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The half-life of this americium nuclide is 470 years. A sample of this nuclide contains 8.0x10^14 atoms. After some time, 6 * 10 ^ 14 americium atoms have decayed.Calculate the time required for this decay

Question

The half-life of this americium nuclide is 470 years. A sample of this nuclide contains 8.0x10^14 atoms. After some time, 6 * 10 ^ 14 americium atoms have decayed.Calculate the time required for this decay

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Solution

To solve this problem, we need to use the formula for radioactive decay, which is:

N = N0 * (1/2)^(t/T)

where: N is the final quantity of the substance N0 is the initial quantity of the substance t is the time that has passed T is the half-life of the substance

In this case, we know that: N0 = 8.0x10^14 atoms N = N0 - 6.0x10^14 atoms = 2.0x10^14 atoms T = 470 years

We need to solve for t. Plugging in the known values, we get:

2.0x10^14 = 8.0x10^14 * (1/2)^(t/470)

Solving for t, we get:

(1/2)^(t/470) = 2.0x10^14 / 8.0x10^14 (1/2)^(t/470) = 0.25 t/470 = log2(0.25) t = 470 * log2(0.25)

Using a calculator, we find that log2(0.25) is approximately -2. So:

t = 470 * -2 = -940 years

However, time cannot be negative. This means that our initial assumption that the nuclide has not yet reached its half-life is incorrect. In fact, it has passed its half-life twice. Therefore, the correct time is:

t = 940 years

This problem has been solved

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