One microgram of radioactive sodium $$_{11}N{a^{24}}$$   with a half-life of $$15$$ $$h$$ was injected into a living system for a bio-assay. How long will it take for the radioactivity to fall to $$25\% $$  of the initial value?

Solution :
$$\eqalign{ & {t_{\frac{1}{2}}}\,o{\text{f}}{\,_{11}}N{a^{24}} = 15\,h \cr & {\left[ R \right]_0} = 1.0\,\mu g \cr & \left[ R \right] = 1.0 \times \frac{{25}}{{100}} = \frac{{25}}{{100}} = 0.25\,\mu g \cr & k = \frac{{0.6932}}{{{t_{\frac{1}{2}}}}} = \frac{{0.6932}}{{15}}\,{h^{ - 1}} \cr & \therefore \,\,k = \frac{{2.303}}{t}{\text{log}}\frac{{{{\left[ R \right]}_0}}}{{\left[ R \right]}} \cr & \therefore \,\,\frac{{0.6932}}{{15}} = \frac{{2.303}}{t}{\text{log}}\frac{1}{{0.25}} \cr & \frac{{0.6932}}{{15}} = \frac{{2.303}}{t}{\text{log}}\frac{{100}}{{25}} = \frac{{2.303}}{t}{\text{log}}4 \cr & \frac{{0.6932}}{{15}} = \frac{{2.303}}{t} \times 0.6020 \cr & t = \frac{{15 \times 2.303 \times 0.6020}}{{0.6932}} \cr & \,\,\, = \frac{{20.7995}}{{0.6932}} \cr & \,\,\, = 30.00\,h \cr} $$