A piece of wood from a recently cut tree shows 20 decays per...

We know that,

The rate of decay at the time $t$ is

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

$2 = 20 \cdot {e}^{\left(- \lambda t\right)}$

$∴ \dfrac{1}{10} = {e}^{\left(- \lambda t\right)}$

$∴ {e}^{\left(\lambda t\right)} = 10$

$⇒ \ln \left({e}^{\lambda t}\right) = \ln 10$

$⇒ \lambdat = \ln 10$

$∵ \lambda = \dfrac{\ln 2}{{T}_{\left(1 / 2\right)}}$

$t = \dfrac{\ln 10}{\ln 2}\left({T}_{1/2}\right)$

$∵ \ln 10 = 2.303$ and $\ln 2 = 0.693$

$t = \dfrac{2.303}{0.693} \times 5730$

$= 3.32 \times 5730$

$\approx 19042years$