A radioactive element has a half life of 200 days. The percentage of original activity remaining after 83…

For first order reaction,

$\mathrm{K}=\frac{2.303}{\mathrm{t}}\log \frac{{\mathrm{N}}_{\mathrm{o}}}{{\mathrm{N}}_{\mathrm{t}}}$ ... $\left(1\right)$

Also, $\mathrm{K}=\frac{0.693}{{\mathrm{t}}_{{\displaystyle \frac{1}{2}}}}$ ... $\left(2\right)$

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

Hence, $\frac{2.303}{\mathrm{t}}\log \frac{{\mathrm{N}}_{\mathrm{o}}}{{\mathrm{N}}_{\mathrm{t}}}=\frac{0.693}{{\mathrm{t}}_{{\displaystyle \frac{1}{2}}}}$

$\frac{2.303}{83}\log \frac{{\mathrm{N}}_{\mathrm{o}}}{{\mathrm{N}}_{\mathrm{t}}}=\frac{0.693}{200}$

$\frac{{\mathrm{N}}_{\mathrm{o}}}{{\mathrm{N}}_{\mathrm{t}}}=\frac{4}{3}$

$\%\frac{{\mathrm{N}}_{\mathrm{t}}}{{\mathrm{N}}_0}=0.75\times 100$

$=75\%$