The half-life of a radioactive isotope is $$3 h.$$ If the initial mass of the isotope was $$300$$  $$g,$$ the mass which remained undecayed after $$18$$ $$h$$ would be

Solution :
$$\eqalign{ & {\text{Given,}} \cr & {\text{Half - life}}\left( {{t_{\frac{1}{2}}}} \right) = 3\,h \cr & {\text{Initial mass}}\left( {{N_0}} \right) = 300\,g \cr & {\text{Total time}}\,\left( T \right) = 18\,h \cr & {\text{Mass left}}\,\left( N \right) = ? \cr & {\text{We know that,}} \cr & \frac{N}{{{N_0}}} = {\left( {\frac{1}{2}} \right)^n} \cr & {\text{where, }}n = {\text{number of half - lives}} \cr & n = \frac{{{\text{Total}}\,{\text{time}}}}{{{\text{Half - life}}}} \cr & \,\,\,\,\, = \frac{{18}}{3} \cr & \,\,\,\,\, = 6 \cr & {\text{So,}}\,\,\frac{N}{{300}} = {\left( {\frac{1}{2}} \right)^6} \cr & {\text{or}}\,\,\,\frac{N}{{300}} = \frac{1}{{64}} \cr & N = \frac{{300}}{{64}} \cr & \,\,\,\,\,\,\, = 4.68\,g \cr} $$