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(B) The radioactive decay follows first-order kinetics: $N_t = N_0 e^{-\lambda t}$.For the first interval: $N_t = N_0 / 10$ at $t = 15 \ min$.So,$1/10

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$15.0$

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(B) The radioactive decay follows first-order kinetics: $N_t = N_0 e^{-\lambda t}$.For the first interval: $N_t = N_0 / 10$ at $t = 15 \ min$.So,$1/10 = e^{-\lambda(15)} \implies \ln(10) = 15\lambda \implies \lambda = \ln(10) / 15$.For the total time $T$ to reach $1/100$ of the original amount: $N_T = N_0 / 100$.$1/100 = e^{-\lambda T} \implies \ln(100) = \lambda T$.Since $\ln(100) = 2 \ln(10)$,we have $2 \ln(10) = (\ln(10) / 15) 	imes T$.Solving for $T$: $T = 2 	imes 15 = 30 \ min$.The additional time required is $T - 15 = 30 - 15 = 15 \ min$.

$A$ radioactive element decays at such a rate that after $15 \ min$ only $1/10$ of the original amount is left. How many more minutes will be needed when only $1/100$ of the original amount will be left? .......... $\min$

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