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The half life of radioactive radon is 3.8 days . The time , at end of which $\;\large\frac{1}{20}^{th}\;$ of radon sample will remain undecayed is

$(a)\;18 \;days\qquad(b)\;16.5 \;days\qquad(c)\;17.6 \;days\qquad(d)\;19.2 \;days$

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Answer : (b) $\;16.5 \;days$
Explanation :
Equation for decay of radioactive nuclei will be
$N=N_{0} e^{-\lambda t}$
Where $\;\lambda\large\frac{ln 2}{t_{\large\frac{1}{2}}}=\large\frac{ln 2}{38}$
$\large\frac{N_{0}}{20}$$=N_{0} e^{-\large\frac{ln2}{.8}\;t}$
answered Feb 27, 2014 by yamini.v
edited Mar 25, 2014 by balaji.thirumalai

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