a
Consider active Nuclei at \(t =0\) is \(N_{0}\) then,

The active nuclei at \(t =10 hr\)

\(N_{1}=N_{0} e^{-10 \lambda}\)

The active nuclei at \(t=11 hr\)

\(N_{2}=N_{0} e^{-11 \lambda}\)

The expression for the decay percentage,

\(\%\) decay \(=\left(\frac{N_{1}-N_{2}}{N_{1}}\right) \times 100\)

\(=\left(\frac{N_{0} e^{-10 \lambda}-N_{0} e^{-11 \lambda}}{N_{0} e^{-10 \lambda}}\right)\)

\(=\left(1-\frac{1}{e^{\lambda}}\right) \times 100\)

\(=\left(1-\frac{1}{e^{\ln 2 / T_{1 / 2}}}\right) \times 100 \times 100\)

The half life of the element is, \(T_{1 / 2}=69.3 hr\) therefore,

\(\%\) decay \(=\left(1-\frac{1}{e^{0.01}}\right) \times 100\)

\(=1 \%\)