Explanation:
C We know that, radioactive law,
$\mathrm{N}=\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-\lambda \mathrm{t}}$
According to question,
$\mathrm{N}=\mathrm{N}_{\mathrm{o}}-\mathrm{N}_{\mathrm{o}} \times \frac{7}{8}=\frac{\mathrm{N}_{\mathrm{o}}}{8}$
On solving equation (i), we get-
$\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\mathrm{e}^{-\lambda \mathrm{t}}$
$\text { Or } \frac{\mathrm{N}_{\mathrm{o}}}{\mathrm{N}}=\mathrm{e}^{\lambda t}$
$\text { Taking natural log on both the side, }$
\(\mathrm{t}=\frac{\ln \left(\frac{\mathrm{N}_{\mathrm{o}}}{\mathrm{N}}\right)}{\lambda}\)
\(\mathrm{t} \propto \ln \left(\frac{\mathrm{N}_{\mathrm{o}}}{\mathrm{N}}\right)\)
\(\text { So, } \quad \frac{\mathrm{t}_1}{\mathrm{t}_2}=\frac{\ln \left(\frac{\mathrm{N}_{\mathrm{o}}}{\mathrm{N}}\right)_1}{\ln \left(\frac{\mathrm{N}_{\mathrm{o}}}{\mathrm{N}}\right)_2}\)
\(\frac{6}{10}=\frac{\ln \left(\frac{\mathrm{N}_{\mathrm{o}}}{\mathrm{N}_{\mathrm{o}} / 8}\right)_1}{\ln \left(\frac{\mathrm{N}_{\mathrm{o}}}{\mathrm{N}_2}\right)_2}\)
\(\frac{3}{5}=\frac{\ln 8}{\ln \left(\frac{\mathrm{N}_{\mathrm{o}}}{\mathrm{N}_2}\right)}\)
\(\ln \left(\frac{\mathrm{N}_{\mathrm{o}}}{\mathrm{N}_2}\right)=\frac{5}{3} \ln 8\)
\(\ln \left(\frac{\mathrm{N}_{\mathrm{o}}}{\mathrm{N}_2}\right)=\ln (8)^{5 / 3}=\ln 32\)
\(\frac{\mathrm{N}_{\mathrm{o}}}{\mathrm{N}_2}=32\)
\(\text { or } \quad \frac{\mathrm{N}_2}{\mathrm{~N}_{\mathrm{o}}}=\frac{1}{32}\)
Decay fraction \(=\mathrm{N}_{\mathrm{o}}-\frac{\mathrm{N}_{\mathrm{o}}}{32}=\frac{31}{32} \mathrm{~N}_{\mathrm{o}}\)