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 $N/N_0 = 1/20$.The decay constant $\lambda$ is related to the half-life $T_{1

Vedclass · Accepted Answer

$16.5$

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(B) The radioactive decay law is given by $N = N_0 e^{-\lambda t}$,where $N/N_0 = 1/20$.The decay constant $\lambda$ is related to the half-life $T_{1/2}$ by $\lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.6931}{3.8 \ days}$.Substituting the values into the decay equation: $\frac{1}{20} = e^{-\lambda t}$,which implies $20 = e^{\lambda t}$.Taking the natural logarithm on both sides: $\ln 20 = \lambda t$.Converting to base $10$ logarithm: $2.303 \log_{10} 20 = \lambda t$.Given $\log_{10} 20 = \log_{10} (2 	imes 10) = 0.3010 + 1 = 1.3010$.Substituting $\lambda = \frac{0.6931}{3.8}$ and $\log_{10} e = 0.4343$,we use the relation $\ln x = 2.303 \log_{10} x$ or $\lambda = \frac{0.6931}{T_{1/2}}$.$t = \frac{\ln 20}{\lambda} = \frac{2.303 	imes 1.3010 	imes 3.8}{0.6931} \approx 16.43 \ days$.Rounding to the nearest provided option,$t \approx 16.5 \ days$.

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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