A) 5yr
B) 2yr
C) 3yr
D) 10 yr
Correct Answer: C
Solution :
\[t=10\,yr,{{t}_{\frac{1}{2}}}=?\] \[\lambda =\frac{2.303}{t}\log \frac{{{N}_{0}}}{{{N}_{t}}}\] Since, radioactivity decreases \[90%\] in \[10\,yr.\] \[\Rightarrow \] \[{{N}_{0}}=100\] and \[{{N}_{t}}=10\] Thus, \[\lambda =\frac{2.303}{10}\log \frac{100}{10}\] or \[\lambda =\frac{2.303}{10}\] Since, \[{{t}_{1/3}}=\frac{0.693}{\lambda }\] \[\Rightarrow \] \[{{t}_{1/2}}=\frac{0.693\times 10}{2.303}\] \[\Rightarrow \] \[=3\,yr\]
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