Physics - Modern Physics Question with Solution | TestHub

$\frac{7}{8}$ fraction decays in $6\mathrm{ }\mathrm{days}$

$\frac{1}{8}$ fraction is active after $6 \mathrm{days}$ 

By using, $\mathrm{N}={\mathrm{N}}_0{\left(\frac{1}{2}\mathrm{ }\right)}^{\mathrm{t}/{\mathrm{T}}_{1/2}}$

$\mathrm{N}=\frac{{\mathrm{N}}_0}{2^{\mathrm{n}}}$

$\frac{{\mathrm{N}}_0}{8}=\frac{{\mathrm{N}}_0}{2^{\mathrm{n}}}$

$\mathrm{n}=3$ 

$6 \mathrm{days} = \mathrm{half} \mathrm{lives}$

${\mathrm{t}}_{\frac{1}{2}}=2\mathrm{ }\mathrm{days}$

$10\mathrm{ }\mathrm{days}=5\mathrm{ }{\mathrm{t}}_{1/2}$

$\mathrm{N}=\frac{{\mathrm{N}}_0}{2^5}=\frac{{\mathrm{N}}_0}{32}$

$\mathrm{So}\mathrm{ }\mathrm{decay}={\mathrm{N}}_0-\frac{{\mathrm{N}}_0}{32}=\frac{31}{32}{\mathrm{N}}_0$