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(C) The amount of radioactive substance remaining after time $t$ is given by $N = N_0 (1/2)^n$, where $n = t/T_{1/2}$ is the number of half-lives.For

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$3 : 7$

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(C) The amount of radioactive substance remaining after time $t$ is given by $N = N_0 (1/2)^n$, where $n = t/T_{1/2}$ is the number of half-lives.For sample $A$: $(1/8) = (1/2)^n_A \implies (1/2)^3 = (1/2)^n_A \implies n_A = 3$.Since $n_A = t/T_A$, we have $3 = 9/T_A \implies T_A = 3 	ext{ hours}$.For sample $B$: $(1/128) = (1/2)^n_B \implies (1/2)^7 = (1/2)^n_B \implies n_B = 7$.Since $n_B = t/T_B$, we have $7 = 9/T_B \implies T_B = 9/7 	ext{ hours}$.The ratio of half-life of $B$ to that of $A$ is $T_B/T_A = (9/7) / 3 = 3/7$.

Two samples $A$ and $B$ contain equal amount of radioactive substances. If $(1/8)^{th}$ of sample $A$ and $(1/128)^{th}$ of sample $B$ remain after $9 \text{ hours}$, then the ratio of half-life period of $B$ to that of $A$ is

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