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

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(B) The radioactive decay law is given by $N(t) = N_0 e^{-\lambda t}$ or $N(t) = N_0 (1/2)^{t/T_{1/2}}$.Given that $10\%$ decays in $5\, days$,the amo

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$65$

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(B) The radioactive decay law is given by $N(t) = N_0 e^{-\lambda t}$ or $N(t) = N_0 (1/2)^{t/T_{1/2}}$.Given that $10\%$ decays in $5\, days$,the amount remaining is $90\%$ of the initial amount $N_0$.So,$\frac{N(5)}{N_0} = 0.90$.We can write $0.90 = (1/2)^{5/T_{1/2}}$.We need to find the amount remaining after $20\, days$,which is $N(20) = N_0 (1/2)^{20/T_{1/2}}$.This can be rewritten as $N(20) = N_0 [(1/2)^{5/T_{1/2}}]^4$.Substituting the value of $(1/2)^{5/T_{1/2}} = 0.90$,we get:$N(20) = N_0 (0.90)^4$.Calculating $(0.90)^4 = 0.9 	imes 0.9 	imes 0.9 	imes 0.9 = 0.6561$.Therefore,the percentage remaining is $0.6561 	imes 100\% = 65.61\%$.Rounding to the nearest integer,the value is approximately $65\%$.

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

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