b
According to radioactive decay law

\(N=N_{e} e^{-\lambda t}\)

where \(N_{0}=\) Number of radioactive nuclei at time \(t=0\)

\(N=\) Number of radioactive nuclei left undecayed at any time \(t\)

\(\lambda=\) decay constant

At time \(t_{2}, \frac{2}{3}\) of the sample had decayed

\(\therefore N=\frac{1}{3} N_{0}\)

\(\therefore \frac{1}{3} N_{0}=N_{0} e^{-\lambda t_{2}}\)  ...... \((i)\)

At time \(t_{1}, \frac{1}{3}\) of the sample had decayed,

\(\therefore N=\frac{2}{3} N_{0}\)

\(\therefore \frac{2}{3} N_{0}=N_{0} e^{-\lambda t_{1}}\)  ...... \((ii)\)

Divide \((i)\) by \((ii)\), we get

\(\frac{1}{2} = \frac{{{e^{ - \lambda {t_2}}}}}{{{e^{ - \lambda {t_1}}}}}\) \( \Rightarrow \frac{1}{2} = {e^{ - \lambda \left( {{t_2} - {t_1}} \right)}}\)

\({\lambda\left(t_{2}-t_{1}\right)=\ln 2} \)

\({t_2} - {t_1} = \frac{{\ln 2}}{\lambda } = \frac{{\ln 2}}{{\left( {\frac{{\ln 2}}{{{T_{1/2}}}}} \right)}}\) \(\left( {\because \lambda  = \frac{{\ln 2}}{{{T_{1/2}}}}} \right)\)

\({=T_{10}=50 \text { days }}\)