$

Step3: Calculate the value of $k$

$k = 6.73\times 10^{-6}\text{ year}^{-1}$

Answer:

$6.73\times 10^{-6}\text{ year}^{-1}$