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The remaining mass of radioactive iodine-131 is given by the formula M=Mo(2)-t/8 where M is the remaining mass in grams, Mo is the initial mass in grams, and t is the time in days. How many days does it take for the 726 grams to be reduced to half?

User Jdesilvio
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Final answer:

To determine the time it takes for the mass of radioactive iodine-131 to reduce by half, we use the decay formula. Since the half-life is 8 days, solving the equation shows that it takes 8 days for the initial mass of 726 grams to be reduced to half.

Step-by-step explanation:

The student has asked how many days it takes for 726 grams of radioactive iodine-131 to be reduced to half its initial mass. The decay of iodine-131 is described by the formula M = Mo(2)^(-t/8), where M is the remaining mass, Mo is the initial mass, and t is the time in days. The half-life of iodine-131 is 8 days, meaning every 8 days the mass of the substance is reduced by half.

To find the time it takes for the mass to reduce to half (363 grams in this case), we set M to Mo/2 and solve for t:

Mo/2 = Mo(2)^(-t/8)

Dividing both sides by Mo:

1/2 = (2)^(-t/8)

By applying the property of logarithms, we can solve for t:

-1 = -t/8 (since (2)^(-1) = 1/2)

Therefore, t = 8 days. This means it takes 8 days for the mass of iodine-131 to reduce to half of its original amount.

User Rochester Oliveira
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