A) \[17s~\]
B) \[7\,s\]
C) \[5\text{ }s\]
D) \[15\text{ }s\]
Correct Answer: B
Solution :
\[Let\text{ }{{N}_{0}}\text{ }=\text{ }initial\text{ }number\text{ }of\text{ }nuclei\] \[~n={{N}_{0}}-{{N}_{0}}{{e}^{-\lambda \,\times \,2}}\] \[\,1.75\,n={{N}_{0}}-{{N}_{0}}{{e}^{-4\lambda }}\] \[\frac{1}{1.75}=\frac{1-{{e}^{-2\lambda }}}{1-{{e}^{-4\lambda }}}\] \[1+{{e}^{-2\lambda }}\,\,=\,\,1.75\] \[{{e}^{2\lambda }}\,\,=\,\,\frac{1}{0.75}\,\,=\,\,1.33\] \[\,2\lambda =\ell n\,(1.33)\] \[\tau \,\,=\frac{1}{\lambda }\,\,=\,\,\frac{2}{\ell n(1.33)}\,\,=\,\,7s\]You need to login to perform this action.
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