\(N_1=N_0-N_0 e^{-\lambda t}\)
Since \(N_1=\frac{N_0}{10}\)
\(N_0 e^{-\lambda t}=\frac{9 N_0}{10}\)
\(\therefore e^{-\lambda t}=\frac{9}{10}\)
Amount left \(=\frac{9 N_0}{10}\)
\(N_2=N_0-N_0 e^{-2 \lambda t}\)
or \(\quad N_2=N_0-\frac{81}{100} N_0\)
or \(\quad N_2=19 \%\) of \(N_0\)
$N _{ A }=6 \times 10^{23}$ આપેલ છે.