Question: 30 Two samples of pure radioactive isotopes $X$ and $Y$ decay with half-lives of $2$ days and $3$ days, respectively.
Both $X$ and $Y$ decay in a single step into different stable isotopes.
Initially the number of atoms of $X$ is twice the number of atoms of $Y$.
After how many days are the expected numbers of atoms of $X$ and $Y$ equal to each other?
A The expected numbers of atoms of $X$ and $Y$ are never equal.
B $2$ days
C $3$ days
D $4$ days
E $6$ days
F $12$ days
top-center: handwritten annotations “$2d$” and “$3d$” above “$X$” and “$Y$”; handwritten “$2N_0$” and “$N_0$” near the sentence “Initially the number of atoms of $X$ is twice the number of atoms of $Y$.”; black underline under “$X$”, “$Y$”, “$2$ days”, “$3$ days”, and “equal to each other?”; graph/diagram: none visible
1. **State the problem:**
We have two radioactive isotopes $X$ and $Y$ with half-lives $T_{1/2,X} = 2$ days and $T_{1/2,Y} = 3$ days respectively.
Initially, the number of atoms of $X$ is twice that of $Y$, i.e., $N_X(0) = 2N_0$ and $N_Y(0) = N_0$.
We want to find the time $t$ (in days) when the expected numbers of atoms of $X$ and $Y$ are equal.
2. **Formula for radioactive decay:**
The number of atoms remaining after time $t$ is given by:
$$N(t) = N(0) \times \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}$$
where $T_{1/2}$ is the half-life.
3. **Write expressions for $X$ and $Y$:**
$$N_X(t) = 2N_0 \times \left(\frac{1}{2}\right)^{\frac{t}{2}}$$
$$N_Y(t) = N_0 \times \left(\frac{1}{2}\right)^{\frac{t}{3}}$$
4. **Set $N_X(t) = N_Y(t)$ and solve for $t$: **
$$2N_0 \times \left(\frac{1}{2}\right)^{\frac{t}{2}} = N_0 \times \left(\frac{1}{2}\right)^{\frac{t}{3}}$$
Divide both sides by $N_0$:
$$2 \times \left(\frac{1}{2}\right)^{\frac{t}{2}} = \left(\frac{1}{2}\right)^{\frac{t}{3}}$$
5. **Divide both sides by $\left(\frac{1}{2}\right)^{\frac{t}{3}}$: **
$$2 \times \cancel{\left(\frac{1}{2}\right)^{\frac{t}{2}}} \div \cancel{\left(\frac{1}{2}\right)^{\frac{t}{3}}} = 1$$
Using the property of exponents:
$$2 \times \left(\frac{1}{2}\right)^{\frac{t}{2} - \frac{t}{3}} = 1$$
Simplify the exponent:
$$\frac{t}{2} - \frac{t}{3} = \frac{3t - 2t}{6} = \frac{t}{6}$$
So:
$$2 \times \left(\frac{1}{2}\right)^{\frac{t}{6}} = 1$$
6. **Divide both sides by 2:**
$$\cancel{2} \times \left(\frac{1}{2}\right)^{\frac{t}{6}} = \frac{1}{2}$$
So:
$$\left(\frac{1}{2}\right)^{\frac{t}{6}} = \frac{1}{2}$$
7. **Since bases are equal, equate exponents:**
$$\frac{t}{6} = 1$$
8. **Solve for $t$: **
$$t = 6$$
9. **Interpretation:**
After $6$ days, the expected numbers of atoms of $X$ and $Y$ are equal.
**Final answer: $6$ days (Option E).**