If 10% of a radioactive material decays in 5 | vedclass

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Question

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

Vedclass · Accepted Answer

$65$

Vedclass · Answer

(B) The radioactive decay law is given by $N(t) = N_0 e^{-\lambda t}$.After $t = 5$ days,$10\%$ decays,so $90\%$ remains: $0.9 N_0 = N_0 e^{-5\lambda}$.Taking the natural logarithm on both sides: $5\lambda = -\ln(0.9) = \ln(1/0.9) = \ln(1.111) \approx 0.10536$.So,$\lambda = 0.02107 	ext{ days}^{-1}$.After $t = 20$ days,the remaining amount $N$ is $N = N_0 e^{-\lambda(20)}$.$N = N_0 e^{-(0.02107 	imes 20)} = N_0 e^{-0.4214}$.$N \approx N_0 	imes 0.656$.Thus,approximately $65.6\%$ of the material remains,which is closest to $65\%$.

If $10\%$ of a radioactive material decays in $5$ days,then the amount of original material left after $20$ days is approximately ...............$\%$

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