question_answer

The half-life of a radioactive material is 3 h. If the initial amount is 300 g, then after 18 h, it will remain:  [AIPMT 2000]

A)  4.68 g  

B)    46.8 g C)       9.375 g           D)              93.75 g Correct Answer: A Solution : Number of half-lives

\[n=\frac{t}{T}=\frac{18}{3}=6\]

Amount remained after n half-lives

\[M={{M}_{0}}{{\left( \frac{1}{2} \right)}^{n}}\]

Given,            \[{{M}_{0}}=300\,g\]

\[\therefore \]   \[M=300\,{{\left( \frac{1}{2} \right)}^{6}}\]

\[=300\times \frac{1}{64}=4.68\,g\]

Alternative: Total time of decay given

\[t=\frac{2.303}{\lambda }{{\log }_{10}}\left( \frac{300}{M} \right)\]

but     \[\lambda =\frac{0.693}{T}\]

\[=\frac{0.693}{3}=0.231/h\]

\[\therefore \]   \[t=\frac{2.303}{0.231}{{\log }_{10}}\left( \frac{300}{M} \right)\]

Given,            \[t=18\,h\]

So,       \[18=\frac{2.303}{0.231}{{\log }_{10}}\left( \frac{300}{M} \right)\]

or \[{{\log }_{10}}\left( \frac{300}{M} \right)=\frac{0.231}{2.303}\times 18\]

or         \[\frac{300}{M}={{(10)}^{1.8}}\]

or \[M=\frac{300}{{{(10)}^{1.8}}}=4.68\,g\]