Given \(A_{0}=10\,A\)
\(\lambda=\frac{0.693}{t_{1 / 2}}=\frac{0.693}{30}\)
\(t_{1 / 2}=\frac{2.303}{\lambda} \log \frac{A_{0}}{A}\)
\(=\frac{2.303}{0.693 / 30} \log \frac{10\,A}{A}\)
\(=\frac{2.303 \times 30}{0.693} \times \log 10\)
\(=100\) days .
| $p ( mm Hg )$ | $50$ | $100$ | $200$ | $400$ |
| સાપેક્ષ $t _{1 / 2}( s )$ | $4$ | $2$ | $1$ | $0.5$ |
પ્રક્રિયાનો ક્રમ શોધો.
$\mathop {2{N_2}{O_5}}\limits_{{\rm{(in}}\,\,{\rm{CC}}{{\rm{l}}_4}{\rm{)}}} \to \mathop {4N{O_2}}\limits_{{\rm{(in}}\,\,{\rm{CC}}{{\rm{l}}_4}{\rm{)}}} + {O_2}$