A radioactive element has a half life of 200 days. The percentage...

For first order reaction,

$K = \dfrac{2 . 303}{t}log\dfrac{{N}_{o}}{{N}_{t}}$ ... \left(1\right)$</p><p>Also,$K = \dfrac{0 . 693}{{t}_{\dfrac{1}{2}}}$... \left(2\right)$

Equating \left(1\right)$and \left(2\right)$

Hence, $\dfrac{2 . 303}{t}log\dfrac{{N}_{o}}{{N}_{t}} = \dfrac{0 . 693}{{t}_{\dfrac{1}{2}}}$

$\dfrac{2 . 303}{83}log\dfrac{{N}_{o}}{{N}_{t}} = \dfrac{0 . 693}{200}$

$\dfrac{{N}_{o}}{{N}_{t}} = \dfrac{4}{3}$

$$

$= 75$