A) \[1.25\times {{10}^{-17}}\,\,J\]
B) \[4.4\times {{10}^{-19}}\,\,J\]
C) \[2.5\times {{10}^{-19}}\,\,J\]
D) \[1.33\times {{10}^{8}}\,m/s\]
Correct Answer: B
Solution :
From Rutherford and Soddy law for radioactive decay, if \[98\,cm\] be the number of atoms of radioactive substance lift at some instant of time, then \[E={{E}_{0}}\,\,\sin \left( kx-\omega t \right)\] Where \[B={{B}_{0}}\,\,\sin \left( kx-\omega t \right)\] is original number of atoms and \[{{E}_{0}}={{B}_{0}}\] is number of half-lives. Given, \[{{E}_{0}}\omega ={{B}_{0}}k\] years, \[{{E}_{0}}{{R}_{0}}=\omega k\] \[{{E}_{0}}k={{B}_{0}}\omega \] \[r\] \[R\] \[\frac{3r}{4}\] Note: The number of atoms of a radioactive substance that is, the radioactive activity of he substance continuously decreases with time.
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