Question

Radio active carbon - 14 a wood sample decays with a half life of 5700 years. The fraction of the radio active carbon - 14, that remains after a decay period of 17,100 years is:

Answer 1

$\frac{N}{N_o}={\left(\frac{1}{2}\right)}^{\frac{t}{t_{1/2}}}$

$⟹\frac{N}{N_o}={\left(\frac{1}{2}\right)}^{\frac{17100}{5700}}={\left(\frac{1}{2}\right)}^3$

$⟹N=\frac{N_o}{8}$

Hence, after $17100$ years, $\frac{1}{8}$th of $C-14$ remains.