The half-life of radioactive Radon is 3.8 da | vedclass

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(B) The radioactive decay law is given by $N = N_0 e^{-\lambda t}$,where $N/N_0 = 1/20$.Taking the natural logarithm on both sides: $\ln(1/20) = -\lam

Vedclass · Accepted Answer

$16.5$

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(B) The radioactive decay law is given by $N = N_0 e^{-\lambda t}$,where $N/N_0 = 1/20$.Taking the natural logarithm on both sides: $\ln(1/20) = -\lambda t$,which implies $\ln(20) = \lambda t$.We know that $\lambda = \frac{\ln 2}{t_{1/2}}$,where $t_{1/2} = 3.8 \ days$.Substituting $\lambda$: $\ln(20) = \frac{\ln 2}{3.8} 	imes t$.$t = 3.8 	imes \frac{\ln 20}{\ln 2} = 3.8 	imes \log_2(20)$.Using the change of base formula: $t = 3.8 	imes \frac{\log_{10} 20}{\log_{10} 2} = 3.8 	imes \frac{\log_{10} (2 	imes 10)}{\log_{10} 2} = 3.8 	imes \frac{1 + \log_{10} 2}{\log_{10} 2}$.Given $\log_{10} 2 \approx 0.3010$,then $t = 3.8 	imes \frac{1 + 0.3010}{0.3010} = 3.8 	imes \frac{1.3010}{0.3010} \approx 3.8 	imes 4.322 = 16.42 \ days$.Rounding to the nearest option,the answer is $16.5 \ days$.

The half-life of radioactive Radon is $3.8 \ days$. The time at the end of which $1/20^{th}$ of the Radon sample will remain undecayed is ............ $days$ (Given $\log_{10}e = 0.4343$)

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