In a radioactive material, fraction of active material remaining after time t is 9/16. The fraction that was remaining after t/2 is
Solution
First order decay
<br><br>N(t) = N<sub>0</sub>e<sup>-$\lambda$t</sup>
<br><br>Given ${{N\left( t \right)} \over {{N_0}}} = {9 \over {16}} =$ e<sup>-$\lambda$t</sup>
<br><br>N(t/2) = N<sub>0</sub>e<sup>-$\lambda$(t/2)</sup>
<br><br>${{N\left( {t/2} \right)} \over {{N_0}}} = \sqrt {{e^{ - \lambda t}}}$ = $\sqrt {{9 \over {16}}}$ = ${3 \over 4}$
<br><br>$\Rightarrow$ $N\left( {t/2} \right) = {3 \over 4}{N_0}$
About this question
Subject: Physics · Chapter: Atoms and Nuclei · Topic: Radioactivity
This question is part of PrepWiser's free JEE Main question bank. 184 more solved questions on Atoms and Nuclei are available — start with the harder ones if your accuracy is >70%.