A) \[3.91\times {{10}^{3}}\text{ }se{{c}^{-1}}\]
B) \[5.81\times {{10}^{4}}\text{ }se{{c}^{-1}}\]
C) \[6.22\times {{10}^{5}}\text{ }se{{c}^{-1}}\]
D) None of these
Correct Answer: A
Solution :
[a] If \[{{A}_{0}}\] is the initial activity of radioactive sample then activity at any time \[A={{A}_{0}}{{e}^{-\lambda t}}\] or \[1\times {{10}^{6}}=4\times {{10}^{6}}{{e}^{-\lambda \times 20}}\] or \[{{e}^{-20\lambda }}=\frac{1}{4}\] The count rate after 100 hour is given by \[A'={{A}_{0}}{{e}^{-\lambda \times 100}}={{A}_{0}}{{e}^{-100\lambda }}={{A}_{0}}{{[{{e}^{-20\lambda }}]}^{5}}\] \[=4\times {{10}^{6}}{{\left[ \frac{1}{4} \right]}^{5}}=3.91\times {{10}^{3}}\]per second
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