A) 1442
B) 1554
C) 1652
D) 1458
Correct Answer: D
Solution :
From \[N={{N}_{0}}{{e}^{-\lambda t}}\] \[1800=2000{{e}^{-\lambda \times 2}}\] \[\frac{9}{10}={{e}^{-2\lambda }}\] \[{{e}^{-\lambda }}={{\left( \frac{9}{10} \right)}^{1/2}}\] Number of nuclei left after 6s, \[N={{N}_{0}}{{e}^{-\lambda t'}}\] \[=2000{{e}^{-\lambda \times 6}}\] \[=2000{{({{e}^{-\lambda }})}^{6}}\] \[=2000{{\left[ {{\left( \frac{9}{10} \right)}^{1/2}} \right]}^{6}}\] \[=2000{{\left( \frac{9}{10} \right)}^{3}}\] \[=2000\times \frac{729}{1000}\] \[=1458\]
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