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(C) Let the initial number of atoms of $X$ be $N_0$. At time $t$,let the number of atoms of $X$ remaining be $N_X$ and the number of atoms of $Y$ form

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between $4 	ext{ hours}$ and $6 	ext{ hours}$

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(C) Let the initial number of atoms of $X$ be $N_0$. At time $t$,let the number of atoms of $X$ remaining be $N_X$ and the number of atoms of $Y$ formed be $N_Y$. Since $X$ converts to $Y$,the total number of atoms remains constant: $N_0 = N_X + N_Y$. Given the ratio $N_X : N_Y = 1 : 4$,we can write $N_Y = 4N_X$. Substituting this into the conservation equation: $N_0 = N_X + 4N_X = 5N_X$. Thus,the fraction of $X$ remaining is $\frac{N_X}{N_0} = \frac{1}{5}$. Using the radioactive decay law $\frac{N_X}{N_0} = \left(\frac{1}{2}ight)^{\frac{t}{T_{1/2}}}$,where $T_{1/2} = 2 	ext{ hours}$: $\frac{1}{5} = \left(\frac{1}{2}ight)^{\frac{t}{2}}$. Taking the logarithm or comparing powers: Since $\frac{1}{2^2} = \frac{1}{4}$ and $\frac{1}{2^3} = \frac{1}{8}$,and we know $\frac{1}{8} we have $\left(\frac{1}{2}ight)^3 Since the base is less than $1$,the inequality reverses for the exponents: $2 Multiplying by $2$,we get $4 < t < 6$.

$A$ radioactive element $X$ converts into another stable element $Y$. The half-life of $X$ is $2 \text{ hours}$. Initially,only $X$ is present. After a time $t$,if the ratio of atoms of $X$ to $Y$ is $1:4$,then the value of $t$ is

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