A radioactive substance decays by 10% in 5 da | vedclass

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(B) The law of radioactive decay is given by $N = N_0 e^{-\lambda t}$.After $5$ days,$10\%$ has decayed,so $90\%$ remains: $0.9 N_0 = N_0 e^{-5\lambda

Vedclass · Accepted Answer

$65$

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(B) The law of radioactive decay is given by $N = N_0 e^{-\lambda t}$.After $5$ days,$10\%$ has decayed,so $90\%$ remains: $0.9 N_0 = N_0 e^{-5\lambda}$.Taking the natural logarithm on both sides: $\ln(0.9) = -5\lambda$,or $5\lambda = -\ln(0.9) = \ln(1/0.9)$.After $20$ days,the remaining amount $N$ is $N = N_0 e^{-20\lambda}$.Substituting $20\lambda = 4 	imes (5\lambda) = 4 \ln(1/0.9)$:$N = N_0 e^{-4 \ln(1/0.9)} = N_0 (1/0.9)^{-4} = N_0 (0.9)^4$.Calculating $(0.9)^4 = 0.81 	imes 0.81 = 0.6561$.Thus,$N \approx 0.656 N_0$,which is approximately $65.6\%$. The closest option is $65\%$.

$A$ radioactive substance decays by $10\%$ in $5$ days. What percentage of the original substance will remain after $20$ days (in $\%$)?

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