The half-life of Po-214 is 0.001 seconds. How much of a 10g sample is left after the 0.003 seconds. Also, create an exponential function to model its decay.

Final answer:

After 0.003 seconds, we would have 1.25g of the initial 10g Po-214 left. The decay process can be modeled with the exponential function: N = 10 * , where N is the final amount, N0 is the initial amount, t is time, and T is the half-life.

Explanation:

The subject matter pertains to the concept of half-life in physics which interprets the decay of a radioactive material. In radioactive decay, the number of atoms in a radioactive sample decreases exponentially over time. This time is taken into consideration in the form of their half-lives. In the case of Po-214, the half-life is given as 0.001 seconds.

Before finding how much of Po-214 is left after 0.003 seconds, let's first understand the calculation. After one half-life (0.001 second), half of the sample will be left, i.e., 10g/2 = 5g. After another half-life (another 0.001 second), half of this 5g will be left, i.e.,5g/2 = 2.5g. After the third half-life (the remaining 0.001 second), we will have 2.5g/2 = 1.25g left. So, after 0.003 seconds, we would have 1.25g of Po-214 remaining from the initial 10g.

The decay process can be modeled with an exponential function N = N0 * , where N is the final amount, N0 is the initial amount, t is time, and T is the half-life. For this problem, the equation would be: N = 10 * (1/2)^(0.003/0.001).

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