A radioactive sample contains two different isotopes, A and B. A has a half-life of 3 days, and B has a half-life of 6 days. Initially in the sample there are twice as many atoms of A as of B. At what time will the ratio
of the number of atoms of A to B be reversed?
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The number of atoms of a radioactive isotope decreases exponentially with time. The formula for the number of atoms remaining after a certain time is:
N = N0 * (1/2)^(t/T)
where:
N is the number of atoms remaining,
N0 is the initial number of atoms,
t is the time elapsed, and
T is the half-life of the isotope.
Let's denote the initial number of atoms of A as 2x and of B as x. We want to find the time t when the number of atoms of A is equal to the number of atoms of B, i.e., when:
2x * (1/2)^(t/3) = x * (1/2)^(t/6)
We can simplify this equation by dividing both sides by x and multiplying both sides by 2:
2^(t/3) = 2^(t/6)
This equation is true when t/3 = t/6, which gives t = 0. However, we are interested in the time when the ratio of the number of atoms of A to B is reversed, i.e., when there are more atoms of B than of A. This happens when:
2x * (1/2)^(t/3) < x * (1/2)^(t/6)
We can simplify this equation by dividing both sides by x and multiplying both sides by 2:
2^(t/3) < 2^(t/6)
This inequality is true when t/3 > t/6, which gives t > 0. The exact time when the ratio of the number of atoms of A to B is reversed can be found by solving the equation:
2^(t/3) = 2^(t/6)
for t. This gives:
t/3 = t/6
2t = t
t = 0
However, this is not possible as we are looking for a time when the ratio is reversed. Therefore, the ratio of the number of atoms of A to B will never be reversed.
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