A) \[9.6\times {{10}^{10}}\]
B) \[4.2\times {{10}^{10}}\]
C) \[5\times {{10}^{10}}\]
D) \[~1.95\times {{10}^{10}}\]
Correct Answer: C
Solution :
According to Rutherford and Soddy law for radioactive decay. Number of atoms remained undecayed after time t \[=N={{N}_{0}}{{e}^{-\lambda t}}\] Or \[\lambda =\frac{2.303}{t}\log \frac{{{N}_{0}}}{N}\] ?.. (i) and \[\lambda =\frac{0.693}{{{T}_{1/2}}}=\frac{0.693}{1.25\times {{10}^{10}}}\] ... (ii) From Eqs. (i) and (ii), we get \[t=\frac{2.303\times 1.25\times {{10}^{10}}}{0.693}\log \frac{16}{1}\] \[=5\times {{10}^{10}}yr.\]You need to login to perform this action.
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